Amherst College, DEPARTMENT OF MATHEMATICS Math 11, Final Examination, May 14, Answer Key. x 1 x 1 = 8. x 7 = lim. 5(x + 4) x x(x + 4) = lim

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1 Amherst College, DEPARTMENT OF MATHEMATICS Math, Final Eamination, May 4, Answer Key. [ Points] Evaluate each of the following limits. Please justify your answers. Be clear if the limit equals a value, + or, or Does Not Eist. (a) lim = = = by the Direct Substitution Property. (b) lim = + because RHL= LHL. 7 7 RHL: lim = + = + LHL: lim ( 7) = = + = (c) lim + ( + 7)( ) ( )( ) + 7 = 8 = (d) lim ( + ) ( + 4)( + ) ( + 4) ( + 4) + + ( + 4)( + ) 4 ( + 4) = 4 ( )( + 4) = ( + 4)( + ). [ Points] Compute each of the following derivatives. Do not simplify your answers. (a) d d ( ) e 7 tan = tan ( ) ( ) e e 7 sec (tan ) (b) dy d, if sin + e y = 7 + ln(y). d ( First we implicit differentiate both sides w.r.t.. sin + e y) = d (7 + ln(y)). d d Then sin cos + e y dy d = ( dy ) y d + y = dy y d + ( As a result, e y ) dy y d = sin cos

2 Finally, dy d = sin cos e y y. (c) d d ( 7 e t ) lnt dt = d ( e t ) d 7 lnt dt = e ln by FTC Part I [ ( )] (d) d ln 7 + e sin d ( + 9) (hint: you might want to simplify first) = d [ ln ] d lne sin ln(( + 9) ) = d [ ] ln(( 7 + ) + sin ln( + 9) d = d [ ] ( ) ( ) d ln(7 + ) + sin ln( + 9) = cos (e) f (), where f() = e cos + 7. f () = e cos ( sin ) 7 8 f () = e cos ( cos ) sine cos ( sin) = e cos cos + sin e cos (f) f (), where f() =. We can solve this two ways: first try Logarithmic Differentiation and using the properties of logs, Let y =, so that lny = ln( ) = ln Net use implicit differentation to differentiate both sides w.r.t. d d (lny) = d d (ln) Then ( ) dy y d = + ln = + ln. As a result, dy = y( + ln). d Finally, dy d = ( + ln). The second option is to rewrite y = = e ln() = e ln. d Then differentiate, d ( ) = d (e ln ) ( ( ) ) = e ln + ln d = e ln() ( + ln) = ( + ln).

3 . [ Points] Compute each of the following integrals. (a) ( ) + = ln + C + + d = d = ( ) d = + + ln + C (b) π 6 π sin()cos() d = u du = u ( ) = 4 ( ) = ( 4 4 ) 4 = 8 Here u = sin() du = cos() d and du = cos()d = π u = sin π 6 = = π 6 u = sin π = (c) Here e 7 + e d = u du = ln u + C = ln 7 + e + C u = 7 + e du = e d du = e d (d) e d = e u du = e u + C = e + C Here u = du = d du = d (e) e 4 e ln d 4. [ Points] Give an ε-δ proof that lim 4 = 9. Scratchwork: we want f() L = ( 4) ( 9) < ε f() L = ( 4) ( 9) = = + = ( + ) = + = ( ) (want < ε) ( ) < ε means ( ) < ε

4 So choose δ = ε to restrict < ( ) < δ. That is < ( ) < ε. Proof: Let ε > be given. Choose δ = ε. Given such that < ( ) < δ, then as desired f() L = ( 4) ( 9) = + = ( + ) = ( ) = ( ) < ε = ε.. [ Points] Let f() =. Calculate f (), using the limit definition of the derivative. f f( + h) f() ( + h) () h h h h ( + h) ( + h) + (( + h) ) ( ) h h ( + h) + h h( ( + h) + ) h + h + h( ( + h) + ) h = h ( + h) + h h( ( + h) + ) Note: Can double check this by applying the Quotient Rule. 6. [ Points] Suppose ( that f() = cos (e ). Write the equation of the tangent line to the curve π y = f() when = ln. ) First the slope f () = sin(e ) e. ( ( π Then f ((ln )) π «) = sin e ln Then the y-value is f π «e ln ( ( ( π ln = cos e )) ln π ( π ( π ( π = sin = () = ) ) ) π. «) = cos π = Therefore, the equation of the tangent line through the point y = π ( ( π ln or y = )) π + π ( π ) ln. ( ( π ) ln, with slope ) π, is 7. [ Points] Let f() = e. For this function, discuss domain, vertical and horizontal asymptote(s), interval(s) of increase or decrease, local etreme value(s), concavity, and inflection point(s). Then use this information to present a detailed and labelled sketch of the curve. Take my word for it that lim e = and 4 lim e =

5 f() has domain (, ) so No Vertical Asymptotes. Vertical asymptotes: none Horizontal asymptotes: at y = towards, since lim f() =. First Derivative Information: f () = e ( ) + e = e ( + ) The critical points occur where f is undefined (never here) or zero. Also note that the eponential function is always non-zero, which implies that + = As a result, = is the critical number. Using sign testing/analysis for f, f f ր ց local ma Therefore, f is increasing on (, ) and decreasing on (, ) with local ma at (, f()) = (, e ). Second Derivative Information f () = e ( ) + ( + )e ( ) = e ( + ) = e ( ) Possible inflection points occur when f is undefined (never here) or zero ( = ) (again note the eponential piece is non-zero). Using sign testing/analysis for f, f f infl. point Therefore, f is concave down on (, ), whereas f is concave up on (, ) with I.P. at (, f()) = (, e ). Piece the first and second derivative information together f ր ց ց f local ma infl. pt.

6 8. [ Points] Suppose the top of a foot ladder is sliding down a vertical wall at a rate of feet per second. Consider the angle formed between the ground and the base of the ladder. At what rate is this angle changing when the top of the ladder is feet above the ground? Diagram y θ Variables Let = distance between bottom of ladder and wall at time t Let y = distance between top of ladder and ground at time t Let θ = angle formed by the ground and base of ladder at time t =? when y = ft and dy ft = dt sec Equation relating the variables: Find dθ dt We have sinθ = y. Differentiate both sides w.r.t. time t. d dt (sin θ) = d ( y ) = cos θ dθ dt dt = dy (Related Rates!) dt Substitute Key Moment Information (now and not before now!!!): We re not given θ for this problem, but we can still compute cosθ from trig. relations on the diagram s triangle with cosθ = adj hyp. When y =, we can use the Pyth. Theorem to compute 6

7 = () () = 4. Finally, cosθ = 4 = 4. 4 dθ dt = ( ) Solve for the desired quantity: dθ dt = 4 = rad 4 sec Answer the question that was asked: The angle is decreasing at a rate of 4 radians every second. 9. [ Points] A bo with a square base and a (flat) top is to be made to hold a volume of 7 cubic feet. Determine the dimensions that minimize the amount of material used. (Remember to state the domain of the function you are computing etreme values for.) y The Volume of this bo, which is fied, is given as V = y = 7 = y = 7. Then the Amount of Material must be minimized: M = material for base + material for top + material for 4 sides = + + 4y = + 4y = + 4 ( ) 7 = + 8 The (common-sense-bounds)domain of M are { : > }. Net M = 4 8. Setting M = we solve for = 7 = as the critical number. Sign-testing the critical number does indeed yield a minimum for the materials function. V V ց ր MIN Since = then y = 7 9 feet a cube. =. As a result, the bo of largest volume will measure, each in 7

8 . [ Points] (a) Compute the area bounded by y =, y =. Sketch the region. The curves intersect at = ±. We will integrate from = to =, and double that area using symmetry. ) Area= top bottom d = ( ) d = d = ( (( = ) ) ( ) = ( ) 7 = (b) Consider the region in the plane bounded by y = e +, y =, = and = ln 4. Compute the volume of the -dimensional object obtained by rotating the region about the ais. Sketch the region. 8

9 Volume= = π ln 4 ln 4 π[(outer radius) (inner radius) ] d = (e + e + ) d = π ln 4 ln 4 π[(e + ) ] d ( ) ln 4 e + e d = π e + e = π (( e ln 4 + e ln 4) ( e + e )) = π (( eln ) ( + )) = π (( 6 + 8) ) = π ( (8 + 8) ) ( ) = π 6 7π =. [ Points] Consider an object moving on the number line, starting at position, such that its acceleration at time t is a(t) = feet per square second. Also assume that the object has initial velocity equaling 6 feet per second. (a) Compute the velocity function v(t) and position function s(t). v(t) = t v() = t 6 s(t) = t 6t + s() = t 6t, because we know the initial condition s() =. (b) Compute the total distance that it travels between time t = and t = 4 seconds. Therefore, Total Distance= t + 6 dt + = = 4 4 v(t) dt = t 6 dt = t + 6t 4 t 6 dt = + t 6t 4 (t 6) dt + 4 t 6 dt = ( 9 + 8) ( ) + (6 4) (9 8) 9