Math 115 Second Midterm November 12, 2018

Transcription

1 EXAM SOLUTIONS Math 5 Second Midterm November, 08. Do not open this eam until you are told to do so.. Do not write your name anywhere on this eam. 3. This eam has 3 pages including this cover. There are problems. Note that the problems are not of equal difficulty, so you may want to skip over and return to a problem on which you are stuck. 4. Do not separate the pages of this eam. If they do become separated, write your UMID (not name) on every page and point this out to your instructor when you hand in the eam. 5. Note that the back of every page of the eam is blank, and, if needed, you may use this space for scratchwork. Clearly identify any of this work that you would like to have graded. 6. Please read the instructions for each individual problem carefully. One of the skills being tested on this eam is your ability to interpret mathematical questions, so instructors will not answer questions about eam problems during the eam. 7. Show an appropriate amount of work (including appropriate eplanation) for each problem, so that graders can see not only your answer but how you obtained it. 8. The use of any networked device while working on this eam is not permitted. 9. You may use any one calculator that does not have an internet or data connection ecept a TI-9 (or other calculator with a qwerty keypad). However, you must show work for any calculation which we have learned how to do in this course. You are also allowed two sides of a single 3 5 notecard. 0. For any graph or table that you use to find an answer, be sure to sketch the graph or write out the entries of the table. In either case, include an eplanation of how you used the graph or table to find the answer.. Include units in your answer where that is appropriate.. Problems may ask for answers in eact form. Recall that = is a solution in eact form to the equation =, but = is not. 3. Turn off all cell phones, smartphones, and other electronic devices, and remove all headphones, earbuds, and smartwatches. Put all of these items away. 4. You must use the methods learned in this course to solve all problems. Problem Points Score Problem Points Score Total 00

2 Math 5 / Eam (November, 08) page. [7 points] Let g() and h() be two functions. The graphs of g () and h () are shown below. y At right is the graph of y = g (), the derivative of g(). Assume that g() is a continuous function. Use the graph to answer the questions below. Circle all correct answers. 3 y = g () a. [ points] At which of the following values of is g() not differentiable? = = 4 = 5 = 6 = 7 none of these b. [ points] For which of the following values of does g() have a local maimum? = = 4 = 5 = 6 = 7.5 none of these c. [ points] For which of the following values of does g() have an inflection point? = = 3 = 4 = 5 = 7.5 none of these d. [ points] On which of the following intervals is g() linear? (0, ) (4, 6) (6, 7) (6, 8) (7, 8) none of these e. [ points] For which of the following values of does g() attain a global maimum on the interval [, 7]? = = 4 = 5 = 6 = 7 none of these

3 Math 5 / Eam (November, 08) page 3 y Use the graph of y = h (), the second derivative of h(), to answer the questions below. Circle all correct answers. 3 y = h () f. [ points] Over which of the following intervals is h() concave up on the entire interval? (0, ) (, 3) (, 4) (4, 5) none of these g. [ points] On which of the following intervals is the function h () (the derivative of h()) decreasing over the entire interval? (0, ) (, 3) (, 3) (4, 5) none of these h. [3 points] If h (4) = 0, which of the following statements must be true? A. = 4 is a local maimum of h(). B. = 4 is a local minimum of h(). C. = 4 is an inflection point of h (). D. = 4 is a critical point of h(). E. = 4 is an inflection point of h(). F. none of these

4 Math 5 / Eam (November, 08) page 4. [6 points] Let A and B be constants and 3 + B for 0 < < k() = B + A 3 for. Find the values of A and B that make the function k() differentiable on (0, ). Show all your work to justify your answers. If there are no such values of A and B, write none. k() will only be differentiable at = if it is also continuous at =. In order for this to happen, we plug = in to both formulas of the original function and set them equal as well: 3 + B = B + A From this second equation, we can subtract B from both sides to find A = 3. The function 3 + B is differentiable on (0, ), and the function B + A 3 is differentiable on (, ), so we just need values of A and B that will make k() differentiable at =. We can compute the derivative: k 3 B () = for 0 < < B + 3A for <. In order for k() to be differentiable at =, we must have k( + h) k() k( + h) k() lim = lim h 0 h h 0 + h d d (3 + B ) = d = d (B + A 3 ) = 3 B = = B + 3A = 3 B = B + 3A In addition, Now we plug this value in for A in the earlier equation, giving us 3 B = B + 9 Solving for B, we get 3B = 6, so B =. Answer: A = 3 B =

5 Math 5 / Eam (November, 08) page 5 3. [ points] Assume the function h(t) is invertible and h (t) is differentiable. Some of the values of the function y = h(t) and its derivatives are shown in the table below t h(t) h (t) h (t) Use the values in the table to compute the eact value of the following mathematical epressions. If there is not enough information provided to find the value, write NI. If the value does not eist, write dne. Show all your work. a. [3 points] Let a(t) = h(t 5). Find a (3). Since a (t) = th (t 5), then a (3) = 6h (4) = 6(5) = 30 Answer: 30 b. [3 points] Let b(t) = h(t) t. Find b (4). Since b (t) = h (t)t th(t) t 4 then b (4) = 6h (4) 8h(4) 6(5) 8(8) = = = 6. Answer: 6 c. [3 points] Let c(y) = h (y). Find c (). Since c (y) = h (h (y)) then c () = h (h ()) = h =. Answer: () d. [3 points] Let g(t) = ln( + h (t)). Find g (0). Since g (t) = h (t) + h (t) then g (0) = h (0) + h (0) = (6) + (3.5) = 8 =.5 Answer:.5

6 Math 5 / Eam (November, 08) page 6 4. [ points] Suppose r() is a differentiable function defined for all real numbers. The derivative and second derivative of r() are given by r () = ( + ) 3 ( ) 4/5 and r () = ( + ) (9 6) 5( ) /5 a. [6 points] Find the -coordinates of all critical points of r() and all values of at which r() has a local etremum. Use calculus to find and justify your answers, and be sure to show enough evidence to demonstrate that you have found all local etrema. For each answer blank below, write none if appropriate. The critical points occur at values of where r () = 0 or r () does not eist. r () eists for all values of, so the only critical points are at points where r () = ( + ) 3 ( ) 4/5 = 0. That is when = and =. Now we need to know the sign of r () on three intervals. We will consider the sign of each factor of in order to do this. 0 sign of ( + ) 3 sign of ( ) 4/5 sign of r () < < + ( )(+) = < < + + (+)(+) = + < < + + (+)(+) = + This tells us that r() is decreasing for < and increasing for >. Therefore we have a local minimum at = and no local maes. Answer: Critcal points at =, Local ma(es) at =NONE Local min(s) at = b. [6 points] Find the -coordinates of all inflection points of r(). If there are none, write none. Use calculus to find and justify your answers, and be sure to show enough evidence to demonstrate that you have found all inflection points. The only potential inflection points occur when r () = 0 or r () does not eist. From the formula, we see that r () = 0 when = and = 6 9, and r () does not eist at =. Again, we need to make a table of the sign of r () on intervals between these points, because inflection points only occur when r() changes concavity, which happens when r () changes sign. 0 sign of ( + ) sign of (9 6) sign of ( ) /5 sign of r () < < + + < < < < + + < < Therefore, r () changes concavity at = 6 9 Answer: Inflection points at = 6 9, and =.

7 Math 5 / Eam (November, 08) page 7 5. [9 points] In each of the following questions, draw a graph satisfying all the properties listed. There may be many correct answers. Make sure that your graph clearly shows all of the properties listed. a. [5 points] The function f() satisfies each of the following properties: i) f() is continuous on (0, 8). ii) f() has a local maimum at = 5. iii) f () < 0 on (4, 7). iv) lim f () = and lim f () = 0 + y 4 3 y = f() b. [4 points] The function g() satisfies each of the following properties: i) g() is defined on (0, 8). ii) g() has an inflection point at = 3. iii) g() is discontinuous at = 6. iv) g() has a local maimum at = 6. v) g() has global maima only at = and = 5. y 4 3 y = g()

8 Math 5 / Eam (November, 08) page 8 6. [4 points] The graph of the function f() is shown below. Note that f() has a vertical tangent line at = 5. y = f() a. [ points] On which of the following intervals does the function f() satisfy the hypotheses of the Mean Value Theorem? Circle the correct answer(s). [0,] [,3] [,4] [3,5] none of these b. [ points] On the interval [8, ] the hypotheses of the Mean Value Theorem are true for the function f(). What does the conclusion of this theorem say in this interval? Answer: There is some c on the interval (8, ) such that f (c) = f() f(8) 8 =. 7. [5 points] Yi is constructing a cardboard bo. The base of the bo will be a square of width w inches. The height of the bo will be h inches. Yi will use gray cardboard for the sides of the bo and brown cardboard for the bottom (the bo does not have a top). Gray cardboard costs $0.05 per square inch, while brown cardboard costs $0.03 per square inch. Yi wants to spend $0 on the cardboard for his bo. h Write a formula for h in terms of w. w w The area covered by the gray and the brown cardboard are A g = 4wh and A b = w respectively. Then the cost of the cardboard, in dollars, used in the cardboard is C = 0.05A g + 0, 03A b. Hence w and h satisfy C = 0 = 0.05(4wh) w = 0.wh w. Then h = w. 0.w Answer: h = w 0.w

9 Math 5 / Eam (November, 08) page 9 8. [7 points] Yi is working on a second bo (not the one from the previous problem). This bo will need a wire frame. The base of the bo will be a square with width w inches. The height of the bo will be h inches. Once the bo is completed, its volume, V, in cubic inches will be V = w h. Yi has to use thicker wire for the edges along the top and bottom of the bo. Let M be the total mass in grams of the wire frame. The equation for M is M = 8w + 6h. Yi has a total of 540 grams of metal to make the wire frame. What values of w and h will maimize the volume of his bo? Use calculus to find and justify your answer, and be sure to show enough evidence that the values you find do in fact maimize the volume. Solving for w in terms of h: V = w = 30 3 h ( 30 3 h ) h Now find the critical points of V. V = 9 h3 0h + 900h dv dh = 3 h 40h Critical points occur when V = 0 so h = 30 or h = 90. Since w and h should both be positive, we get h > 0 and 30 3h > 0 so the possible values of h satisfy 0 < h < 90. Since V is continuous on this interval and has only one critical point, h = 30, we only have to test the behavior of V = 9 h3 0h close to the endpoints h = 0 and h = 90 and the critical point h = 30. lim V h 0 + h = 30 lim h 90 0 V(30)=,000 0 Answer: The volume of the bo is maimum when w = 0 h = 30 Other approach in this case (after finding the critical points of V in 0 < h < 90): In 0 < h < 30 and 30 < h < 90 we pick h = 0 and h = 50 to find the sign of dv in these intervals. We get dh dv dh h= and dv dh h= < h < < h < 90 dv dh + So, h = 30 is local ma. Since V is continuous on this interval and has only one critical point, h = 30 is also the global ma. When h = 30 we get w = 0.

10 Math 5 / Eam (November, 08) page 0 9. [3 points] A curve C is defined implicitly by the equation Note that the curve C satisfies (y 4y) + 5 = 5. dy d = y + 8y 5. 4y 8 a. [4 points] Find all points on C with a vertical tangent line. Give your answers as ordered pairs (coordinates). Justify your answer algebraically. Write none if no such points eist. dy C will have a vertical tangent line when the denominator of is 0, that is, d 4y 8 = 0. Factoring, we can rewrite this as 4(y ) = 0. This happens when = 0 or y =. Now we need to find points on C where = 0 or y =. Plugging = 0 in to the equation for C, we get 0 (y 4y) = 5, which reduces to 0 = 5. Since this is not true for any value of y, there are no points (0, y) satisfying this equation. Net we plug in y =. This gives ( 4 ) + 5 = 5 ( 4) + 5 = 5 3 = 5 = 5 So the denominator is 0 at the point (, y) = ( 5, ). To make sure the numerator is not also 0, we plug y = in to the numerator: = = 3 0 Answer: ( 5, ) b. [4 points] There is one point on C with a coordinate (k, 0). Find the value of k and write the equation of the tangent line to C at the point (k, 0). Your equation should not include the letter k. To find the value of k, we plug (k, 0) into the equation for C. k(0 4 0) + 5k = k = 5 k = 3 So k = 3. Now we plug (3, 0) into the formula for dy to find the slope of the tangent line at d this point: = 5. Finally, we use point-slope form to get the equation for the tangent line: y = 5 4 ( 3). Answer: k = 3 Equation of tangent line: y = 5 ( 3) 4

11 Math 5 / Eam (November, 08) page c. [5 points] Another curve D is defined implicitly by the equation cos(w y 3 + V y) = 6 where W and V are constants. Find a formula for dy d in terms of, y, W, and V. To earn credit for this problem, you must compute this by hand and show every step of your work clearly. Differentiating both sides with respect to and then solving for dy d gives sin(w y 3 + V y)(3w y dy d + V y + V dy d ) = 0 (3W y + V ) dy d + V y = 0 dy d = V y 3W y + V Alternatively, we can start by taking arccos of both sides, giving us W y 3 + V y = arccos( 6 ), and then use implicit differentiation to get (3W y + V ) dy d + V y = 0; from here, we solve as in above. Answer: dy d = V y 3W y + V

12 Math 5 / Eam (November, 08) page 0. [7 points] The function g has the property that g(), g (), and g () are defined for all real numbers. The quadratic approimation of g() at = is Q() = 4( + ) + ( + ) 5. a. [5 points] Find the eact value of each of the following quantities. If there is not enough information to answer the question, write NI. g( ) = 5 g ( ) = 0.5 g ( ) = 8 g(0) = NI Q (0) = 6.5 b. [ points] Write a formula for L(), the tangent approimation of g() near =. The quadratic approimation near = satisfies Q() = 4( + ) + L(). Answer: L() = ( + ) 5

13 Math 5 / Eam (November, 08) page 3. [8 points] Consider the function for 8 0 for 8 < < 0 f() = and its derivative f 3 3 () = 5e for > 0. 5( 0.5)e 0.5 for > 0. Find the -coordinates of the global maimum and the global minimum of the function f() for 8. If one of them does not eist, write none in the answer line below. Use calculus to find your answers, and be sure to show enough evidence that the point(s) you find are indeed global etrema. Critical points on the interval ( 8, 0) occur when the epression is 0 or undefined. The 3 3 numerator is equal to 0 at = 8 7. The denominator is equal to zero when = 0, which is not in the interval, so the only critical point in the interval ( 8, 0) is at = 8 7. Critical points on the interval (0, ) occur when the epression 5( 0.5)e 0.5 is 0 or undefined. Since e 0.5 is always positive, the only critical point is when 0.5 = 0, which occurs when =. In addition to critical points, we must also check for global etrema by checking the end behavior for each piece of f(). f f( 8) = 4 ( 8) ( 8) /3 = 4 = 8 ( 8 ) ( = 4 8 ) ( 8 ) /3 = = 04 7 f(0) = = 4 lim f() = lim + 4) = = (5e 0.5 f() = 5 e + 4 = 0 e lim f() = lim e = = 4 None of the end behavior is infinite. The highest number in the calculations above is 8, which occurs at = 8. The lowest number is 04 7, which occurs at = 8 7 Answer: Global maimum(s) at = 8 Global minimum(s) at = 8 7