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Question about the Derivation of the cable equation for neurites

Question about the Derivation of the cable equation for neurites



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I read in Wikipedia how the cable equation was derived (here) and had a specific problem regarding one of its equation: At the start of the derivation it states that we first need to pretend that the membrane is perfectly sealed (Rm=∞) and that the capacitance of the membrane equal zero. Then we can say that the change in voltage over distance depends only on the resistance of the axoplasm, by the equation: and from that, the current equal:

But later when they derive the last equation, , they set as in the equation above, which seems to me incorrect, since the resistance of the membrane also affect . What did I get wrong?

Thanks!


The part you are quoting is about helping you conceptualize the cable equation by starting from some approximations that let you remove most of the terms. See the sentence beginning (emphasis mine):

To better understand how the cable equation is derived, first simplify the theoretical neuron even further and pretend

They then walk you through how, because these approximations are just pretend, one needs to include terms to account for the actual permeability and capacitance of the membrane. This means that you do not have a constant internal current with distance as you travel along; instead, the current decays with distance because of leak through the membrane and membrane capacitance.

This is what the last equation you quote from says: the change in internal current over distance x (written as dil/dx) is equal to the current lost to the membrane, which includes two parts: leak through the membrane (V/rm) and charging the membrane capacitor (cm*dV/dt).


Another correction that might be tripping you up: you write

change in voltage over time

in your question, but that is not what the equations are representing, they are representing change in voltage over distance x.


Pulley Mass System | Definition, Examples in Physics – Laws of Motion

We are giving a detailed and clear sheet on all Physics Notes that are very useful to understand the Basic Physics Concepts.

Pulley Mass System Physics:
Determine the acceleration of the masses and the tension in the string.

(i) When unequal masses m1 and m2 are suspended from a pulley (m1 > m2)

m1g – T = m1a, and T – m2g = m2a
On solving equations, we get
(a=frac-m_<2> ight)>+m_<2> ight)> g,)
(T=frac <2 m_<1>m_<2>>+m_<2> ight)> g)

(ii) When a body of mass m2 is placed on a frictionless horizontal surface, then

Mass Pulley System acceleration, a = (frac g>+m_<2> ight)>)

Tension in a Pulley System with Two Masses, T = (frac m_ <2>g>+m_<2> ight)>)

(iii) When a body of mass m2 is placed on a rough horizontal surface, then

Tension in string, T = (frac m_<2>(1+mu) g>+m_<2> ight)>)

(iv) When two masses m1 and m2 (m1 > m2) are connected to a single mass M as shown in figure, then


m1g – T1 = m1a ……(i)
T2 – m2g = m2a ……(ii)
T1 – T2 = Ma ……(iii)

(v) Motion on a smooth inclined plane, then

m1g – T = m1a ……(i)
T – m2g sin θ = m2a ……(ii)
Acceleration, a = (left(frac-m_ <2>sin heta>+m_<2>> ight) g)
Tension, T = (frac m_<2>(1+sin heta) g>+m_<2> ight)>)

(vi) Motion of two bodies placed on two inclined planes having different angle of inclination, then

Acceleration, a = (frac sin heta_<1>-m_ <2>sin heta_<2> ight) g>+m_<2>>)
Tension, T = (frac m_<2>>+m_<2>>left(sin heta_<1>+sin heta_<2> ight) g)

Laws of Motion:
There are various laws in Physics that define the motion of the object. When an object is in motion whether it is linear or circular there is some force which is always imposed on it.


Physics Questions

On this page I put together a collection of physics questions to help you understand physics better. These questions are designed to challenge and inspire you to think about physics at a deeper level. In addition to being challenging, these questions are fun and interesting. This page is a good resource for students who want good quality problems to practice with when studying for tests and exams.

To see the questions click on the category you are interested in:

High School Physics Questions

Do heavier objects fall more slowly than lighter objects?

Why do objects float in liquids denser than themselves?

A particle is moving around in a circle and its position is given in polar coordinates as x = Rcosθ, and y = Rsinθ, where R is the radius of the circle, and θ is in radians. From these equations derive the equation for centripetal acceleration.

How come in free fall you feel weightless even though gravity is pulling down on you? (ignore air resistance when answering this question).

What is the difference between centripetal acceleration and centrifugal force?

What is the difference between energy and power?

Two identical cars collide head on. Each car is traveling at 100 km/h. The impact force on each car is the same as hitting a solid wall at:

Why is it possible to drive a nail into a piece of wood with a hammer, but it is not possible to push a nail in by hand?

An archer pulls back 0.75 m on a bow which has a stiffness of 200 N/m. The arrow weighs 50 g. What is the velocity of the arrow immediately after release?

When a moving car encounters a patch of ice the brakes are applied. Why is it desirable to keep the wheels rolling on the ice without locking up?

Solutions For High School Physics Questions

No. If an object is heavier the force of gravity is greater, but since it has greater mass the acceleration is the same, so it moves at the same speed (if we neglect air resistance). If we look at Newton's second law, F = ma. The force of gravity is F = mg, where m is the mass of the object and g is the acceleration due to gravity.

Equating, we have mg = ma. Therefore, a = g.

If there was no air resistance, a feather would fall at the same speed as an apple.

If an object were completely immersed in a liquid denser than it, the resulting buoyant force would exceed the weight of the object. This is because the weight of the liquid displaced by the object is greater than the weight of the object (since the liquid is denser). As a result, the object cannot remain completely submerged and it floats. The scientific name for this phenomenon is Archimedes Principle.

Without loss of generality, we only need to look at the equation for the x-position, since we know that centripetal acceleration points towards the center of the circle. Thus, when θ = 0, the second derivative of x with respect to time must be the centripetal acceleration.

The first derivative of x with respect to time t is:

dx/dt = -Rsinθ(dθ/dt)

The second derivative of x with respect to time t is:

d 2 x/dt 2 = -Rcosθ(dθ/dt) 2 −Rsinθ(d 2 θ/dt 2 )

In both of the above equations the chain rule of Calculus is used and by assumption θ is a function of time. Therefore, θ can be differentiated with respect to time.

Now, evaluate the second derivative at θ = 0.

d 2 x/dt 2 = -R(dθ/dt) 2

The term dθ/dt is usually called the angular velocity, which is the rate of change of the angle θ. It has units of radians/second.

For convenience we can set w ≡ dθ/dt.

d 2 x/dt 2 = -Rw 2

This is the well-known form for the centripetal acceleration equation.

The reason you feel weightless is because there is no force pushing against you, since you are not in contact with anything. Gravity is pulling equally on all the particles in your body. This creates a sensation where no forces are acting on you and you feel weightless. It would be the same sensation as if you were floating in space.

Centripetal acceleration is the acceleration an object experiences as it travels a certain velocity along an arc. The centripetal acceleration points towards the center of the arc.

Centrifugal force is the imaginary force an unrestrained object experiences as it moves around an arc. This force acts opposite to the direction of centripetal acceleration. For example, if a car makes a sharp right turn the passengers would tend to slide in their seats away from the center of the turn, towards the left (if they are not wearing their seat belts, that is). The passengers would feel as if they are experiencing a force. This is defined as centrifugal force.

Power is the rate of energy being generated or consumed. For example, if an engine produces 1000 Watts of power (where Watts is Joules/second), then after an hour the total energy produced by the engine is 1000 Joules/second × 3600 seconds = 3,600,000 Joules.

Since the collision is head on and each car is identical and traveling at the same speed, the force of impact experienced by each car is equal and opposite. This means that the impact is the same as hitting a solid wall at 100 km/h.

When you swing a hammer you increase its kinetic energy, so that by the time it strikes the nail it imparts a large force which drives the nail into the wood.

The hammer is basically an energy reservoir to which you are adding energy during the course of the swing, and which is released all at once upon impact. This results in the impact force greatly exceeding the maximum force you can exert by just pushing on the nail.

This can be solved using an energy method.

We can solve this by equating the potential energy of the bow to the kinetic energy of the arrow.

The bow can be treated as a type of spring. The potential energy of a spring is:

(1/2)kx 2 , where k is the stiffness and x is the amount the spring is stretched, or compressed.

Therefore, the potential energy PE of the bow is:

PE = (1/2)(200)(0.75) 2 = 56.25 J

The kinetic energy of a particle is:

(1/2)mv 2 , where m is the mass and v is the velocity.

The arrow can be treated as a particle since it is not rotating upon release.

Therefore, the kinetic energy KE of the arrow is:

If we assume energy is conserved, then

Solving for the velocity of the arrow v we get

Static friction is greater than kinetic friction.

Static friction exists if the wheels keep rolling on the ice without locking up, resulting in maximum braking force. However, if the wheels lock up then kinetic friction takes over since there is relative slipping between wheel and ice. This reduces the braking force and the car takes longer to stop.

Anti-lock braking systems (ABS) on a vehicle prevent the wheels from locking up when the brakes are applied, thus minimizing the amount of time it takes for the vehicle to reach a complete stop. Also, by preventing the wheels from locking up you have greater control of the vehicle.


College and University Physics Questions (mostly first year level)


Extra Challenging Physics Questions


The 20 physics questions given below are both interesting and highly challenging. You will likely have to take some time to work through them. These questions go beyond the typical problems you can expect to find in a physics textbook. Some of these physics questions make use of different concepts, so (for the most part) there is no single formula or set of equations that you can use to solve them. These questions make use of concepts taught at the high school and college level (mostly first year).

It is recommended that you persist through these physics questions, even if you get stuck. It's not a race, so you can work through them at your own pace. The result is that you will be rewarded with a greater understanding of physics.

A crank drive mechanism is illustrated below. A uniform linkage BC of length L connects a flywheel of radius r (rotating about fixed point A) to a piston at C that slides back and forth in a hollow shaft. A variable torque T is applied to the flywheel such that it rotates at a constant angular velocity. Show that for one full rotation of the flywheel, energy is conserved for the entire system consisting of flywheel, linkage, and piston (assuming no friction).

Note that gravity g is acting downwards, as shown.

Even though energy is conserved for the system, why is it a good idea to make the components of the drive mechanism as light as possible (with the exception of the flywheel)?

An engine uses compression springs to open and close valves, using cams. Given a spring stiffness of 30,000 N/m, and a spring mass of 0.08 kg, what is the maximum engine speed to avoid “floating the valves”?

During the engine cycle the spring is compressed between 0.5 cm (valve fully closed) and 1.5 cm (valve fully open). Assume the camshaft rotates at the same speed as the engine.

Floating the valves occurs when the engine speed is high enough so that the spring begins to lose contact with the cam when the valve closes. In other words, the spring doesn’t extend quickly enough to maintain contact with the cam, when the valve closes.

For simplicity, you may assume that Hooke’s Law applies to the spring, where the force acting on the spring is proportional to its amount of compression (regardless of dynamic effects).

You may ignore gravity in the calculations.

An object is traveling in a straight line. Its acceleration is given by

where C is a constant, n is a real number, and t is time.

Find the general equations for the position and velocity of the object as a function of time.

In archery, when an arrow is released it can oscillate during flight. If we know the location of the center of mass of the arrow (G) and the shape of the arrow at an instant as it oscillates (shown below), we can determine the location of the nodes. The nodes are the “stationary” points on the arrow as it oscillates.

Using a geometric argument (no equations), determine the location of the nodes.

Assume that the arrow oscillates in the horizontal plane, so that no external forces act on the arrow in the plane of oscillation.

A gyroscope wheel is spinning at a constant angular velocity ws while precessing about a vertical axis at a constant angular velocity wp. The distance from the pivot to the center of the front face of the spinning gyroscope wheel is L, and the radius of the wheel is r. The rod connecting the pivot to the wheel makes a constant angle θ with the vertical.

Determine the acceleration components normal to the wheel, at points A, B, C, D labeled as shown.

When a vehicle makes a turn, the two front wheels trace out two arcs as shown in the figure below. The wheel facing towards the inside of the turn has a steering angle that is greater than that of the outer wheel. This is necessary to ensure that both front wheels smoothly trace out two arcs, which have the same center, otherwise the front wheels will skid on the ground during the turn.

During a turn, do the rear wheels necessarily trace out the same arcs as the front wheels? Based on your answer, what are the implications for making a turn close to the curb?

A horizontal turntable at an industrial plant is continuously fed parts into a slot (shown on the left). It then drops these parts into a basket (shown on the right). The turntable rotates 180° between these two stages. The turntable briefly stops at each 1/8 th of a turn in order to receive a new part into the slot on the left.

If the rotational speed of the turntable is w radians/second, and the outer radius of the turntable is R2, what must be the inner radius R1 so that the parts fall out of the slot and into the basket, as shown?

• The angular speed w of the turntable can be treated as constant and continuous which means you can ignore the brief stops the turntable makes at each 1/8 th of a turn.

• The location of the basket is 180° from the feed location.

• The slots are very well lubricated so that there is no friction between the slot and part.

• The parts can be treated as particles, which means you can ignore their dimensions in the calculation.

• The slots are aligned with the radial direction of the turntable.

A flywheel for a single piston engine rotates at an average speed of 1500 RPM. During half a rotation the flywheel has to absorb 1000 J of energy. If the maximum permissible speed fluctuation is ± 60 RPM, what is the minimum rotational inertia of the flywheel? Assume there is no friction.

An aluminum extrusion process is simulated numerically with a computer. In this process, a punch pushes an aluminum billet of diameter D through a die of smaller diameter d. In the computer simulation, what is the maximum punch velocity Vp so that the net dynamic force (predicted by the simulation) acting on the aluminum during extrusion is at most 5% of the force due to deformation of the aluminum? Evaluate for a specific case where D = 0.10 m, d = 0.02 m, and the density of aluminum is ρ = 2700 kg/m 3 .

The force due to deformation of the aluminum during extrusion is given by

The extrusion of the aluminum through the die is analogous to fluid flowing through a pipe which transitions from a larger diameter to a smaller diameter (e.g. water flowing through a fireman’s hose). The net dynamic force acting on the fluid is the net force required to accelerate the fluid, which occurs when the velocity of the fluid increases as it flows from the larger diameter section to the smaller diameter section (due to conservation of mass).

A child on a horizontal merry-go-round gives an initial velocity Vrel to a ball. Find the initial direction and velocity Vrel of the ball relative to the merry-go-round so that, relative to the child, the ball goes around in a perfect circle as he’s sitting on the merry-go-round. Assume there is no friction between merry-go-round and ball.

The merry-go-round is rotating at a constant angular velocity of w radians/second, and the ball is released at a radius r from the center of the merry-go-round.

A heavy pump casing with a mass m is to be lifted off the ground using a crane. For simplicity, the motion is assumed to be two-dimensional, and the pump casing is represented by a rectangle having side dimensions ab (see figure). A cable of length L1 is attached to the crane (at point P) and the pump casing (at point O). The crane pulls up vertically on the cable with a constant velocity Vp.

The center of mass G of the pump casing is assumed to lie in the center of the rectangle. It is located at a distance L2 from point O. The right side of the pump casing is located at a horizontal distance c from the vertical line passing though point P.

Find the maximum cable tension during the lift, which includes the part of the lift before the pump casing loses contact with the ground, and after the pump casing loses contact with the ground (lift off). In this stage the pump casing swings back and forth.

Evaluate for a specific case where:

IG = 9 kg-m 2 (rotational inertia of pump casing about G)

• The friction between the pump casing and ground is high enough so that the pump casing does not slide along the ground (towards the right), before lift off occurs.

• Before lift off occurs, dynamic effects are negligible.

• The velocity Vp is fast enough so that the bottom of the pump casing swings clear of the ground after lift off occurs.

• For purposes of approximating the cable tension, you can model the system as a regular pendulum during swinging (you can ignore double pendulum effects).

• The mass of the cable can be neglected.


A linkage arrangement is shown below. The pin joints O1 and O2 are attached to a stationary base and are separated by a distance b. The linkages of identical color have the same length. All linkages are pin jointed and allow for rotation. Determine the path traced by the end point P as the blue linkage of length b rotates back and forth.

Why is this result interesting?

A conveyor belt carrying aggregate is illustrated in the figure below. A motor turns the top roller at a constant speed, and the remaining rollers are allowed to spin freely. The belt is inclined at an angle θ. To keep the belt in tension a weight of mass m is suspended from the belt, as shown.

Find the point of maximum tension in the belt. You don’t have to calculate it, just find the location and give a reason for it.

A quality test has determined that a pump impeller is too heavy on one side by an amount equal to 0.0045 kg-m. To correct this imbalance it is recommended to cut out a groove around the outer circumference of the impeller, using a milling machine, on the same side as the imbalance. This will remove material with the intent of correcting the imbalance. The dimension of the groove is 1 cm wide and 1 cm deep. The groove will be symmetric with respect to the heavy spot. How far around the outer circumference of the impeller should the groove be? Specify the answer in terms of θ. Hint: Treat the groove as a thin ring of material.

The outer radius of the impeller, at the location of the groove, is 15 cm.

The impeller material is steel, with a density of ρ = 7900 kg/m 3 .

As part of a quality check, an axisymmetric container is placed over a very well lubricated fixed mandrel, as shown below. The container is then given an initial pure rotation w, with no initial translational motion. What do you expect to see if the center of mass of the container is offset from the geometric center O of the container?

A stream of falling material hits the plate of an impact weigher and the horizontal force sensor allows the mass flow rate to be calculated from this. If the speed of the material just before it strikes the plate is equal to the speed of the material just after it strikes the plate, determine an equation for the mass flow rate of the material, based on the horizontal force readout on the sensor. Ignore friction with the plate.

Hint: This can be treated as a fluid flow problem.

The SunCatcher is a Stirling engine that is powered by solar energy. It uses large parabolic mirrors to focus sunlight onto a central receiver, which powers a Stirling engine. In the parabolic mirror you can see the reflection of the landscape. Why is the reflection upside down?

On a cold, dry winter day your glasses fog up when you go indoors after being outside for a while. Why is that?

And if you go back outside with your glasses still fogged up, they quickly clear up. Why is that?

In an astronaut training exercise, an airplane at high altitude travels along a circular arc in order to simulate weightlessness for its passengers. Explain how this is possible.

A rope is wrapped around a pole of radius R = 3 cm. If the tension on one end of the rope is T = 1000 N, and the coefficient of static friction between the rope and pole is μ = 0.2, what is the minimum number of times the rope must be wrapped around the pole so that it doesn’t slip off?

Assume that the minimum number of times the rope must be wrapped around the pole corresponds to a tension of 1 N on the other end of the rope.

I created solutions for the 20 physics questions given above. The solutions are given in an ebook, in PDF format. They are available through this link.


Population Change

In addition to analyzing velocity, speed, acceleration, and position, we can use derivatives to analyze various types of populations, including those as diverse as bacteria colonies and cities. We can use a current population, together with a growth rate, to estimate the size of a population in the future. The population growth rate is the rate of change of a population and consequently can be represented by the derivative of the size of the population.

Definition

If is the number of entities present in a population, then the population growth rate of is defined to be .

Estimating a Population

The population of a city is tripling every 5 years. If its current population is 10,000, what will be its approximate population 2 years from now?

Solution

Let be the population (in thousands) years from now. Thus, we know that and based on the information, we anticipate . Now estimate , the current growth rate, using

.

By applying (Figure) to , we can estimate the population 2 years from now by writing

thus, in 2 years the population will be approximately 18,000.

The current population of a mosquito colony is known to be 3,000 that is, . If , estimate the size of the population in 3 days, where is measured in days.

Solution

Use .


No, it doesn't quite work like that. When you derive the characteristic impedance in transmission line theory sure, you use a lumped-model to get the math process started but then, as you approach the solution, you purposely make the "dimensions" of the lump approach zero and this means that what you refer to as the cut-off frequency rises to infinity and we don't care about that. For instance, a simple derivation is to consider this lumped line model: -

Each small section (having an input impedance $Z_0$ ) will join on to the next small section. So then, if you do the "fairly straightforward" math you get this impedance looking into the left port: -

$Z_0 = R + jomega L + Z_0||dfrac<1>$

$Z_0cdot left[1 + Z_0(G+jomega C) ight] = [R+jomega L][1 + Z_0(G+jomega C)]+Z_0$

$Z_0 + Z_0^2(G+jomega C) = R+jomega L + Z_0[(R+jomega L)(G+jomega C)]+Z_0$

$Z_0^2(G+jomega C) = R+jomega L + Z_0[(R+jomega L)(G+jomega C)]$

The important thing next is to recognize that $(R+jomega L)(G+jomega C)$ is insignificant as the "lump" approaches zero length and we are left with: -

If you looked at the equation above for high frequencies, it becomes this: -

And is basically resistive in nature.

But when you take a signal generator with the same impedance as the transmission line and feed it into a transmission line with an open termination you can see that the reflection of the square wave has "slower" edges than the square wave itself

You have to be careful here because any capacitive connection on the t-line output can screw things up - even the oscilloscope input capacitance can cause this when placed at either end. Simulation of 5 pF capacitor loading the end of an unterminated 50 Ω line: -

You can see that the edges are slower but this is wholly because of the 5 pF loading at the end of the 10 metre transmission line.


NCERT Solutions for Class 11 Science Math Chapter 11 - Conic Sections

NCERT Solutions for Class 11 Science Math Chapter 11 Conic Sections are provided here with simple step-by-step explanations. These solutions for Conic Sections are extremely popular among Class 11 Science students for Math Conic Sections Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Science Math Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 11 Science Math are prepared by experts and are 100% accurate.

Page No 241:

Question 1:

Find the equation of the circle with centre (0, 2) and radius 2

Answer:

The equation of a circle with centre (h, k) and radius r is given as

( &ndash h) 2 + (y ­&ndash k) 2 = r 2

It is given that centre (h, k) = (0, 2) and radius (r) = 2.

Therefore, the equation of the circle is

(x &ndash 0) 2 + (y &ndash 2) 2 = 2 2

x 2 + y 2 + 4 ­&ndash 4 y = 4

x 2 + y 2 ­&ndash 4y = 0

Page No 241:

Question 2:

Find the equation of the circle with centre (&ndash2, 3) and radius 4

Answer:

The equation of a circle with centre (h, k) and radius r is given as

( &ndash h) 2 + (y ­&ndash k) 2 = r 2

It is given that centre (h, k) = (&ndash2, 3) and radius (r) = 4.

Therefore, the equation of the circle is

(x + 2) 2 + (y &ndash 3) 2 = (4) 2

x 2 + 4x + 4 + y 2 &ndash 6y + 9 = 16

x 2 + y 2 + 4x &ndash 6y &ndash 3 = 0

Page No 241:

Question 3:

Find the equation of the circle with centreand radius

Answer:

The equation of a circle with centre (h, k) and radius r is given as

( &ndash h) 2 + (y ­&ndash k) 2 = r 2

It is given that centre (h, k) = and radius (r) =.

Therefore, the equation of the circle is

Page No 241:

Question 4:

Find the equation of the circle with centre (1, 1) and radius

Answer:

The equation of a circle with centre (h, k) and radius r is given as

( &ndash h) 2 + (y ­&ndash k) 2 = r 2

It is given that centre (h, k) = (1, 1) and radius (r) =.

Therefore, the equation of the circle is

Page No 241:

Question 5:

Find the equation of the circle with centre (&ndasha, &ndashb) and radius

Answer:

The equation of a circle with centre (h, k) and radius r is given as

( &ndash h) 2 + (y ­&ndash k) 2 = r 2

It is given that centre (h, k) = (&ndasha, &ndashb) and radius (r) =.

Therefore, the equation of the circle is

Page No 241:

Question 6:

Find the centre and radius of the circle (x + 5) 2 + (y &ndash 3) 2 = 36

Answer:

The equation of the given circle is (x + 5) 2 + (y &ndash 3) 2 = 36.

(x + 5) 2 + (y &ndash 3) 2 = 36

&rArr <x &ndash (&ndash5)> 2 + (y &ndash 3) 2 = 6 2 , which is of the form (x &ndash h) 2 + (y &ndash k) 2 = r 2 , where h = &ndash5, k = 3, and r = 6.

Thus, the centre of the given circle is (&ndash5, 3), while its radius is 6.

Page No 241:

Question 7:

Find the centre and radius of the circle x 2 + y 2 &ndash 4x &ndash 8y &ndash 45 = 0

Answer:

The equation of the given circle is x 2 + y 2 &ndash 4x &ndash 8y &ndash 45 = 0.

x 2 + y 2 &ndash 4x &ndash 8y &ndash 45 = 0

&rArr (x 2 &ndash 4x) + (y 2 &ndash 8y) = 45

&rArr (x &ndash 2) 2 + (y &ndash4) 2 = 65

&rArr (x &ndash 2) 2 + (y &ndash4) 2 = , which is of the form (x &ndash h) 2 + (y &ndash k) 2 = r 2 , where h = 2, k = 4, and .

Thus, the centre of the given circle is (2, 4), while its radius is.

Page No 241:

Question 8:

Find the centre and radius of the circle x 2 + y 2 &ndash 8x + 10y &ndash 12 = 0

Answer:

The equation of the given circle is x 2 + y 2 &ndash 8x + 10y &ndash 12 = 0.

x 2 + y 2 &ndash 8x + 10y &ndash 12 = 0

&rArr (x 2 &ndash 8x) + (y 2 + 10y) = 12

&rArr (x &ndash 4) 2 + (y + 5) 2 = 53

, which is of the form (x &ndash h) 2 + (y &ndash k) 2 = r 2 , where h = 4, k = &ndash5, and .

Thus, the centre of the given circle is (4, &ndash5), while its radius is.

Page No 241:

Question 9:

Find the centre and radius of the circle 2x 2 + 2y 2 &ndash x = 0

Answer:

The equation of the given circle is 2x 2 + 2y 2 &ndash x = 0.

, which is of the form (x &ndash h) 2 + (y &ndash k) 2 = r 2 , where h = , k = 0, and .

Thus, the centre of the given circle is, while its radius is .

Page No 241:

Question 10:

Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

Answer:

Let the equation of the required circle be (x &ndash h) 2 + (y &ndash k) 2 = r 2 .

Since the circle passes through points (4, 1) and (6, 5),

(4 &ndash h) 2 + (1 &ndash k) 2 = r 2 &hellip (1)

(6 &ndash h) 2 + (5 &ndash k) 2 = r 2 &hellip (2)

Since the centre (h, k) of the circle lies on line 4x + y = 16,

From equations (1) and (2), we obtain

(4 &ndash h) 2 + (1 &ndash k) 2 = (6 &ndash h) 2 + (5 &ndash k) 2

&rArr 16 &ndash 8h + h 2 + 1 &ndash 2k + k 2 = 36 &ndash 12h + h 2 + 25 &ndash 10k + k 2

&rArr 16 &ndash 8h + 1 &ndash 2k = 36 &ndash 12h + 25 &ndash 10k

On solving equations (3) and (4), we obtain h = 3 and k = 4.

On substituting the values of h and k in equation (1), we obtain

(4 &ndash 3) 2 + (1 &ndash 4) 2 = r 2

Thus, the equation of the required circle is

(x &ndash 3) 2 + (y &ndash 4) 2 =

x 2 &ndash 6x + 9 + y 2 ­&ndash 8y + 16 = 10

x 2 + y 2 &ndash 6x &ndash 8y + 15 = 0

Page No 241:

Question 11:

Find the equation of the circle passing through the points (2, 3) and (&ndash1, 1) and whose centre is on the line x &ndash 3y &ndash 11 = 0.

Answer:

Let the equation of the required circle be (x &ndash h) 2 + (y &ndash k) 2 = r 2 .

Since the circle passes through points (2, 3) and (&ndash1, 1),

(2 &ndash h) 2 + (3 &ndash k) 2 = r 2 &hellip (1)

(&ndash1 &ndash h) 2 + (1 &ndash k) 2 = r 2 &hellip (2)

Since the centre (h, k) of the circle lies on line x &ndash 3y &ndash 11 = 0,

From equations (1) and (2), we obtain

(2 &ndash h) 2 + (3 &ndash k) 2 = (&ndash1 &ndash h) 2 + (1 &ndash k) 2

&rArr 4 &ndash 4h + h 2 + 9 &ndash 6k + k 2 = 1 + 2h + h 2 + 1 &ndash 2k + k 2

&rArr 4 &ndash 4h + 9 &ndash 6k = 1 + 2h + 1 &ndash 2k

&rArr 6h + 4k = 11 &hellip (4)

On solving equations (3) and (4), we obtain.

On substituting the values of h and k in equation (1), we obtain

Thus, the equation of the required circle is

Page No 241:

Question 12:

Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

Answer:

Let the equation of the required circle be (x &ndash h) 2 + (y &ndash k) 2 = r 2 .

Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

Now, the equation of the circle becomes (x &ndash h) 2 + y 2 = 25.

It is given that the circle passes through point (2, 3).

When h = &ndash2, the equation of the circle becomes

(x + 2) 2 + y 2 = 25

x 2 + 4x + 4 + y 2 = 25

x 2 + y 2 + 4x &ndash 21 = 0

When h = 6, the equation of the circle becomes

(x &ndash 6) 2 + y 2 = 25

x 2 &ndash 12x +36 + y 2 = 25

x 2 + y 2 &ndash 12x + 11 = 0

Page No 241:

Question 13:

Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

Answer:

Let the equation of the required circle be (xh) 2 + (yk) 2 = r 2 .

Since the circle passes through (0, 0),

The equation of the circle now becomes (xh) 2 + (yk) 2 = h 2 + k 2 .

It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,

From equation (1), we obtain

From equation (2), we obtain

Thus, the equation of the required circle is

Page No 241:

Question 14:

Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

Answer:

The centre of the circle is given as (h, k) = (2, 2).

Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).

Thus, the equation of the circle is

Page No 241:

Question 15:

Does the point (&ndash2.5, 3.5) lie inside, outside or on the circle x 2 + y 2 = 25?

Answer:

The equation of the given circle is x 2 + y 2 = 25.

&rArr (x &ndash 0) 2 + (y &ndash 0) 2 = 5 2 , which is of the form (x &ndash h) 2 + (y &ndash k) 2 = r 2 , where h = 0, k = 0, and r = 5.

&there4Centre = (0, 0) and radius = 5

Distance between point (&ndash2.5, 3.5) and centre (0, 0)

Since the distance between point (&ndash2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (&ndash2.5, 3.5) lies inside the circle.

Page No 246:

Question 1:

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y 2 = 12 x

Answer:

The given equation is y 2 = 12 x .

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with y 2 = 4 ax , we obtain

4 a = 12 &rArr a = 3

&there4 Coordinates of the focus = ( a , 0) = (3, 0)

Since the given equation involves y 2 , the axis of the parabola is the x-axis.

Equation of direcctrix , x = &ndash a i.e., x = &ndash 3 i.e., x + 3 = 0

Length of latus rectum = 4 a = 4 × 3 = 12

Page No 246:

Question 2:

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x 2 = 6 y

Answer:

The given equation is x 2 = 6 y .

Here, the coefficient of y is positive. Hence, the parabola opens upwards.

On comparing this equation with x 2 = 4 ay , we obtain

&there4 Coordinates of the focus = (0, a ) =

Since the given equation involves x 2 , the axis of the parabola is the y-axis.

Length of latus rectum = 4 a = 6

Page No 246:

Question 3:

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y 2 = &ndash 8 x

Answer:

The given equation is y 2 = &ndash8 x .

Here, the coefficient of x is negative. Hence, the parabola opens towards the left.

On comparing this equation with y 2 = &ndash4 ax , we obtain

&there4 Coordinates of the focus = (&ndash a , 0) = (&ndash2, 0)

Since the given equation involves y 2 , the axis of the parabola is the x-axis.

Equation of directrix , x = a i.e., x = 2

Length of latus rectum = 4 a = 8

Page No 246:

Question 4:

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x 2 = &ndash 16 y

Answer:

The given equation is x 2 = &ndash16 y .

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with x 2 = &ndash 4 ay, we obtain

&ndash 4 a = &ndash16 &rArr a = 4

&there4 Coordinates of the focus = (0, &ndash a ) = (0, &ndash4)

Since the given equation involves x 2 , the axis of the parabola is the y-axis.

Equation of directrix , y = a i.e., y = 4

Length of latus rectum = 4 a = 16

Page No 246:

Question 5:

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y 2 = 10 x

Answer:

The given equation is y 2 = 10 x .

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with y 2 = 4 ax , we obtain

&there4 Coordinates of the focus = ( a , 0)

Since the given equation involves y 2 , the axis of the parabola is the x-axis.

Length of latus rectum = 4 a = 10

Page No 246:

Question 6:

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x 2 = &ndash9 y

Answer:

The given equation is x 2 = &ndash9 y .

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with x 2 = &ndash4 ay , we obtain

Since the given equation involves x 2 , the axis of the parabola is the y-axis.

Length of latus rectum = 4 a = 9

Page No 247:

Question 7:

Find the equation of the parabola that satisfies the following conditions: Focus (6, 0) directrix x = &ndash6

Answer:

Focus (6, 0) directrix, x = &ndash6

Since t he focus lies on the x -axis, the x- axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form y 2 = 4 ax or

It is also seen that the directrix, x = &ndash6 is to the left of the y -axis, while the focus (6, 0) is to the right of the y -axis. Hence, the parabola is of the form y 2 = 4 ax .

Thus, the equation of the parabola is y 2 = 24 x .

Page No 247:

Question 8:

Find the equation of the parabola that satisfies the following conditions: Focus (0, &ndash3) directrix y = 3

Answer:

Focus = (0, &ndash3) directrix y = 3

Since the focus lies on the y -axis, the y- axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form x 2 = 4 ay or

It is also seen that the directrix, y = 3 is above the x -axis, while the focus

(0, &ndash3) is below the x -axis. Hence, the parabola is of the form x 2 = &ndash4 ay .

Thus, the equation of the parabola is x 2 = &ndash12 y .

Page No 247:

Question 9:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) focus (3, 0)

Answer:

Since the vertex of the parabola is (0, 0) and the focus lies on the positive x -axis, x -axis is the axis of the parabola, while the equation of the parabola is of the form y 2 = 4 ax .

Since the focus is (3, 0) , a = 3.

Thus, the equation of the parabola is y 2 = 4 × 3 × x , i.e., y 2 = 12 x

Page No 247:

Question 10:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) focus (&ndash2, 0)

Answer:

Since the vertex of the parabola is (0, 0) and the focus lies on the negative x -axis, x -axis is the axis of the parabola, while the equation of the parabola is of the form y 2 = &ndash4 ax .

Since the focus is ( &ndash2, 0), a = 2.

Thus, the equation of the parabola is y 2 = &ndash4(2) x , i.e., y 2 = &ndash8 x

Page No 247:

Question 11:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) passing through (2, 3) and axis is along x -axis

Answer:

Since the vertex is (0, 0) and the axis of the parabola is the x -axis, the equation of the parabola is either of the form y 2 = 4 ax or y 2 = &ndash4 ax .

The parabola passes through point (2, 3), which lies in the first quadrant.

Therefore, t he equation of the parabola is of the form y 2 = 4 ax , while point

(2, 3) must satisfy the equation y 2 = 4 ax .

Thus, the e quation of the parabola is

Page No 247:

Question 12:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0), passing through (5, 2) and symmetric with respect to y -axis

Answer:

Since the vertex is (0, 0) and the parabola is symmetric about the y -axis, the equation of the parabola is either of the form x 2 = 4 ay or x 2 = &ndash4 ay .

T he parabola passes through point (5, 2), which lies in the first quadrant.

Therefore, t he equation of the parabola is of the form x 2 = 4 ay , while point

(5, 2) must satisfy the equation x 2 = 4 ay .

Thus, the equation of the parabola is

Page No 255:

Question 1:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

Answer:

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the x -axis, while the minor axis is along the y -axis.

On comparing the given equation with , we obtain a = 6 and b = 4.

The coordinates of the foci are .

The coordinates of the vertices are (6, 0) and (&ndash6, 0).

Length of major axis = 2 a = 12

Length of minor axis = 2 b = 8

Page No 255:

Question 2:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

Answer:

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the y -axis, while the minor axis is along the x -axis.

On comparing the given equation with , we obtain b = 2 and a = 5.

The coordinates of the foci are .

The coordinates of the vertices are (0, 5) and (0, &ndash5)

Length of major axis = 2 a = 10

Length of minor axis = 2 b = 4

Page No 255:

Question 3:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

Answer:

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the x -axis, while the minor axis is along the y -axis.

On comparing the given equation with , we obtain a = 4 and b = 3.

The coordinates of the foci are .

The coordinates of the vertices are .

Length of major axis = 2 a = 8

Length of minor axis = 2 b = 6

Page No 255:

Question 4:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

Answer:

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the y -axis, while the minor axis is along the x -axis.

On comparing the given equation with , we obtain b = 5 and a = 10.

The coordinates of the foci are .

The coordinates of the vertices are (0, ± 10).

Length of major axis = 2 a = 20

Length of minor axis = 2 b = 10

Page No 255:

Question 5:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

Answer:

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the x -axis, while the minor axis is along the y -axis.

On comparing the given equation with , we obtain a = 7 and b = 6.

The coordinates of the foci are .

The coordinates of the vertices are ( ± 7, 0).

Length of major axis = 2 a = 14

Length of minor axis = 2 b = 12

Page No 255:

Question 6:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse

Answer:

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the y -axis, while the minor axis is along the x -axis.

On comparing the given equation with , we obtain b = 10 and a = 20.

The coordinates of the foci are .

The coordinates of the vertices are (0, ± 20)

Length of major axis = 2 a = 40

Length of minor axis = 2 b = 20

Page No 255:

Question 7:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 36 x 2 + 4 y 2 = 144

Answer:

The given equation is 36 x 2 + 4 y 2 = 144.

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the y -axis, while the minor axis is along the x -axis.

On comparing equation (1) with , we obtain b = 2 and a = 6.

The coordinates of the foci are .

The coordinates of the vertices are (0, ± 6).

Length of major axis = 2 a = 12

Length of minor axis = 2 b = 4

Page No 255:

Question 8:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 16 x 2 + y 2 = 16

Answer:

The given equation is 16 x 2 + y 2 = 16.

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the y -axis, while the minor axis is along the x -axis.

On comparing equation (1) with , we obtain b = 1 and a = 4.

The coordinates of the foci are .

The coordinates of the vertices are (0, ± 4).

Length of major axis = 2 a = 8

Length of minor axis = 2 b = 2

Page No 255:

Question 9:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 4 x 2 + 9 y 2 = 36

Answer:

The given equation is 4 x 2 + 9 y 2 = 36.

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the x -axis, while the minor axis is along the y -axis.

On comparing the given equation with , we obtain a = 3 and b = 2.

The coordinates of the foci are .

The coordinates of the vertices are ( ± 3, 0).

Length of major axis = 2 a = 6

Length of minor axis = 2 b = 4

Page No 255:

Question 10:

Find the equation for the ellipse that satisfies the given conditions: Vertices (±5, 0), foci (±4, 0)

Answer:

Vertices ( ± 5, 0), foci ( ± 4, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form , where a is the semi-major axis.

Accordingly, a = 5 and c = 4.

Thus, the equation of the ellipse is .

Page No 255:

Question 11:

Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ±13), foci (0, ±5)

Answer:

Here, the vertices are on the y-axis.

Therefore, the equation of the ellipse will be of the form , where a is the semi-major axis.

Accordingly, a = 13 and c = 5.

Thus, the equation of the ellipse is .

Page No 255:

Question 12:

Find the equation for the ellipse that satisfies the given conditions: Vertices (±6, 0), foci (±4, 0)

Answer:

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form , where a is the semi-major axis.

Accordingly, a = 6, c = 4.

Thus, the equation of the ellipse is .

Page No 255:

Question 13:

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (±3, 0), ends of minor axis (0, ±2)

Answer:

Ends of major axis (±3, 0), ends of minor axis (0, ±2)

Here, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form , where a is the semi-major axis.

Accordingly, a = 3 and b = 2.

Thus, the equation of the ellipse is .

Page No 255:

Question 14:

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis , ends of minor axis (±1, 0)

Answer:

Ends of major axis , ends of minor axis (±1, 0)

Here, the major axis is along the y-axis.

Therefore, the equation of the ellipse will be of the form , where a is the semi-major axis.

Accordingly, a = and b = 1.

Thus, the equation of the ellipse is .

Page No 255:

Question 15:

Find the equation for the ellipse that satisfies the given conditions: Length of major axis 26, foci (±5, 0)

Answer:

Length of major axis = 26 foci = (±5, 0).

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form , where a is the semi-major axis.

Accordingly, 2 a = 26 &rArr a = 13 and c = 5.

Thus, the equation of the ellipse is .

Page No 255:

Question 16:

Find the equation for the ellipse that satisfies the given conditions: Length of minor axis 16, foci (0, ±6)

Answer:

Length of minor axis = 16 foci = (0, ±6).

Since the foci are on the y-axis, the major axis is along the y-axis.

Therefore, the equation of the ellipse will be of the form , where a is the semi-major axis.

Accordingly, 2 b = 16 &rArr b = 8 and c = 6.

Thus, the equation of the ellipse is .

Page No 255:

Question 17:

Find the equation for the ellipse that satisfies the given conditions: Foci (±3, 0), a = 4

Answer:

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form , where a is the semi-major axis.

Accordingly, c = 3 and a = 4.

Thus, the equation of the ellipse is .

Page No 255:

Question 18:

Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre at the origin foci on the x axis.

Answer:

It is given that b = 3, c = 4, centre at the origin foci on the x axis.

Since the foci are on the x-axis, the major axis is along the x-axis.

Therefore, the equation of the ellipse will be of the form , where a is the semi-major axis.

Accordingly, b = 3, c = 4.

Thus, the equation of the ellipse is .

Page No 255:

Question 19:

Find the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), major axis on the y -axis and passes through the points (3, 2) and (1, 6).

Answer:

Since the centre is at (0, 0) and the major axis is on the y -axis, the equation of the ellipse will be of the form

The ellipse passes through points (3, 2) and (1, 6). Hence,

On solving equations (2) and (3), we obtain b 2 = 10 and a 2 = 40.

Thus, the equation of the ellipse is .

Page No 255:

Question 20:

Find the equation for the ellipse that satisfies the given conditions: Major axis on the x -axis and passes through the points (4, 3) and (6, 2).

Answer:

Since the major axis is on the x -axis, the equation of the ellipse will be of the form

The ellipse passes through points (4, 3) and (6, 2). Hence,

On solving equations (2) and (3), we obtain a 2 = 52 and b 2 = 13.

Thus, the equation of the ellipse is .

Page No 262:

Question 1:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola

Answer:

On comparing this equation with the standard equation of hyperbola i.e. we obtain a = 4 and b = 3.

We know that a 2 + b 2 = c 2 .

The coordinates of the foci are (±5, 0).

The coordinates of the vertices are (±4, 0).

Page No 262:

Question 2:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola

Answer:

On comparing this equation with the standard equation of hyperbola i.e. we obtain a = 3 and.

We know that a 2 + b 2 = c 2 .

The coordinates of the foci are (0, ±6).

The coordinates of the vertices are (0, ±3).

Page No 262:

Question 3:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9y 2 &ndash 4x 2 = 36

Answer:

The given equation is 9y 2 &ndash 4x 2 = 36.

9y 2 &ndash 4x 2 = 36

On comparing equation (1) with the standard equation of hyperbola i.e. we obtain a = 2 and b = 3.

We know that a 2 + b 2 = c 2 .

The coordinates of the foci are.

The coordinates of the vertices are.

Page No 262:

Question 4:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 16x 2 &ndash 9y 2 = 576

Answer:

The given equation is 16x 2 &ndash 9y 2 = 576.

16x 2 &ndash 9y 2 = 576

On comparing equation (1) with the standard equation of hyperbola i.e. we obtain a = 6 and b = 8.

We know that a 2 + b 2 = c 2 .

The coordinates of the foci are (±10, 0).

The coordinates of the vertices are (±6, 0).

Page No 262:

Question 5:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y 2 &ndash 9x 2 = 36

Answer:

The given equation is 5y 2 &ndash 9x 2 = 36.

On comparing equation (1) with the standard equation of hyperbola i.e. we obtain a = and b = 2.

We know that a 2 + b 2 = c 2 .

Therefore, the coordinates of the foci are.

The coordinates of the vertices are.

Page No 262:

Question 6:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 49y 2 &ndash 16x 2 = 784

Answer:

The given equation is 49y 2 &ndash 16x 2 = 784.

It can be written as
49y 2 &ndash 16x 2 = 784

On comparing equation (1) with the standard equation of hyperbola i.e. we obtain a = 4 and b = 7.

We know that a 2 + b 2 = c 2 .

The coordinates of the foci are.

The coordinates of the vertices are (0, ±4).

Page No 262:

Question 7:

Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci (±3, 0)

Answer:

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form.

Since the vertices are (±2, 0), a = 2.

Since the foci are (±3, 0), c = 3.

We know that a 2 + b 2 = c 2 .

Thus, the equation of the hyperbola is.

Page No 262:

Question 8:

Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±5), foci (0, ±8)

Answer:

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form.

Since the vertices are (0, ±5), a = 5.

Since the foci are (0, ±8), c = 8.

We know that a 2 + b 2 = c 2 .

Thus, the equation of the hyperbola is.

Page No 262:

Question 9:

Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±3), foci (0, ±5)

Answer:

Here, the vertices are on the y-axis.

Therefore, the equation of the hyperbola is of the form.

Since the vertices are (0, ±3), a = 3.

Since the foci are (0, ±5), c = 5.

We know that a 2 + b 2 = c 2 .

Thus, the equation of the hyperbola is.

Page No 262:

Question 10:

Find the equation of the hyperbola satisfying the give conditions: Foci (±5, 0), the transverse axis is of length 8.

Answer:

Foci (±5, 0), the transverse axis is of length 8.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form.

Since the foci are (±5, 0), c = 5.

Since the length of the transverse axis is 8, 2a = 8 &rArr a = 4.

We know that a 2 + b 2 = c 2 .

Thus, the equation of the hyperbola is.

Page No 262:

Question 11:

Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±13), the conjugate axis is of length 24.

Answer:

Foci (0, ±13), the conjugate axis is of length 24.

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form.

Since the foci are (0, ±13), c = 13.

Since the length of the conjugate axis is 24, 2b = 24 &rArr b = 12.

We know that a 2 + b 2 = c 2 .

Thus, the equation of the hyperbola is.

Page No 262:

Question 12:

Find the equation of the hyperbola satisfying the give conditions: Foci, the latus rectum is of length 8.

Answer:

Foci, the latus rectum is of length 8.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form.

Since the foci are, c =.

Length of latus rectum = 8

We know that a 2 + b 2 = c 2 .

&rArr a 2 + 4a &ndash 45 = 0

&rArr a 2 + 9a &ndash 5a &ndash 45 = 0

Since a is non-negative, a = 5.

&there4b 2 = 4a = 4 × 5 = 20

Thus, the equation of the hyperbola is.

Page No 262:

Question 13:

Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus rectum is of length 12

Answer:

Foci (±4, 0), the latus rectum is of length 12.

Here, the foci are on the x-axis.

Therefore, the equation of the hyperbola is of the form.

Since the foci are (±4, 0), c = 4.

Length of latus rectum = 12

We know that a 2 + b 2 = c 2 .

&rArr a 2 + 6a &ndash 16 = 0

&rArr a 2 + 8a &ndash 2a &ndash 16 = 0

Since a is non-negative, a = 2.

&there4b 2 = 6a = 6 × 2 = 12

Thus, the equation of the hyperbola is.

Page No 262:

Question 14:

Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0),

Answer:

Here, the vertices are on the x-axis.

Therefore, the equation of the hyperbola is of the form.

Since the vertices are (±7, 0), a = 7.

We know that a 2 + b 2 = c 2 .

Thus, the equation of the hyperbola is.

Page No 262:

Question 15:

Find the equation of the hyperbola satisfying the give conditions: Foci, passing through (2, 3)

Answer:

Here, the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form.

Since the foci are, c =.

We know that a 2 + b 2 = c 2 .

&rArr b 2 = 10 &ndash a 2 &hellip (1)

Since the hyperbola passes through point (2, 3),

From equations (1) and (2), we obtain

In hyperbola, c > a, i.e., c 2 > a 2

&rArr b 2 = 10 &ndash a 2 = 10 &ndash 5 = 5

Thus, the equation of the hyperbola is.

Page No 264:

Question 1:

If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Answer:

The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive x-axis.

This can be diagrammatically represented as

The equation of the parabola is of the form y 2 = 4ax (as it is opening to the right). Since the parabola passes through point A (5, 10),

Therefore, the focus of the parabola is (a, 0) = (5, 0), which is the mid-point of the diameter.

Hence, the focus of the reflector is at the mid-point of the diameter.

Page No 264:

Question 2:

An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

Answer:

The origin of the coordinate plane is taken at the vertex of the arch in such a way that its vertical axis is along the negative y-axis.

This can be diagrammatically represented as

The equation of the parabola is of the form x 2 = - 4ay (as it is opening downwards).

It can be clearly seen that the parabola passes through point 5 2 , - 10 .

Therefore, the arch is in the form of a parabola whose equation is x 2 = - 5 8 y .

When y = - 2, x 2 = - 5 8 - 2


Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23 m.

Page No 264:

Question 3:

The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.

Answer:

The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y-axis. This can be diagrammatically represented as

Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadway, 18 m from the middle.

Here, AB = 30 m, OC = 6 m, and.

The equation of the parabola is of the form x 2 = 4ay (as it is opening upwards).

The coordinates of point A are (50, 30 &ndash 6) = (50, 24).

Since A (50, 24) is a point on the parabola,

&there4Equation of the parabola,or 6x 2 = 625y

The x-coordinate of point D is 18.

DF = DE + EF = 3.11 m + 6 m = 9.11 m

Thus, the length of the supporting wire attached to the roadway 18 m from the middle is approximately 9.11 m.

Page No 264:

Question 4:

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Answer:

Since the height and width of the arc from the centre is 2 m and 8 m respectively, it is clear that the length of the major axis is 8 m, while the length of the semi-minor axis is 2 m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis. Hence, the semi-ellipse can be diagrammatically represented as

The equation of the semi-ellipse will be of the form, where a is the semi-major axis

Accordingly, 2a = 8 &rArr a = 4

Therefore, the equation of the semi-ellipse is

Let A be a point on the major axis such that AB = 1.5 m.

The x-coordinate of point C is 2.5.

On substituting the value of x with 2.5 in equation (1), we obtain

Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.

Page No 264:

Question 5:

A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Answer:

Let AB be the rod making an angle &theta with OX and P (x, y) be the point on it such that AP = 3 cm.

Then, PB = AB &ndash AP = (12 &ndash 3) cm = 9 cm [AB = 12 cm]

From P, draw PQ&perpOY and PR&perpOX.

Thus, the equation of the locus of point P on the rod is.

Page No 264:

Question 6:

Find the area of the triangle formed by the lines joining the vertex of the parabola x 2 = 12y to the ends of its latus rectum.

Answer:

The given parabola is x 2 = 12y.

On comparing this equation with x 2 = 4ay, we obtain 4a = 12 &rArr a = 3

&there4The coordinates of foci are S (0, a) = S (0, 3)

Let AB be the latus rectum of the given parabola.

The given parabola can be roughly drawn as

At y = 3, x 2 = 12 (3) &rArr x 2 = 36 &rArr x = ±6

&there4The coordinates of A are (&ndash6, 3), while the coordinates of B are (6, 3).

Therefore, the vertices of &DeltaOAB are O (0, 0), A (&ndash6, 3), and B (6, 3).

Thus, the required area of the triangle is 18 unit 2 .

Page No 264:

Question 7:

A man running a racecourse notes that the sum of the distances from the two flag posts form him is always 10 m and the distance between the flag posts is 8 m. find the equation of the posts traced by the man.

Answer:

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man. Accordingly, PA + PB = 10.

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis is 10 m, while points A and B are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis along the x-axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form, where a is the semi-major axis

Accordingly, 2a = 10 &rArr a = 5

Distance between the foci (2c) = 8

On using the relation, we obtain

Thus, the equation of the path traced by the man is.

Page No 264:

Question 8:

An equilateral triangle is inscribed in the parabola y 2 = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Answer:

Let OAB be the equilateral triangle inscribed in parabola y 2 = 4ax.

Let AB intersect the x-axis at point C.

From the equation of the given parabola, we have

&there4The respective coordinates of points A and B are

Since OAB is an equilateral triangle, OA 2 = AB 2 .

Thus, the side of the equilateral triangle inscribed in parabola y 2 = 4 ax is .


An Explicit Numerical Method for the Fractional Cable Equation

An explicit numerical method to solve a fractional cable equation which involves two temporal Riemann-Liouville derivatives is studied. The numerical difference scheme is obtained by approximating the first-order derivative by a forward difference formula, the Riemann-Liouville derivatives by the Grünwald-Letnikov formula, and the spatial derivative by a three-point centered formula. The accuracy, stability, and convergence of the method are considered. The stability analysis is carried out by means of a kind of von Neumann method adapted to fractional equations. The convergence analysis is accomplished with a similar procedure. The von-Neumann stability analysis predicted very accurately the conditions under which the present explicit method is stable. This was thoroughly checked by means of extensive numerical integrations.

1. Introduction

Fractional calculus is a key tool for solving some relevant scientific problems in physics, engineering, biology, chemistry, hydrology, and so on [1–6]. A field of research in which the fractional formalism has been particularly useful is that related to anomalous diffusion processes [1, 7–13]. This kind of process is singularly abundant and important in biological media [14–16]. In this context, the electrodiffusion of ions in neurons is an anomalous diffusion problem to which the fractional calculus has recently been applied. The precise origin of the anomalous character of this diffusion process is not clear (see [17] and references therein), but in any case the consideration of anomalous diffusion in the modeling of electrodiffusion of ions in neurons seems pertinent. This problem has been addressed recently by Langlands et al. [17, 18]. An equation that plays a key role in their analysis is the following fractional cable (or telegrapher's or Cattaneo) equation (model II):

, is the Riemann-Liouville fractional derivative. Here

is the difference between the membrane potential and the resting membrane potential,

is the exponent characterizing the anomalous flux of ions along the nerve cell, and

is the exponent characterizing the anomalous flux across the membrane [17, 18]. Some earlier fractional cable equations were discussed in [19, 20].

A variety of analytical and numerical methods to solve many classes of fractional equations have been proposed and studied over the last few years [10, 21–30]. Of the numerical methods, finite difference methods have been particularly fruitful [31–38]. These methods can be broadly classified as explicit or implicit [39]. An implicit method for dealing with (1.1) has recently been considered by Liu et al. [38]. Although implicit methods are more cumbersome than explicit methods, they usually remain stable over a larger range of parameters, especially for large timesteps, which makes them particularly suitable for fractional diffusion problems. Nevertheless, explicit methods have some features that make them widely appreciated [32, 39]: flexibility, simplicity, small computational demand, and easy generalization to spatial dimensions higher than one. Unfortunately, they can become unstable in some cases, so that it is of great importance to determine the conditions under which these methods are stable. In this paper we will discuss an explicit finite difference scheme for solving the fractional cable equation, which is close to the methods studied in [32, 33]. We shall address two main questions: (i) whether this kind of method can cope with fractional equations involving different fractional derivatives, such as the fractional cable equation (ii) whether the von Neumann stability analysis put forward in [32, 34] is suitable for this kind of equation.

2. The Numerical Method

Henceforth, we will use the notation

is the numerical estimate of the exact solution

In order to get the numerical difference algorithm, we discretize the continuous differential and integro-differential operators as follows. For the discretization of the fractional Riemann-Liouville derivative we use the Grünwald-Letnikov formula

. These coefficients come from the generating function [40]

To discretize the integer derivatives we use standard formulas: for the second-order spatial derivative we employ the three-point centered formula

and for the first-order time derivative we use the forward derivative

Inserting (2.1), (2.5), and (2.7) into (1.1), one gets

where, as can easily be proved, the truncating error

Neglecting the truncating error we get the finite difference scheme we are seeking:

To test this algorithm, we solved (1.1) in the interval

, with absorbing boundary conditions,

, and initial condition given by a Dirac's delta function centered at

. The exact solution of this problem for

function [10, 41]. In our numerical procedure, the exact initial condition

The explicit difference scheme (2.12) is tested by comparing the analytical solution with the numerical solution for several cases of the problem described following (2.13) with different values of

. We have computed the analytical solution by means of (2.14) truncating the series at

function was evaluated by means of the series expansion described in [10, 41] truncating the infinite series after the first 50 terms. In Figures 1 and 2 we show the analytical and numerical solution for two values of

. The differences between the exact and the numerical solution are shown in Figures 3 and 4. One sees that, except for very short times, the agreement is quite good. The large value of the error for small times is due in part to the approximation of the Dirac's delta function at

by (2.15). This is clearly appreciated when noticing the quite different scales of Figures 3 and 4: the error is much smaller for

we used a smaller value of

and, simultaneously, a larger value of

in order to keep the numerical scheme stable. This issue will be discussed in Section 3.


(filled symbols) of the fractional cable equation for


(filled symbols) of the fractional cable equation for


of the numerical method for the problems considered in Figures 1 and 2 at


of the numerical method for the problem considered in Figures 1 and 2 at

3. Stability

As usual for explicit methods, the present explicit difference scheme (2.12) is not unconditionally stable, that is, for any given set of values of

for which the method is unstable. Therefore, it is important to determine the conditions under which the method is stable. To this end, here we shall employ the fractional von Neumann stability analysis (or Fourier analysis) put forward in [32] (see also [33–35]). The question we address is to what extent this procedure is valid for fractional diffusion equations that involve fractional derivatives of different order.

Proceeding as [32], we start by recognizing that the solution of our problem can be written as the linear combination of subdiffusive modes,

, where the sum is over all the wave numbers

supported by the lattice. Therefore, following the von Neumann ideas, we reduce the problem of analyzing the stability of the solution to the problem of analyzing the stability of a single generic subdiffusion mode,

. Inserting this expression into (2.12) one gets

The stability of the mode is determined by the behavior of

and assuming that the amplification factor

of the subdiffusive mode is independent of time, we get

, the temporal factor of the solution grows to infinity [c.f., (3.2)], and the mode is unstable. Considering the extreme value

, one gets from (3.3) that the numerical method is stable if this inequality holds:

, we find that a sufficient condition for the present method to be stable is that

. In Figures 5 and 6 we show two representative examples of the problem considered in Figure 2 but for two values of

, respectively, larger and smaller than the stability bound provided by (3.7). One sees that the value of

that one chooses is crucial: when

one is inside the stable region and gets a sensible numerical solution (Figure 5) otherwise one gets an evidently unstable and nonsensical solution (Figure 6).


Exact solution (lines) and numerical solution (symbols) provided by our method for the fractional cable equation with

for different numbers of timesteps when

. This case is inside the stability region because

is smaller than the stability limit


. Note that this value is larger than the stability limit

4. Numerical Check of the Stability Analysis

In this section we describe a comprehensive check of the validity of our stability analysis by using many different values of the parameters

and testing whether the stability of the numerical method is as predicted by (3.7). Without loss of generality, we assume

in all cases. We proceed in the following way. First, we choose a set of values of

and integrate the corresponding fractional cable equation. If

within the first 1000 integrations, then we say the method is unstable otherwise, we label the method as stable. We generated Figure 7 by starting the integration for values of

well below the theoretical stability limit given by (3.7) and kept increasing its value by 0.001 until condition (4.1) was first reached. The last value for which the method was stable is recorded and plotted in Figure 7. The limit value

is arbitrary, but choosing any other reasonable value does not significantly change these plots.


5. Convergence Analysis

In this section we show that the present numerical method is convergent, that is, that the numerical solution converges towards the exact solution when the size of the spatiotemporal discretization goes to zero. Let us define

as the difference between the exact and numerical solutions at the point

. Taking into account (2.9) and (2.11), one gets the equation that describes how this difference evolves:

As we did in the previous section for

as a combination of (sub) diffusion modes,

, and analyze the convergence of a single but generic

-mode [36, 39, 42]. Therefore, replacing

Now we will prove by induction that

satisfies the initial condition by construction, so that

. Taking into account (2.4), using the value

is bounded (in fact, it is smaller than 2). Using this result in (5.3), together with

Therefore the amplitude of the subdiffusive modes goes to zero when the spatiotemporal mesh goes to zero. Employing the Parseval relation, this means that the norm of the error

go to zero. This is what we aimed to prove.

6. Conclusions

An explicit method for solving a kind of fractional diffusion equation that involves several fractional Riemann-Liouville derivatives, which are approximated by means of the Grünwald-Letnikov formula, has been considered. The method was used to solve a class of equations of this type (fractional cable equations) with free boundary conditions, Dirac's delta initial condition, and different fractional exponents. The error of the numerical method is compatible with the truncating error, which is of order

. It was also proved that the method is convergent. Besides, it was also found that a fractional von-Neumann stability analysis, which provides very precise stability conditions for standard fractional diffusion equations, leads also to a very accurate estimate of the stability conditions for cable equations.

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Copyright

Copyright © 2011 J. Quintana-Murillo and S. B. Yuste. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


Shortest Distance between a Point and a Circle

What is the distance between a circle C with equation x 2 + y 2 = r 2 which is centered at the origin and a point P ( x 1 , y 1 ) ?

The ray O P &rarr , starting at the origin O and passing through the point P , intersects the circle at the point closest to P . So, the distance between the circle and the point will be the difference of the distance of the point from the origin and the radius of the circle.

Using the Distance Formula , the shortest distance between the point and the circle is | ( x 1 ) 2 + ( y 1 ) 2 &minus r &thinsp | .

Note that the formula works whether P is inside or outside the circle.

If the circle is not centered at the origin but has a center say ( h , k ) and a radius r , the shortest distance between the point P ( x 1 , y 1 ) and the circle is | ( x 1 &minus h ) 2 + ( y 1 &minus k ) 2 &minus r &thinsp | .

What is the shortest distance between the circle x 2 + y 2 = 9 and the point A ( 3 , 4 ) ?

The circle is centered at the origin and has a radius 3 .

So, the shortest distance D between the point and the circle is given by

That is, the shortest distance between them is 2 units.

What is the shortest distance between the circle x 2 + y 2 = 36 and the point Q ( &minus 2 , 2 ) ?

The circle is centered at the origin and has a radius 6 .

So, the shortest distance D between the point and the circle is given by

That is, the shortest distance between them is about 3.17 units.

What is the shortest distance between the circle ( x + 3 ) 2 + ( y &minus 3 ) 2 = 5 2 and the point Z ( &minus 2 , 0 ) ?

Compare the given equation with the standard form of equation of the circle,

( x &minus h ) 2 + ( y &minus k ) 2 = r 2 where ( h , k ) is the center and r is the radius.

The given circle has its center at ( &minus 3 , 3 ) and has a radius of 5 units.

Then, the shortest distance D between the point and the circle is given by

That is, the shortest distance between them is about 1.84 units.

What is the shortest distance between the circle x 2 + y 2 &minus 8 x + 10 y &minus 8 = 0 and the point P ( &minus 4 , &minus 11 ) ?

Rewrite the equation of the circle in the form ( x &minus h ) 2 + ( y &minus k ) 2 = r 2 where ( h , k ) is the center and r is the radius.

x 2 + y 2 &minus 8 x + 10 y &minus 8 = 0 x 2 &minus 8 x + 16 + y 2 + 10 y + 25 = 8 + 16 + 25 ( x &minus 4 ) 2 + ( y + 5 ) 2 = 49 ( x &minus 4 ) 2 + ( y + 5 ) 2 = 7 2

So, the circle has its center at ( 4 , &minus 5 ) and has a radius of 7 units.

Then, the shortest distance D between the point and the circle is given by


Discharge Equations and Curves for an RC Circuit

Once a capacitor is charged, we can replace the supply by a short circuit and investigate what happens capacitor voltage and current as it discharges. This time current flows out of the capacitor in the reverse direction. In the circuit below, we take KVL around the circuit in a clockwise direction. Since current flows anticlockwise, the potential drop across the resistor is positive. The voltage across the capacitor "points the other way" to the clockwise direction we&aposre taking KVL, so its voltage is negative.

So this gives us the equation:

Note that we could have taken KVL anti-clockwise around the loop, then the equation would be: Vc(t) - i(t)R = 0 . The important thing is that we assume a current direction first, then stay consistent. So voltage sources "pointing" and more positive in the direction we take KVL are positive and current flowing in the direction we take KVL cause negative PD across resistors.

Again the expression for voltage and current can be found by working out the solution to the differential equation for the circuit.

RC circuit capacitor discharge.

Equations for discharge current and voltage for an RC circuit.

Graph of discharge current through a capacitor in an RC circuit.

Voltage on a capacitor in an RC circuit as it discharges through the resistor R

An RC circuit is used to produce a delay. It triggers a second circuit when it&aposs output voltage reaches 75% of it&aposs final value. If the resistor has a value of 10k (10,000 ohms), and triggering must occur after an elapsed time of 20ms, calculate a suitable value of capacitor.

We know the voltage on the capacitor is Vc(t) = Vs (1 - e -t/RC )

75% of the final voltage is 0.75 Vs

So triggering of the other circuit occurs when:

Dividing both sides by Vs and replacing R by 10 k and t by 20ms gives us:

(1 - e -20 x 10^-3/(10^4 x C) ) = 0.75

e -20 x 10^-3/(10^4 x C) = 1 - 0.75 = 0.25

Take the natural log of both sides:

ln(e -2 x 10 ^ -7 / C ) = ln (0.25)


Cryptography

Cryptography involves encrypting data so that a third party can not intercept and read the data.

In the early days of satellite television, the video signals weren't encrypted and anyone with a satellite dish could watch whatever was being shown. Well, this didn't work because all of the networks using satellites didn't want the satellite dish owners to be able to receive their satellite feed for no cost while cable subscribers had to pay for the channel, they were losing money. So, they started encrypting the video signal with a system called Videocipher (later replaced by Videocipher II).

What the Videocipher encryption system did was to convert the signal into digital form, encrypt it, and send the data over the satellite. If the satellite dish owner had a Videocipher box, and paid for the channel, then the box would descramble (unencrypt) the signal and return it to its original, useful form.

This was done by using a key that was invertible. It was very important that they key be invertible, or there would be no way to return the encrypted data to its original form.

The same thing can be done using matrices.

Encryption Process

  1. Convert the text of the message into a stream of numerical values.
  2. Place the data into a matrix.
  3. Multiply the data by the encoding matrix.
  4. Convert the matrix into a stream of numerical values that contains the encrypted message.

Example

Consider the message "Red Rum"

A message is converted into numeric form according to some scheme. The easiest scheme is to let space=0, A=1, B=2, . Y=25, and Z=26. For example, the message "Red Rum" would become 18, 5, 4, 0, 18, 21, 13.

This data was placed into matrix form. The size of the matrix depends on the size of the encryption key. Let's say that our encryption matrix (encoding matrix) is a 2x2 matrix. Since I have seven pieces of data, I would place that into a 4x2 matrix and fill the last spot with a space to make the matrix complete. Let's call the original, unencrypted data matrix A.

18 5
A = 4 0
18 21
13 0

There is an invertible matrix which is called the encryption matrix or the encoding matrix. We'll call it matrix B. Since this matrix needs to be invertible, it must be square.

This could really be anything, it's up to the person encrypting the matrix. I'll use this matrix.

The unencrypted data is then multiplied by our encoding matrix. The result of this multiplication is the matrix containing the encrypted data. We'll call it matrix X.

67 -21
X = A B = 16 -8
51 27
52 -26

The message that you would pass on to the other person is the the stream of numbers 67, -21, 16, -8, 51, 27, 52, -26.

Decryption Process

  1. Place the encrypted stream of numbers that represents an encrypted message into a matrix.
  2. Multiply by the decoding matrix. The decoding matrix is the inverse of the encoding matrix.
  3. Convert the matrix into a stream of numbers.
  4. Conver the numbers into the text of the original message.

Example

The message you need to decipher is in the encrypted data stream 67, -21, 16, -8, 51, 27, 52, -26.

The encryption matrix is not transmitted. It is known by the receiving party so that they can decrypt the message. Other times, the inverse is known by the receiving party. The encryption matrix can not be sent with the data, otherwise anyone could grab the data and decode the information. Also, by not having the decoding matrix, someone intercepting the message doesn't know what size of matrix to use.

The receiving end gets the encrypted message and places it into matrix form.

67 -21
X = 16 -8
51 27
52 -26

The receiver must calculate the inverse of the encryption matrix. This would be the decryption matrix or the decoding matrix.

The receiver then multiplies the encrypted data by the inverse of the encryption matrix. The result is the original unencrypted matrix.

18 5
A = X B -1 = 4 0
18 21
13 0

The receiver then takes the matrix and breaks it apart into values 18, 5, 4, 0, 18, 21, 13, 0 and converts each of those into a character according to the numbering scheme. 18=R, 5=E, 4=D, 0=space, 18=R, 21=U, 13=M, 0=space.

Trailing spaces will be discarded and the message is received as intended: "RED RUM"


Watch the video: Cable Equation Derivation (August 2022).