Homework_8_Solutions - Math 1110 Name: Homework 8 Due 10/20...

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Unformatted text preview: Math 1110 Name: Homework 8 Due 10/20 or 10/21 in class Please print out these pages. Write your answers to the “text exercises” on separate paper and staple it to these pages. You should include computational details. These problems will be assessed for completeness. Always write neatly and legibly. Please answer the “presentation problems” in the spaces provided. Include full explanations and write your answers in complete, mathematically and grammatically correct sentences. Your answers will be assessed for style and accuracy, and you will be given written feedback on these problems. GRADES Text exercises / 20 Pres probs / 20 Staple /1 Text exercises. Please do the following problems from the book. §3.8 #2, 5, 8, 10, 16, 23, 28, 32, 36, 44, 50, 58, 68, 74, 81, 94, 96, 98 §3.9 #4, 8, 10, 14, 15, 20, 22, 27, 32, 40, 48, §3.10 #2, 8, 14, 18, 23, 26, 30, 33, 37, 41 Question 1. (8 points) Derive the formula dy 1 = dx 1 + x2 for the derivative of y = tan−1(x) by differentiating both sides of the equivalent equation tan(y) = x. We begin by differentiating both sides of the equation with respect to x, to get sec2(y) dy = 1. dx Solving for dy , we get dy = cos2(y). To get the equation in terms of x, we solve the right triangle dx dx % "#$ $ ! " we get dy 1 =√ . dx 1 + x2 We see that cos(y) = √ 1 1 + x2 . Substituting back in, Math 1110 (Fall 2011) HW8 Presentation Problems 2 Question 2. (12 points) Water is slowly leaking out of a bowl at a rate of π in3/min, as shown in the figure below. The bowl is a perfect hemisphere with radius 10 inches. If the water level is y π inches, the volume of water in a hemispherical bowl of radius R is V = y2(3R − y). 3 *,-. *!+ " + !"#$$!$%&'(#) /0%12#34 (a) At what rate is the water level changing when the water is 8 inches deep? dV , and we have an equation for the water level in terms of volume, dt so we should start by differentiating that equation with respect to time. We know the value of π2 y (3R − y) 3 π V = y2(30 − y) 3 dV π dy dy = (2y(30 − y) − y2 ) dt 3 dt dt π dy 2 dy −π = (2 · 8(30 − 8) −8 ) 3 dt dt dy −3π = 8π(44 − 8) dt dy −3 = 8 · 36 dt 1 dy −= 96 dt V= Remembering units, we have that 1 96 in/min. dy 1 =− in/min, or that it is decreasing at at a rate of dt 96 Math 1110 (Fall 2011) HW8 Presentation Problems 3 (b) What is the radius r of the water’s surface when the water is y inches deep? Looking at the right triangle in the figure, we see that (10 − y)2 + r2 = 102 and r= ￿ 20y − y2 in (c) When the water level is 4 inches deep, the water level is changing at a rate of minute. At what rate is the radius of the water changing at that time? 1 64 inches per We take our equation from part (b) and differentiate both sides with respect to time, and then substitute in the values given. dr dt dr dt dr dt dr dt 1 1 dy = (20y − y2)− 2 (20 − 2y) 2 dt 1 1 dy = (20 · 4 − 42)− 2 (20 − 2 · 4) 2 dt 1 −1 −1 = (64) 2 (12) 2 64 −3 = 256 So the radius is decreasing at a rate of 3 256 in/min. Homework 8 Book Problem Answers: Section 3.8: Section 3.9 Section 3.10: ...
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