Calculus II Class 05/02

Problem Number 1

What is the interval of convergence for the series x / 3 + x^2 / 12 + x^3 / 48 + ... ?

The interval of convergence for a power series is found by taking the limit of a(n+1) / a(n), where a(n) is the coefficient of x^n.

The coefficients are 1/3, 1/12, 1/48, ... .

The pattern is that each coefficient is 1/4 of the preceding coefficient.

An expression for the nth coefficient is a(n) = 1 / 4^(n-1) * 1/3 = 1 / ( 3 * 4^(n-1) ).

So a(n+1) / a(n) = 1 / (3 * 4^(n+1 -1) ) / (1 / (3 * 4^(n-1) ) = 1/4.

The radius of convergence is therefore R = 1 / (1/4) =4.

The power series is in the form a(n) * ( x - 0 )^n so the center of the interval of convergence is x = 0.

The interval of convergence is therefore -R < x < R, or | x | < R.

Problem Number 2

Use the Taylor polynomial of a simpler function to find the Taylor polynomial of degree 4 for the function f(x) = sin( 4 x^-1.002 ), expanding about x=0.

The Taylor polynomial for sin(z) is

• z - z^3/3! + z^5/5! -z^7/7! + ... . 

The degree 4 polynomial is just

• P4(z) = z - z^3 / 3!.

Letting z = 4 x^-1.002 we find that the degree-4 polynomial for f(x) is

• deg 4 polynomial for sin(4x^-1.002) = (4x^-1.002) - (4x^-1.002)^3 / 3!.

Problem Number 3

Sketch the direction field for the differential equation dy / dx = .62 x / y.   Sketch some solution curves corresponding to this field.

Problem Number 4

In a predator-prey model we assume that in the absence of predators the prey will grow exponentially with an annual growth rate of .56, while in the absence of this prey the predators will decline with an annual growth rate of -.8201.  If the interaction rate between predator and prey is the product of the number of predators and the number of prey, then we assume that the rate at which prey are consumed is .0008 times the interaction rate.  We assume also that the annual growth rate of the predators increases by .0000078 times the number of prey.  Write the differential equations for the rates of change of the populations of predators and prey.

If there are initially 1058 predators and 98320 prey,

• find the equilibrium populations of predators and prey
• for populations of predators and prey up to about three times the equilibrium populations, sketch the slope field for this model
• sketch the approximate trajectory of the populations
• sketch on one set of coordinate axes the approximate populations of the predators and the prey for one cycle, using a different vertical scale for each species, with the scales chosen so that the curves occupy approximately the same region of the graph.

This wasn't covered this term, due to snow days and other hassles the instructor had the uncharacteristic mercy to omit the last 4 sections of the text.

Problem Number 5

If a capacitor with capacitance 5.2 farads and an inductor with inductance 8.9 henry form an oscillating circuit, then if the initial charge on the capacitor is 0 and the initial current dQ/dt is 3.7 Coulombs/second, what function gives the charge Q as a function of clock time?  The voltage across the inductor is L Q'', the voltage across the capacitor is C Q and the total voltage around the circuit is 0.

This is from one of the omitted sections.  Too bad, this is pretty neat stuff.

Problem Number 6 Problem Number 7

The charge on a capacitor changes at a rate proportional to the difference between the charge and a maximum charge of 7.7 Coulombs.  The charge is initially 0, and after 26 seconds the charge is 2.28 Coulombs.  What function models the charge as a function of time?  After how long will the charge be within .1% of the maximum?

Interpret one phrase at a time.

'The charge changes at a rate proportional to .... ' means that if we let Q be the charge, dQ/dt is proportional to something.

So what's dQ/dt proportional to?  Read on.  It's proportional to the difference between the charge and 7.7 C (max charge).  This difference is | Q - 7.7 |, suppressing the unit C (which stands for Coulombs).

The proportinality is therefore

dQ/dt = k | Q - 7.7 |.

This is a differential equation.  We might as well solve it before we read on.

Since k isn't specified as positive or negative we can just write

dQ/dt = k ( Q - 7.7).

Separating variables we get

dQ/(Q-7.7) = k dt.

The integration is straightforward (substitute u = Q - 7.7, etc.) and we get

ln | Q - 7.7 | = k t + c.

The usual steps lead us to

• Q - 7.7 = A e^(kt), where A cannot be 0.

We write the final solution as

• Q(t) = 7.7 + A e^(kt).

Now we can read more.

We see that the initial charge is 0.  That means Q(0) = 0 so

• 0 = 7.7 + A e^(k*0),

which tells us that

• A = -7.7

so that

• Q(t) = 7.7 - 7.7 e^(kt) = 7.7 ( 1 - e^(kt))

Let's read a little more.

We see that ' after 26 seconds the charge is 2.28 Coulombs ' , which we interpret at Q(26) = 2.28, giving us

• Q(26) = 7.7 ( 1 - e^(k * 26) ) = 2.28

so that

• e^(k*26) = 1 - 2.28 / 7.7  = .7039.

Solving for k we take the ln of both sides to get

• 26 k = ln(.7039),

so that

• k = ln(.7039)/26 = -.0135.

Our function is therefore

• Q(t) = 7.7 ( 1 - e^(-.0135 t) ).

The charge is increasing from 0 to max charge 7.7 C.  It's within 1% of the max when it's equal to .99 * 7.7 C = 7.623 C, so we solve for t:

• 7.7 ( 1 - e^(-.0135 t) ) = 7.623

Get the equation into the form e^(-.0135 t) = something and take the ln of both sides, etc..

We end up with t = 341, approx.

Problem Number 8

Give the first three Fourier approximations to the sawtooth-wave function f(x) = [-x if -`pi <= x <= 0, x if 0 < x < =`pi].  What fraction of the energy of the function is contained in the constant term and the first three harmonics?

We see that the integral of the function itself from -pi to pi is double the area of a triangle whose base and altitude are both pi.  The area of one such triangle is pi^2/2 so the total area is 2 * pi^2/2 = pi^2.

The first coefficient is a(0) = pi^2 / (2 pi) = pi / 2.

Since f(x) is even and sin(kx) is odd for any k, the integral of f(x) sin(kx) from x = -pi to x = pi must be zero.

The integral from -pi to pi of x cos(kx) can be found by integration by parts.  The result is 2 / k^2, so the coefficient a(k) is 2 / (pi k^2).

Problem Number 7

Assuming that k and b are constants, solve the differential equation dR / dt = k (R - b).

We separate variables to get

• dR / (R - b) = k dt

and after substituting u = R - b, etc., we get

• ln | R - b | = k t + c

with general solution

• R = b + A e^(kt), where A is a nonzero constant.

Problem Number 2

Find the approximate solution of the differential equation dy / dx = 1.6 x / y on the interval ( .5, 1.5), if y( .5) = .97. Use increment .25.

Think of a slope field.  We start at the point (.5, 1.5) and follow the slope for an x increment of .25, then repeat starting at the new point, etc. until we reach x = 1.5.

At (.5, 1.5) we have x = .5 and y = 1.5 so that dy/dx = 1.6 * .5 / 1.5 = .533... .  Following this slope for `dx = .25 we get `dy = .533 * .25 = .133.  So the new y will be 1.5 + .133 = 1.633 and our new point is (.75, 1.633).

Starting at (.75, 1.633) we have dy/dx = 1.6 * .75 / 1.633 = .735 approx..  Following this slope for `dx = .25 we get `dy = .735 * .25 = .181 approx..  So the new y will be 1.633 + .181 = 1.813 approx. and our new point is (1.00, 1.813).

Find the Fourier series for f(x) = | x |, -pi < x < pi.

2·COS(pi·k)/k^2 + 2·pi·SIN(pi·k)/k - 2/k^2