Here is quiz 4.

## 175 – Integrating products of secants and tangents

October 9, 2009(In what follows, I will write for and for )

Recall that

that

and that

The formulas below make use of these identities repeatedly.

We want a series of methods and reduction formulas that allow us to evaluate any expression of the form

for and integers,

## 175 – Quiz 3

October 2, 2009Here is quiz 3.

**Problem 1** is Exercise 9.3.16 from the book. Here is a graph showing the cardioid and the circle. The easiest way to compute the required area is by subtracting the area of the circle from that of the cardioid. (Note that the circle is completely contained in the cardioid as, for each we have that This shows that in the portion of the circle in the first quadrant is within the portion of the cardiod there. The same holds in the fourth quadrant by symmetry. Of course, all of this also follows directly from the graph.)

The area of the cardioid is given by and the area of the circle is (note that as varies from to the whole circle is traveled once).

By symmetry, so the area we want is given by

Recall that so the integral is

Now note from the Cartesian graph of that so the area we want is just

**Problem 2** is Exercise 9.5.50 from the book. Recall that the standard polar equation of a line not going through the origin is given by

where is the distance from the line to the origin, and is the angle of the point in the line that realizes this distance, i.e., are the polar coordinates of the point on the line closest to the origin.

There are at least two ways we can proceed:

1) We can directly find the distance from the line to the origin. To do this, we recall that the line that goes through the origin and is perpendicular to where is given by The point in closest to the origin is the intersection of these two lines. In the case that interests us, this is the intersection of and so or or so We have found that the point is closest in to the origin. Its distance is Its angle is on the fourth quadrant and satisfies i.e.,

Putting this together, is the desired equation.

2) The other method we saw in lecture is based in the fact that

and consists in first writing the Cartesian equation of the line directly in polar coordinates, and then using the resulting expression to find and We have so or This gives us

We find a constant such that and are the cosine and the sine of some angle, precisely We then have

To find this we use that for any so must satisfy or so The value of such that and is This means that the equation we are looking for is

just as with the previous method.

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## 175 – Midterm 1

September 25, 2009Here is the midterm.

1. Hooke’s law says that the force required to stretch a spring a distance of units from its natural length is given by where is a constant that only depends on the spring.

We are told that for the spring of problem 1 we have where units of distance are measured in meters, and forces in Newtons. This means that or The work required to stretch the spring 10 m beyond its natural length is then

2. We are given a curve with equation and asked to rotate about the -axis the segment where To find the area of the resulting surface, we can consider splitting the surface into little shells. A typical `thin’ shell, at height has surface area where is the radius of the shell, in this case just the value of corresponding to i.e., Also, is here the arc length element, given by

We have: so

Hence the requested area is given by

This expression reduces to

3. We are given the region bounded by the curve and the lines and We rotate it about the -axis, and are asked to find the volume of the resulting solid. A first attempt is to use the washer method. A typical thin washer consists of a slice of the solid for a fixed value of and with thickness Its volume is Here, is the exterior radius, which in this case is always 2, and is the interior radius, given by This means the washer contributes of the full volume, and adding all of them together we obtain The upper limit of 4 comes from observing that the curves and cross precisely at the point

Unfortunately, this integral is not one of the given expressions.

So we attempt a different approach, using now the shell method. Now a typical thin shell is obtained by fixing a height and rotating about the -axis the slice of the figure of thickness and height Its volume is where is the radius of the shell, in this case and is the length of the shell, in this case We thus proceed to express in terms of Since then Hence the thin shell contributes of the full volume, and the volume is obtained by adding all these contributions, obtaining This is expression (c).

4. The length of the curve for is given by the integral In this case, so and so the integral reduces to which is expression (b).

[There is a small problem with this integral, though, because the denominator vanishes at both endpoints. Later, in Chapter 7, we will learn how to handle integrals of this kind. Note that if the question had been to actually find the length, there is an easier method: The curve is half a circumference of radius 1, so the length is ]

5. (a) The differential equation can be rewritten as which is clearly separable. (T)

(b) The area swept by for is four times the area swept by Intuitively, we are making lengths twice as long, and areas carry two dimensions of length. Think of the area of a square of side 2, for example. Formally, we can check that this is the case: The area swept by the first curve is given by while the area swept by the second curve is given by (T)

(c) The half life of a given element is 3 days. This means that 50% of the given amount of the element has decayed after 3 days. After 3 more days, another 25% would have decayed. After 3 more days, another 12.5%. After 3 more days, another 6.25%. Since the given statement is clearly true.

A longer way of checking the same is by recalling that the total amount of the element that remains after time if originally the amount is is given by where is a given constant. Measure in days. We are told that so Then and (T)

## 175 – Dandelin spheres

September 22, 2009Here is a link to the Wikipedia page on Dandelin spheres, giving us the “modern” proof of the equivalence between the definition of conic sections as regions of intersection of planes and cones, with the standard definition in terms of distance to foci. The links on the Wikipedia page provide further explanations and nice graphics illustrating the argument.

## 175 – Quiz 2

September 19, 2009Here is quiz 2.

**Problem 1** is (simplification of part of) exercise 6.6.16 from the book.

To solve the question, use coordinates as in the accompanying figure in the book, so the origin is at ground level, and *increases* downwards. The units of are feet. For a fix with the thin slice of water in the tank at depth and of tickness has volume and weighs This is a constant force, so the work required to remove it to ground level is just where is the depth at which the slice is located, i.e., Hence, The total work is obtained by adding all these contributions, i.e., ft-lb.

**Problem 2** is exercise 9.2.16 from the book.

The curve is Since the graph is symmetric about the -axis (because whenever is in the graph, then so is ).

Since the graph is symmetric about the origin (because whenever is in the graph, then so is ).

Since the graph is symmetric about both the origin and the -axis, it is also symmetric about the -axis.

To sketch the curve, look first at Here so which is impossible, so there is nothing to graph here. Consider now what happens when As increases, increases, from to So the same occurs with This means that increases from to , and decreases from to The part with gives us a curve in the second quadrant, and the part with gives us its reflection about the origin. This part of the curve is in the fourth quadrant. Their reflections on the -axis complete the curve, which can be seen here.

Note that is undefined. This corresponds to the fact that at the origin the tangent to the curve is the -axis, as can be seen from the graph.

## 175 – Quiz 1

September 4, 2009Here is quiz 1.

**Problem 1** is exercise 6.1.16 from the book. (The graph in the book has a typo. The maximum height of the graph is rather than 1.)

Using the disk method, the volume can be expressed as Here, If this expression were not squared, to find the integral would be easier, by a direct substitution. Being squared, we need to work harder. One possible approach is to use some trigonometric identities. For example:

- so Again, the fact that the sine expression is squared makes things difficult. We use another trigonometric identity:
- so Now we can proceed to integrate:

which simplifies to

There are other approaches. For example, other trigonometric identities could be used as well. Also, the book includes a formula for the integral of powers of sine and cosine; we will study this formula later. Trying to use the shell method leads to rather messy expressions, I didn’t work out the full details. I believe that the argument above is perhaps the most efficient.

**Problem 2** is exercise 6.3.14 from the book. Since is given as a function of the most efficient route seems to be to use as the parameter, so the expression for the length of the curve takes the form where the derivative of is with respect to

We have so

The expression for the length then reduces to