Math 50A November 16, Name: Answer Key D. Arnold. Midterm #2

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1 Math 50A November 6, 206 Midterm #2 Name: Answer Key D. Arnold Instructions. (90 points) This midterm eam is open book, open notes. All work must be your own. You may not ask for help on any of the questions. Answer each of the following questions on your own paper. Arrange your solutions in order, then staple this eam atop your solution, then sign the honor pledge. I promise that all of the solutions submitted were done entirely by me. I received no help from colleagues, friends, or tutors. I also did not share any of my solutions with any of my classmates. Signature: You are encouraged to check the answers to your problems using Mathematica. If they differ, work to find your mistakes. Attach your notebook to the back of your eam. Although you may use Mathematica to check your work, just providing a Mathematica answer for any question is not acceptable. No credit will be awarded if you just provide an answer. You must show steps that indicate you understand the process for each question. (5 pts ea. ). Find the derivatives of each of the following functions. Simplify your answers as much as possible; e.g., place your answer in factored form. (a) y (5 + 2) 3 (3 3) 5 Solution: Use the product rule: y (5 + 2) [ 3 5(3 3) 3 ( 3) ] + 3(5 + 2) 2 (5)(3 3) 5 5(5 + 2) 3 (3 3) 4 + 5(5 + 2) 2 (3 3) 5 Factor. 5(5 + 2) 2 (3 3) 4 [ (5 + 2) + (3 3)] 5(5 + 2) 2 (3 3) 4 ( ) 5(5 + 2) 2 (3 3) 4 ( 8) (b) y 2 + Solution: Use the quotient rule. y ( + )D 2 2 D ( + ) ( + ) 2 ( + )2 ln 2 2 ( + ) 2

2 Math 50A/Midterm #2 Page 2 of 9 Name: Answer Key Now we can factor. (c) y cos 5 2 [( + ) ln 2 ] ( + ) 2 Solution: Use the chain rule. y cos 5 y (cos 5) /2 y 2 (cos 5) /2 D cos 5 y 2 (cos 5) /2 ( sin 5)(5) y 5 sin 5 2 cos 5 (d) y tan 3 ( ) Solution: Use the product rule and the chain rule. Now we can factor. (e) y ln(cos ) y /2 (tan /2 ) 3 y [ /2 3(tan /2 ) 2 D tan /2] + 2 /2 (tan /2 ) 3 ( ) y 3 /2 (tan /2 ) 2 (sec 2 /2 ) 2 /2 + 2 /2 (tan /2 ) 3 y 3 2 tan2 sec /2 tan 3 y 2 /2 tan 2 [ 3 /2 sec 2 + tan ] y tan2 [3 sec 2 + tan ] 2

3 Math 50A/Midterm #2 Page 3 of 9 Name: Answer Key Solution: Use the chain rule. y ln(cos ) y cos D cos y cos 2 y 2 cos (f) y ln Solution: Use the chain rule. Also, recall that D ln /. y ln y (32 4) y Here is a bit more eplanation. Let y ln u where u Then, because /du /u, we can write: (g) y π tan du du u du d ( ) (32 4) Solution: Use the chain rule. Recall that D a a ln a. y π tan y π tan ln π D ( tan ) y π tan ln π [D tan + tan D ] y π tan ln π [ sec 2 + tan ]

4 Math 50A/Midterm #2 Page 4 of 9 Name: Answer Key (h) y sin2 Solution: First, take the logarithm of both sides y sin2 ln y ln sin2 ln y (sin 2 ) ln Now, differentiate. y y (sin 2 )D ln + ln D sin 2 2 ( ) y y (sin 2 ) + (2 sin cos )(ln ) Multiply both sides by y, then replace y with sin2. [ ] sin y 2 y + 2 sin cos ln [ ] y sin2 sin sin cos ln Now we can factor out sin. y sin2 sin [sin + 2 cos ln ] y sin2 sin [sin + 2 cos ln ] (0 pts ) 2. Given 3 + tan y e y, find / using implicit differentiation. Solution: Note that it is implied that y is a function of and differentiate. 3 + tan y e y y 2 + tan y e y + y 2 ey 32 tan y ( ) + y 2 ey 32 tan y 32 tan y + y 2 ey

5 Math 50A/Midterm #2 Page 5 of 9 Name: Answer Key Multiply numerator and denominator by tan y e y + y 2 Multiply numerator and denominator by + y 2. (32 + tan y)( + y 2 ) ( + y 2 )e y (0 pts ) 3. Given 3 + y 3, find d 2 y/ 2 using implicit differentiation. To receive full credit, you must simplify your answer as much as possible, which means you will have to substitute / and perhaps the fact that 3 + y 3. Solution: Note that it is implied that y is a function of and differentiate. 3 + y y 2 0 3y y 2 Now we need to take the derivative with respect to once more. We ll need to use the quotient rule. ( d 2 y 2 (2) 2 2y ) y 2 y y 2y2 y 4 Now we need to insert / 2 /y 2. d 2 y y 24 y Multiply numerator and denominator by y. ( 2 y 2 y 4 2y2 y y 3 y 5 ) 2y 2

6 Math 50A/Midterm #2 Page 6 of 9 Name: Answer Key Now, recall that 3 + y 3, so: d 2 y 2(3 + y 3 ) 2 y 5 2() y 5 2 y 5 (0 pts ) 4. The curve defined implicitly by the equation 2( 2 + y 2 ) 2 25( 2 y 2 ) is called a lemniscate. Use implicit differentiation to help find the equation of the tangent line at the point (3, ). Use Mathematica s ContourPlot to sketch the curve, the tangent line, and the point of tangency and label it with its coordinates. Use Frame False, Aes True, AesLabel to label the aes, and PlotLegends to indicate the equations of the curve and the tangent line. Use a separate Mathematica notebook for this problem, print the image, and slide the page after your pencil and paper work for this problem. Solution: Assume y is a function of and differentiate implicitly. ( 4( 2 + y 2 ) 2 + 2y 8( 2 + y 2 ) + 8y( 2 + y 2 ) [ 8y( 2 + y 2 ) + 50y ] 50 8(2 + y 2 ) 2( 2 + y ( 2 y 2 ) ) 25 ( 2 2y 50 50y 50 8(2 + y 2 ) 8y( 2 + y 2 ) + 50y 25 4(2 + y 2 ) 4y( 2 + y 2 ) + 25y To find the slope at the point (3, ), enter 3 and y. 25(3) 4(3)( ) (,y)(3,) 4()( ) + 25() 75 2(0) 4(0) )

7 Math 50A/Midterm #2 Page 7 of 9 Name: Answer Key Enter the slope m 9/3 and the point ( 0, y 0 ) (3, ) into: y y 0 m( 0 ) y 9 ( 3) 3 y y Enter the equation of the lemniscate, the point (3, ), and the tangent line y (9/3)+ 40/3 into Mathematica. Which produces this image.

8 Math 50A/Midterm #2 Page 8 of 9 Name: Answer Key (0 pts ) 5. Use logarithmic differentiation to find the derivative of the following function. y (2 8) / Solution: Start by taking the logarithm of both sides. Now differentiate. y (2 8) / ln y ln (2 8) / ln y 3 ln(2 8) + 2 ln(3 + ) ln( ) y y (2) (32 ) (65 7) [ ] y 2 y 3( 2 8) ( 3 + ) y (2 8) /3 [ ] ( 2 8) ( 3 + )

9 Math 50A/Midterm #2 Page 9 of 9 Name: Answer Key (0 pts ) 6. Grain pouring from a chute at the rate of 8 ft 3 /min forms a conical pile whose altitude is always twice its radius. How fast is the altitude of the pile increasing at the instant when the pile is 6 ft high? Include an etremely detailed sketch with your solution. Solution: A sketch is helpful. We re given a conical pile whose volume is changing at a rate dv/dt 8 ft 3 /min and whose height h is always twice its radius r; i.e., h 2r. h r The volume of the cone is: V 3 πr2 h Because the height is twice the radius, we can replace r with h/2. V 3 π ( h 2 ) 2 h V 2 πh3 Now we can differentiate with respect to time. dv dt ( 2 π 3h 2 dh ) dt dv dt dh πh2 4 dt Now, we need to find dh/dt at a time when h 6 feet and dv/dt 8 ft 3 /min. So, substitute these numbers into our last equation. 8 dh π(6)2 4 dt 8 9π dh dt dh dt 8 9π Hence, the height is changing at a rate dh dt 8 9π ft/min.