1. Find a power series representation of

This function differentiates to a fraction we can interpret as the sum of a geometric series.

To complete Step 4, note that for x=0, = 0 and the series also equals 0, so C must be 0. The radius of convergence for the series obtained in Step 2 is 1, with an interval of (-1, 1). When you integrate the interval of convergence can gain or lose endpoints, but that’s a small thing to check after this fairly quick conversion.

2. Find a power series representation of

Instead of differentiating, this time we integrate to get the sum of a geometric series.

In Step 3 note that from the perspective of the derivative operator is constant. There’s no constant of integration to find so once you’ve rounded the bases you’re done – well, except for thinking about the radius and interval of convergence.

One more after the jump.

3. Find a power series representation for

This time we differentiate twice.

Once again you do not have to find C at the end, but in this case it is because C stays undefined: the series is intended to equal an indefinite integral, not a specific function. C does not affect the radius or interval of convergence of the series.

We can use the series above to find a power series for the definite integral Since plugging 0 into the series obtained in Step 6 gives 0, the answer is the result of plugging in 1:

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Imagine when you have f(g(x)) that x is a little sealed box. It is inside a second sealed box marked g, and that box is in a third sealed box marked f.

To differentiate you open the box marked f, and remove and open the box marked g, and remove and open the box marked x. The open boxes are the derivatives of the closed boxes, and containment still represents composition.

All put together, the drawing above represents the derivative of f(g(x)):

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	optimization	related rates
sample keywords	maximize, minimize, most, smallest	how fast, rate of change/increase/decrease
sample problem	What dimensions maximize the area enclosed in a rectangular fence built with 100 feet of fencing?	A shape is deforming such that it is always a rectangle of perimeter 100 ft, but the width is increasing by 5 ft/min. How fast is the area changing when the width is 20 feet?
goal	maximize area	find rate of change of area
answer form	dimensions in feet	rate in ft2/min
equation		likewise,
equation, unabridged	(area viewed as function of width)	(area and width viewed as related functions of time)
final answer	a 25′ square fence maximizes the area	the area is changing by 50 ft2/min

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A. Make two-word phrase of goal, where first word is minimize or maximize and second word is the name of a quantity.
B. Name quantities and state all relationships they have, as equations. Graphs or other pictures can be helpful.
C. Make the quantity to be optimized a function of a single variable, using other relationship(s) as needed to simplify.
D. Find critical points and extrema for the function from C, and pick out the point(s) at which the goal from A is met.
E. Unpack the information from D to answer the question as stated.

Example 1.
Of all possible pairs of numbers that sum to 100, find those with the largest product.
A. Goal: maximize product.
B. Call the product and the numbers P, x, and y, respectively. We have by definition and from the problem statement.
C. Use the sum equation to simplify the product equation: .
D. . CPs: only when . Is this a maximum? , , so the first derivative test says yes.
E. The pair of numbers with maximum product out of all those that sum to 100 is 50 and 50.

Example 2.
A cylindrical drum must hold 10 m³ of sand. The material for the sides costs $10 per m² and the material for the base and lid cost $15 per m². Find the dimensions of the drum which minimize its cost.
A. Goal: minimize cost.
B. The volume of a cylinder is , which we know equals 10. The cost of the drum is 10*(side area) + 15*(top and bottom area), which we may also put in terms of radius r and height h. Let cost be C; then .
C. To avoid square roots, put everything in terms of r. We have from the volume constraint, which means cost as a function of radius is .
D. . Critical points occur when ; solving for r we obtain , so . To find whether this is a minimum we’ll do the second derivative test. . For positive r (such as our critical point) is positive, so the graph is concave up and we have indeed found a minimum.
E. We must have a drum with a radius of m and a height of m, which simplifies to m. In decimal form, m and m.

Example 3.
Find the dimensions of the largest-area rectangle that can be inscribed in a circle of positive radius r.
Note: inscribed means the corners of the rectangle are on the circle.
A. Goal: maximize area.
B. For simplicity, let us suppose the circle is centered at the origin. Any inscribed rectangle must have a corner in each quadrant, and any corner tells us the location of the other three, by symmetry. Call the location of the first quadrant corner ; the lengths of the sides of the rectangle are and . Now we can wrote relationship equations: area . Since the corners are on the circle, the distance from the origin to must be r: .
C. We must put the area equation in terms of one variable since it is what we are trying to maximize. Not having a numerical value for the radius can make it confusing, but r is a parameter: a stand-in for a constant we don’t wish to actually specify. It does not matter which variable we substitute for since we have perfect symmetry, so solve for . Then , where r is an unspecified constant.
D. . Setting this equal to zero, we obtain , which simplifies to and finally . Is this a maximum? Use the first derivative test. is approximately 1.4; $latex 5/4 = 1.25 < \sqrt{2} < 2$, so test $latex r/2 < r/\sqrt{2} < 4r/5$. $latex A'(r/2) = \frac{-4(r^2/2)}{\sqrt{r^2 - r^2/4}} + 4\sqrt{r^2-r^2/4} = \frac{-r^2}{\sqrt{3r^2/4}} + 4\sqrt{3r^2/4} = \frac{-r^2}{\sqrt{3}/2 r} + 2\sqrt3 r$ $latex = 2\sqrt 3 r - 2r/\sqrt 3$.    $latex 2\sqrt 3 > 2/\sqrt 3$ so this is positive.
, which is negative, and hence is a maximum.
E. When , so does x, by the equation relating both of them to r. Hence a square of side length is the largest possible inscribed rectangle.

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A. Assign variables to your quantities and find relationship equations for them.
B. If relevant, simplify the equations to remove extraneous variables.
C. Express given and asked-for rates of change in terms of d/dt. At what point(s) do you want to know the rate(s) of change?
D. Differentiate relationship equation with respect to t.
E. Plug in known values and solve for desired value.

In C I will be finding the value of all variables at the desired point, although frequently you don’t need them.

Example 1.
The area of a square is increasing by 3 in²/s. What is the rate of change of the length of the sides when the side length is 2 in?
A. Name quantities: area A, side length (same for all sides since it’s a square). Relationship: .
B. This is already as simple as it gets.
C. Rates of change: given: in²/s. We want to know at the point , which is .
D. is .
E. Know and want when ; so . When the side length is 2 in, the sides are changing at a rate of 3/4 in/s.

Example 2.
You are testing walkie-talkie range with a friend. The two of you start at the same point, and then you amble north at 3 mph and your friend strides east at 4 mph. How fast are you moving away from each other when your friend is 3 mi away from the starting point?
A. Since you are moving north, label your position x, and since your friend is moving east, label her position y. You are really interested in the distance between you; call it R for (walkie-talkie) range. That is related to your positions by the equation .
B. Since and are both constant, and hence x and y are constant multiples of each other, we could simplify by putting one in terms of the other but won’t gain much by doing so.
C. We are given and and want . The point at which we want is , which is and .
D. is .
E. We know and , and we want when and ; . When you have gone 2.25 mi and your friend 3 mi, the distance between you is changing at a rate of 5 mph.

Example 3.
25 cubic inches of pizza dough are being spun in the air. Assume the surface is smooth and the outline is circular at all times. If the radius increases at a constant rate of 0.5 in/s, what is the rate of change of the thickness of the dough when the radius is 4 inches?
A. We have thickness, radius, and volume. Since t is taken, we’ll call them h, r, and V, respectively. The relationship is .
B. Since V is constant, we can plug in its value. Since V is by itself on one side of the equation it is roughly equal amounts of work to do that as to leave it as V and plug in 0 for , but since we did not simplify Example 2 we’ll simplify this one: .
C. We are given in/s, and we want at the point , which is .
D. is .
E. We know , and we want to know when and . Hence ; ; in/s. Make sense? If the volume is constant an increase in one dimension must be accompanied by a decrease in the other.

Example 4.
A coffee filter is being held under a running tap. It is cone-shaped, with height 8 cm and top radius 3 cm. Water flows in at the top at a rate of 4 cm³/s and drips out the bottom at 1.5 cm³/s. How fast is the water level in the filter rising when the depth is 2 cm?
A. Let’s call depth h since d is taken, radius r, and volume V. The volume equation for a cone is .
B. Because the filter is a cone, the ratio between depth and radius is the same for any amount of water in the filter. That is, while r is not always 3 and h is not always 8, is always 3/8: . Thus the volume equation simplifies to .
C. We are not given right off, but the increase and decrease of water volume are both constant, so we can compute it: cm³/s. We want to find at the point , which gives .
D. is .
E. We have and , so we obtain , for cm/s. When the water is 2 cm deep, it is rising at a rate of 1.41 cm/s.

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