Solutions to Test 2 Spring = y+x dy dx +0 = ex+y x+y dy. e x = dy dx (ex+y x) = y e x+y. dx = y ex+y e x+y x

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1 12pt 1 Consider the equation e +y = y +10 Solutions to Test 2 Spring 2018 (a) Use implicit differentiation to find dy d d d (e+y ) = d ( (y+10) e+y 1+ dy ) d d = y+ dy d +0 = e+y +y dy +e d = y+ dy d +y dy dy e d d = y e+y = dy d (e+y ) = y e +y = dy d = y e+y e +y (b) Find an equation of the tangent line to the graph e +y = y +10 at the point (3, 3) dy The slope of the tangent line is = 3 e0 d (3, 3) e 0 3 = 4 = 2 Then an equation of the tangent 2 line can be written in the form: y ( 3) = 2( 3) or y = 2 9 8pt 2 Use Logarithmic Differentiation to find y if y = lny = ln( ) = ln d d (lny) = d d ( ln) = 1 +ln 1 = 2+ln y y = 2+ln y = y 2+ln = y = 2 + ln

2 15pt 3 A particle moves in a straight line with its position function s(t) given by the equation s(t) = te 1 t, t 0 (where s is measured in meters and t is measured in seconds) (a) Find v(t), the velocity of the particle at time t v(t) = s (t) = te 1 t ( 1)+1 e 1 t = e 1 t (1 t) (m/s) (b) When is the particle at rest? v(t) = 0 e 1 t (1 t) = 0 1 t = 0 (e 1 t > 0 for all t) Therefore, the particle is at rest when t = 1 (s) (c) When is the particle moving in the positive direction? v(t) > 0 e 1 t (1 t) > 0 1 t > 0 (since e 1 t > 0 for all t) Therefore, the particle is moving in the positive direction when 0 t < 1 (d) Find the total distance traveled by the particle during the first 2 seconds Show your work Distance = s(1) s(0) + s(2) s(1) = 1e e 1 e 0 = e = 2 2 e (m) (e) Find the acceleration a(t) of the particle at time t a(t) = v (t) = d dt (e1 t (1 t)) = e 1 t (0 1)+e 1 t ( 1)(1 t) = e 1 t (t 2) (m/s 2 ) (f) When during the first 2 seconds is the particle speeding up and when is it slowing down? The graph of the velocity v(t) of the particle, shown at right below, will help you answer this question Speeding up on the interval (1,2) (where v < 0 and a < 0) v Slowing down on the interval (0,1) (where v > 0 and a < 0) 1 2 t

3 7pt 4 Let f() = 4 4 (a) Find the linearization L() of f() at a = 16 L() = f(16)+f (16)( 16) f(16) = = 4 2 = 8 and f () = /4 = 1 3/4, so f (16) = /4 = = 1 8 Therefore, L() = ( 16) 8 (b) Use L() from part (a) to estimate the value of f(14) f(14) L(14) = (14 16) = = 31 4 = 775 4pt 5 Find f (θ) and f (θ) of the function f(θ) = ln sec(2θ)+tan(2θ) Simplify your answers f (θ) = sec(2θ)tan(2θ) 2+sec2 (2θ) 2 sec(2θ) + tan(2θ) = 2sec(2θ)(tan(2θ)+sec(2θ)) sec(2θ) + tan(2θ) = 2sec(2θ) f (θ) = 2sec(2θ)tan(2θ) 2 = 4sec(2θ)tan(2θ)

4 8pt 6 Find the absolute maimum and absolute minimum values of the function f() = +cos on the interval [0,π] The absolute maimum and absolute minimum values can only occur at the critical numbers of f in the interval (0,π) or at the endpoints of that interval f () = 1 sin If f () = 0, then sin = 1 There is only one solution of this equation in (0,π): it is = π 2 Thus, π 2 is the only critical number of f in (0,π) We have f( π ) = π 2 2 +cos(π) = π, f(0) = 0+cos0 = 1, and f(π) = π +cosπ = π By comparing these values, we conclude that the absolute minimum of f on [0,π] is f(0) = 1 and the absolute maimum value is f(π) = π 1 8pt 7 Let f() = 1 2 (a) Verify that f satisfies the hypotheses of the Mean Value Theorem on the interval [2,6] by filling the blanks f is continuous on the interval [2, 6]; f is differentiable on the interval (2, 6) (b) Find all numbers c that satisfy the conclusion of the Mean Value Theorem on the interval [2,6] f (c) = 2c 3 = 2 c 3 f (c) = f(6) f(2) 6 2 = = = = c 3 = 1 18 c 3 = 36 c = 3 36

5 13pt 8 In each part indicate whether the ratio represents an indeterminate form 0/0 or / or neither of them Then find the it 5 1 (a) Circle one: 0 0 neither Find the it L H 5 4 = = (b) 6 1 Circle one: 0 0 neither Find the it L H 5 4 = 6 = = (c) 6 1 Circle one: 0 0 neither Find the it = = 1 e e (d) 1 sin(π) Circle one: 0 0 neither Find the it 1 e e sin(π) L H = 1 e πcos(π) = e πcosπ = e π

6 20pt 9 Consider the function f() = ln+ 2 2, > 0 (a) Find f () f () = 1 + 2( 2) 3 = (b) Find the critical number(s) of f (remember, > 0) f () = = = 0 2 = 4 = ±2 = 2 ( > 0) (c) Determine the intervals of increase and decrease of f Show your work For > 0, f () = > 0 when > 2 and f () = < 0 when 0 < < 2 f increases on the interval(s) (2, ) f decreases on the interval(s) (0, 2) (d) For each critical number of f found in part (b), state whether f has a local maimum, local minimum, or neither at that number f has a local minimum at = 2 (e) Find f () and the value(s) of, > 0, such that f () = 0 f () = d ( 1 d 4 ) 3 = ( 1) 2 4( 3) 4 = = f () = = 0 2 = 12 = ± 12 = 2 3 ( > 0) (f) Find the intervals of concavity and the number(s) at which f has its inflection point(s) For > 0, f () = > 0 when 0 < < 2 3 and f () = < 0 when > 2 3 concave up on the interval (0,2 3) concave down on the interval (2 3,) Inflection point at = 2 3 (g) Sketch the graph of f() assuming that 0 +f() = Plot the point (1,f(1)) and label the points of local maimum/minimum and inflection points y 2 IP loc min

7 10pt 10 A bo with an open top is to be constructed from a square piece of cardboard with dimensions 12 in by 12 in by cutting out equal squares of side at each corner and then folding up the sides (a) Epress the volume V of the bo as a function of 12 V () = (12 2) 2 = 4(6 ) 2 = 4( ) or V () = 4( ) 12 (b) Find the value of that maimizes the volume of the bo Be sure to justify your answer Here 0 6 Since V() is continuous on the closed interval [0,6] (as a polynomial, V() is continuous everywhere), V() attains its absolute maimum value either at its critical numbers in (0,6) or at the endpoints of that interval V () = 4( ) = 12( ) V () = = 0 ( 2)( 6) = 0 = 2 is the only critical number of V in (0,6) Comparing the values V(2) = (12 2 2) 2 2 = 128, V(0) = 0, and V(6) = 0, we conclude that the absolute maimum of V occurs at = 2 (Another way to justify that V() attains its absolute maimum value at = 2 is to use the First or the Second Derivative Test to show that V() has a local maimum at = 2)