Solutions to Problem Set 15

Math 211-03

10-12-2017

[Work]

1. How much work is done in lifting a 40 pound box of books a distance of 18 inches?

Since 18 inches is 1.5 feet, the work done is

$$40 \cdot 1.5 = 60\ \hbox{foot-pounds}.\quad\halmos$$


2. An 8-pound force is needed to stretch a spring 2 feet beyond its unstretched length. Find the work done in stretching the spring from an unstretched state to 5 feet beyond its unstretched length.

Using the equation $F = -k x$ , I have

$$\eqalign{ 8 & = -k \cdot 2 \cr -4 & = k \cr}$$

Hence, $F = 4 x$ .

The work done in stretched the spring from an unstretched state ($x = 0$ ) to 5 feet beyond its unstretched length ($x = 5$ ) is

$$W = \int_0^5 F\,dx = \int_0^5 4 x\,dx = \left[2 x^2\right]_0^5 = 50\ \hbox{foot-pounds}.\quad\halmos$$


3. A 10-foot chain hangs below a platform, with a 20-pound rock attached to its end. The chain weighs 2 pounds per foot. How much work is done in lifting the chain and the rock up to the platform?

The work done is the sum of the work done in lifting the chain and the work done in lifting the rock.

The rock weighs 20 pounds and must be lifted a distance of 10 feet. The work done is

$$20 \cdot 10 = 200\ \hbox{foot-pounds}.$$

For the chain, set up vertical coordinates so that $y = 0$ corresponds to initial position of the bottom end of the chain and $y = 10$ corresponds to the platform. Consider a small piece of the chain of length $\Delta y$ located y units above the bottom. The chain weighs 2 pounds per foot, so the weight of the small piece is $2 \cdot
   \Delta y$ . The piece must travel a distance of $10 - y$ feet to get to the platform.

$$\hbox{\epsfysize=1.5in \epsffile{work1.eps}}$$

The work done in lifting the small piece up to the platform is $(10 - y) \cdot 2 \Delta y$ . Hence, the work done in lifting the whole chain is

$$\int_0^{10} (10 - y) \cdot 2\,dy = 2 \left[10 y - \dfrac{1}{2} y^2\right]_0^{10} = 100\ \hbox{foot-pounds}.$$

The total work done in lifting the chain and the rock is

$$200 + 100 = 300\ \hbox{foot-pounds}.\quad\halmos$$


4. A rectangular tank is 6 feet high, and has a base which is 4 feet by 3 feet. The tank is full of water. How much work is done in pumping all the water out of the top?

Note: The weight of water is 62.4 pounds per cubic foot.

Set up vertical coordinates so that $y = 0$ corresponds to the bottom of the tank and $y = 6$ corresponds to the top.

$$\hbox{\epsfysize=1.5in \epsffile{work2.eps}}$$

Divide the water up into rectangular slabs parallel to the base. Consider a slab of thickness $\Delta y$ located y units above the base. Its volume is

$$4 \cdot 3 \cdot \Delta y = 12 \Delta y.$$

Its weight is $62.4 \cdot 12
   \Delta y$ . The slab must be lifted a distance of $6 - y$ feet to reach the top of the tank, so the work done in lifting the slab to the top is $62.4 \cdot 12 (6 - y) \Delta y$ .

The total amount of work is

$$\int_0^6 62.4 \cdot 12 (6 - y)\,dy = 62.4 \cdot 12 \left[6 y - \dfrac{1}{2} y^2\right]_0^6 = 13478.4\ \hbox{foot-pounds}.\quad\halmos$$


5. A circular tank with radius 6 feet is filled with water. Find the work done in pumping all the water out of the top.

Note: The weight of water is 62.4 pounds per cubic foot.

Set up vertical coordinates so that $y = -6$ corresponds to the bottom of the tank and $y = 6$ corresponds to the top.

$$\hbox{\epsfysize=1.75in \epsffile{work3.eps}}$$

Note that y is measured from the center of the sphere (the circle in the cross-sectional picture above). So the center is at $y = 0$ .

Divide the water up into circular slabs parallel to the base. Consider a slab of thickness $\Delta y$ located y units above the base. Its radius is $r
   = \sqrt{36 - y^2}$ .

Its volume is

$$\pi r^2 \cdot \Delta y = \pi (36 - y^2) \Delta y.$$

Its weight is $62.4 \cdot \pi
   (36 - y^2) \Delta y$ . The slab must be lifted a distance of $6
   - y$ feet to reach the top of the tank, so the work done in lifting the slab to the top is $62.4 \cdot \pi (36 - y^2)(6 - y)
   \Delta y$ .

Since the tank is full, the water extends from $y = -6$ to $y = 6$ .

The total amount of work is

$$\int_{-6}^6 62.4 \cdot \pi (36 - y^2)(6 - y)\,dy = 62.4 \pi \int_{-6}^6 \left(y^3 - 6 y^2 - 36 y + 216\right)\,dy =$$

$$62.4 \pi \left[\dfrac{1}{4} y^4 - 2 y^3 - 18 y^2 + 216 y\right]_{-6}^6 = 107827.2 \pi = 338749.13937 \ldots\ \hbox{foot-pounds}.\quad\halmos$$


[Sequences]

6. A sequence is defined by $a_n
   = \dfrac{n}{2 n + 3}$ for $n \ge 1$ . List the first four terms $a_1$ , $a_2$ , $a_3$ , and $a_4$ .

$$a_1 = \dfrac{1}{5}, \quad a_2 = \dfrac{2}{7}, \quad a_3 = \dfrac{3}{9} = \dfrac{1}{3}, \quad a_4 = \dfrac{4}{11}.\quad\halmos$$


7. A sequence is defined recursively by

$$a_1 = 3, \quad a_{n + 1} = a_n^2 + n \quad\hbox{for}\quad n \ge 1.$$

List the first four terms $a_1$ , $a_2$ , $a_3$ , and $a_4$ .

$$a_1 = 3.$$

$$a_2 = a_1^2 + 1 = 3^2 + 1 = 10.$$

$$a_3 = a_2^2 + 2 = 10^2 + 2 = 102.$$

$$a_4 = a_3^2 + 3 = 102^2 + 3 = 10407.\quad\halmos$$


8. Determine whether the sequence $a_n = 4^{-n}$ converges or diverges. If it converges, find the limit.

$$\lim_{n \to \infty} 4^{-n} = 0.$$

The sequence converges to 0.


9. Determine whether the sequence $a_n = \cos 10 n$ converges or diverges. If it converges, find the limit.

$$\lim_{n \to \infty} \cos 10 n \quad\hbox{is undefined}.$$

The sequence diverges.


10. Determine whether the sequence $a_n = \dfrac{4 n^2 + 1}{3 n^2 + n + 1}$ converges or diverges. If it converges, find the limit.

You can use the fact the the highest powers on the top and bottom are both $n^2$ , so the limit is the ratio of their coefficients:

$$\dfrac{4 n^2}{3 n^2} = \dfrac{4}{3}.$$

Alternatively, you can use L'H\^opital's rule:

$$\lim_{n \to \infty} \dfrac{4 n^2 + 1}{3 n^2 + n + 1} = \lim_{n \to \infty} \dfrac{8 n}{6 n + 1} = \lim_{n \to \infty} \dfrac{8}{6} = \dfrac{8}{6} = \dfrac{4}{3}.\quad\halmos$$


11. Determine whether the sequence $a_n = \dfrac{6 n^3 + 5 n + 1}{n^2 + 3}$ converges or diverges. If it converges, find the limit.

Since the power on the top is greater than the power on the bottom,

$$\lim_{n \to \infty} \dfrac{6 n^3 + 5 n + 1}{n^2 + 3} = \infty.\quad\halmos$$

Note: This means that the sequence diverges. Since I can say that the terms go to $\infty$ , it's better to be specific and say this.


12. Determine whether the sequence $a_n = (-1)^n \dfrac{8 n + 3}{3 n + 1}$ converges or diverges. If it converges, find the limit.

Note that

$$\lim_{n \to \infty} \dfrac{8 n + 3}{3 n + 1} = \dfrac{8}{3}.$$

However, $(-1)^n$ oscillates between 1 and -1.

Therefore, the terms of the sequence oscillate between the limits $\dfrac{8}{3}$ and $-\dfrac{8}{3}$ .

Hence, the sequence diverges: The limit of the sequence is undefined.


13. Determine whether the sequence $a_n = (-1)^n \dfrac{7 n - 2}{5 n^2 + 3}$ converges or diverges. If it converges, find the limit.

$(-1)^n$ is either 1 or -1, depending on whether n is even or odd. Hence,

$$-\dfrac{7 n - 2}{5 n^2 + 3} \le (-1)^n \dfrac{7 n - 2}{5 n^2 + 3} \le \dfrac{7 n - 2}{5 n^2 + 3}.$$

Since the largest power on the bottom (in $5 n^2$ ) is bigger than the largest power on the top (in $7 n$ ), I have

$$\lim_{n \to \infty} \dfrac{7 n - 2}{5 n^2 + 3} = 0 \quad\hbox{and}\quad \lim_{n \to \infty} -\dfrac{7 n - 2}{5 n^2 + 3} = 0.$$

The terms on the ends of the inequality above both go to 0. By the Squeezing Theorem,

$$\lim_{n \to \infty} (-1)^n \dfrac{7 n - 2}{5 n^2 + 3} = 0.\quad\halmos$$


14. Determine whether the sequence $a_n = \dfrac{\cos n + 2}{n^2}$ converges or diverges. If it converges, find the limit.

$$\eqalign{ -1 \le & \cos n \le 1 \cr 1 \le & \cos n + 2 \le 3 \cr \noalign{\vskip2pt} \dfrac{1}{n^2} \le & \dfrac{\cos n + 2}{n^2} \le \dfrac{3}{n^2} \cr}$$

Now

$$\lim_{n \to \infty} \dfrac{1}{n^2} = 0 \quad\hbox{and}\quad \lim_{n \to \infty} \dfrac{3}{n^2} = 0.$$

By the Squeezing Theorem,

$$\lim_{n \to \infty} \dfrac{\cos n + 2}{n^2} = 0.$$

The sequence converges to 0.


15. Determine whether the sequence $a_n = n^2 e^{-4 n}$ converges or diverges. If it converges, find the limit.

Use L'H\^opital's rule:

$$\lim_{n \to \infty} n^2 e^{-4 n} = \lim_{n \to \infty} \dfrac{n^2}{e^{4 n}} = \lim_{n \to \infty} \dfrac{2 n}{4 e^{4 n}} = \lim_{n \to \infty} \dfrac{2}{16 e^{4 n}} = 0.\quad\halmos$$


I don't know that I have ever found any satisfactory answers of my own. But every time I ask it, the question is refined. ... questioning as exploration, rather than the search for certainty. - Ta-Nehisi Coates


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