F. S. Ellermeyer
MATH 1190 Exam 2 (Version 2) Solutions October 2, 2006 Name
Instructions. Your work on this exam will be graded according to two criteria: mathematical correctness and clarity of presentation. In other words, you must know what you are doing (mathematically) and you must also express yourself clearly. In particular, write answers to questions using correct notation and using complete sentences where appropriate. Also, you must supply su cient detail in your solutions (relevant calculations, written explanations of why you are doing these calculations, etc.). It is not su cient to just write down an answer with no explanation of how you arrived at that answer. As a rule of thumb, the harder that I have to work to interpret what you are trying to say, the less credit you will get. You may use your calculator on this exam but you may not use any books or notes. 1. Draw the graph of the piecewise linear function f (x) = x jx 3j
and use your graph to explain why this function is not dierentiable at a certain point. (What is this point?) Make sure to write this clearly and in complete sentence form. Also, write a piecewise-dened formula for f 0 (x). Solution: f itself can be written in piecewise form as follows: f (x) = 2x 3 if x 3 3 if 3 < x
from which it can be seen that the graph of f is as shown below:
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-1
0
1
2
3
4
5
6
Graph of f (x) = x
jx
3j
1
Since the graph of f has a sharp corner at the point on the graph corresponding to x0 = 3, we see that f is not dierentiable at this point. f is dierentiable at all other points and, since f is piecewise linear, it is easily seen that the derivative of f is f 0 (x) = 2. A graph of the polynomial function f (x) = 2x4 3 3 x + 5x2 2 5 x+1 2 2 if x < 3 . 0 if 3 < x
is shown below. Use your knowledge of the Power Rule, Constant Multiple, and Sum Rule to nd the slope of the tangent line to the graph of f at the point on the graph corresponding to x0 = 0. Also, write an equation for the tangent line to the graph of f at this point.
y 10
5 -2 -1 0 0 1 x 2
-5
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Graph of f (x) =
2x4
3 3 x 2
+ 5x2
5 x 2
+1
Solution: The derivative function of f is f 0 (x) = 8x3 9 2 x + 10x 2 5 . 2
Since f 0 (0) = 5=2 = 2:5, we see the slope of the tangent line to the graph of f at the point (0; f (0)) = (0; 1) is 2:5. An equation for this tangent line is y or y= 2:5x + 1. Below, we have sketched the graph of f together with the graph of this tangent line. 2 1= 2:5 (x 0)
y
10
5 -2 -1 0 0 1 x 2
-5
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-25
3. A rectangular box (without a top) is to be constructed from an 8 ft 8 ft square cardboard sheet by cutting equal size squares from each corner of the sheet, discarding these squares, and then folding the remaining cardboard into the shape of a box. (Refer to the gure below.) What size squares should be cut from the corners of the sheet in order to construct the box of greatest possible volume? (Your solution must be detailed - including all mathematical details and written explanations of your reasoning.)
3
Solution: With x as shown in the gure, the volume of the box is V (x) = x (8 2x)2 = 4x3 32x2 + 64x.
We want to maximize this function over the domain [0; 4]. Since V 0 (x) = 12x2 64x + 64 = 4 3x2 16x + 16 = 4 (3x 4) (x 4) ,
we see that V 0 (x) = 0 at the values x = 4=3 = 1:3 and x = 4. Obviously when x = 4, the cardboard is completely cut away and there is no box. Thus the maximum volume must be achieved when x = 1:3 ft. The volume achieved in this case is V 4 = 3 2 4 8 2 4 = 37: 926 ft3 . The graph of V shown below veries this conclusion. 3 3
4
V 35 30 25 20 15 10 5 0 0 1 2 3 x 4
Graph of V (x) = 4x
3
32x + 64x
2
4. If you have ever taken Chemistry and done some lab experiments, you might have used a Florence ask (which is also known as a boiling ask) in some of your experiments. In case you have never seen a Florence ask, a picture of one is shown below.
Suppose that we have a Florence that is initially empty and suppose that we start ask lling it with up some liquid at some constant rate (say, for instance, a constant rate of 100 milliliters per minute). Let D be the depth (in millimeters) of the liquid in the 5
ask at time t (in seconds). Thus D is a function of t and we can write D = f (t). Assume that we rst start pouring liquid into the at time t = 0. At time t = T1 , ask the liquid has reached the depth indicated in the gure (half way up the round part of the ask). At time t = T2 , the liquid has reached the depth indicated in the gure (the top of the round part of the ask). At time t = T3 , the is full. ask (a) Draw a graph of the function D = f (t). On your t axis, be sure to label the times t = 0, t = T1 , t = T2 , and t = T3 . It is important that your graph have the right concavity on each of the intervals (0; T1 ), (T1 ; T2 ), and (T2 ; T3 ). (b) (Circle the correct choice. Only one of the three choices is correct in each case): 1. dD=dt is (positive, negative, zero) throughout the interval (0; T1 ). 2. dD=dt is (positive, negative, zero) throughout the interval (T1 ; T2 ). 3. dD=dt is (positive, negative, zero) throughout the interval (T2 ; T3 ). (c) (Circle the correct choices. Only one of the three choices is correct in each case): 1. D is (increasing, decreasing, remaining constant) throughout the interval (0; T1 ). 2. D is (increasing, decreasing, remaining constant) throughout the interval (T1 ; T2 ). 3. D is (increasing, decreasing, remaining constant) throughout the interval (T2 ; T3 ). (d) (Circle the correct choices. Only one of the three choices is correct in each case): 1. d2 D=dt2 is (positive, negative, zero) throughout the interval (0; T1 ). 2. d2 D=dt2 is (positive, negative, zero) throughout the interval (T1 ; T2 ). 3. d2 D=dt2 is (positive, negative, zero) throughout the interval (T2 ; T3 ). (e) (Circle the correct choices. Only one of the three choices is correct in each case): 1. The graph of D is (concave up, concave down, has no concavity) throughout the interval (0; T1 ). 2. The graph of D is (concave up, concave down, has no concavity) throughout the interval (T1 ; T2 ). 3. The graph of D is (concave up, concave down, has no concavity) throughout the interval (T2 ; T3 ). 5. A very strong person standing at the top of the Empire State Building (which is 1200 feet tall) throws a bowling ball upward with an initial speed of 100 ft/sec. The position of the ball at any time t after launching (and befor...