Answer Key-Math - Optional Review Homework For Eam 2. Compute the derivative or each o the ollowing unctions: Please do not simpliy your derivatives here. I simliied some, only in the case that you want to check your answers) 3 + a) ) = 8 2 ) = 2 [3 2 8 9 ] 3 + 8 )2) 4 = 34 8 7 2 4 2 7 4 0 or more simply you can rewrite ) = + 0 and then ) = 0. b) ) = sin 2 3 2 ) ) = 2 sin3 2 ) cos3 2 ) 3 + 2 3 ) ) 3 c) ) = 2 tan sec ) ) ) 3 sec )3 3)sec tan 3 ) = 2 sec 2 sec sec 2 + tan 2 sec d) ) = 3 + 5 + 3) 7 2 7 2 73 2 + 7 5 + 3) 6 5 4 + 3)) 3 + 5 + 3) 7 ) ) = e) ) = cos ) 7 + [ First rewrite ) = cos )] 7 + ) = cos 2 ) [ 7 + sin sin ) = 7 + 2 + 7 8 + ) cos 2 ) 7 + 2 7 = 4 0 7 4 = 2 2 7) 2 7 )] 7 + 2 + 7 8 + ) ) h) = + cos 2 + ) [ h ) = 2 + cos 2 + ) 2 sin2 + ) 2 + )] 2 g) gy) = y 2 cos y g y) = y 2 sin y) 2 y + cos y 2y = y2 sin y 2 + 2y cos y y h) t) = 3t ) 7 4t + 3) 9 t) = 3t ) 7 94t + 3) 8 4) + 4t + 3) 9 73t ) 6 3) = 33t ) 6 4t + 3) 8 [23t ) + 74t + 3)] = 33t ) 6 4t + 3) 8 [64t + 9]
i) w) = + ) 5 4 + w ) = 5 + ) [ 4 4 + + ] 2 4 + 43 ) sin3 j) ) = 2 + 7) cos 3 4) ) = [ cos 3 4) 2 sin3 2 + 7) cos32 + 7)6 + 7) ] sin3 2 + 7)3 cos 2 4) sin4))4 2. Compute y where y + y 3 = 4 2 cos 2 ). cos 6 4) Dierentiating both sides implicitly yields dy dy + y + 3y2 d d = 42 sin 2 ))2) + cos 2 )8) and then dy d + 3y2 ) = 8 3 sin 2 ) + 8cos 2 ) y so that dy d = 83 sin 2 ) + 8cos 2 ) y + 3y 2. 3. Evaluate each o the ollowing limits: sin6) a) lim = 6 0 8 8 lim sin6) = 6 0 6 8 = 6 8 sin6) b) lim 0 cos + = 0 + 0 = 0 3 8 2 c) lim 0 sin4) = lim 0 4 3 = 3 4 d) lim 3 8) sin4) 3 8 2 7 2 + 5 9 = lim = lim 0 sin4) lim 0 3 8) = 4 lim 0 3 8 2 ) 7 2 + 5 9 2 ) = lim 2 9 8 + 7 3 200 9 8 + 7 3 200 e) lim 6 3 + 2 7 = lim + 5 6 3 + 2 7 + 5 2 + 4 200 2 + 4 200 ) lim 6 9 + 2 = lim 6 9 + 2 9 ) 3 8 7 + 5 9 = 8 7 2 3 ) 3 ) = lim 9 ) = lim 4 sin4) lim 0 3 8) = 9 + 7 200 3 28 3 6 + 2 + 5 = 0 24 3 3 + 5 200 9 6 + 7 = 4. Consider the unction ) = 2. Discuss domain, intervals o increase or decrease, local 4 etreme values), concavity, inlection points), and any horizontal and vertical asymptotes. Use this inormation to give a detailed and labelled sketch o the curve. 2
Domain: ) has domain { ±2} Vertical asymptotes: VA at = ±2. Horizontal asymptote: HA at y = 0 or this since lim ) = 0. ± First Derivative Inormation We compute ) = 2 2 and set it equal to 0 and solve or to ind critical numbers. 4) 2 The critical points occur where is undeined or zero. The latter happens when = 0. The derivative is undeined when = ±2, but those values are not in the domain o the original unction. As a result, = 0 is the critical number. Using sign testing/analysis or, or our chart is 2 0 2 ր ր ց ց local ma, 0) 0, ) ) ) ր ց So is increasing on, 0); and is decreasing on 0, ). Moreover, has a local ma at = 0 with maimum value 0) = 4. Second Derivative Inormation Meanwhile, = 232 + 4) 2 4) 3 is never zero. Using sign testing/analysis or around the vertical asymptotes, 2 2 or our chart is, 2) 2, 2) 2, ) ) ) So is concave down on 2, 2) and concave up on, 2) and 2, ) with no inlection points. Piece the irst and second derivative inormation together 3
2 0 2 ր ր ց ց Sketch: local ma 5. Consider the unction ) = 2 ) 2. Discuss domain, intervals o increase or decrease, local etreme values), concavity, inlection points), and any horizontal and vertical asymptotes. Use this inormation to give a detailed and labelled sketch o the curve. Domain: ) has domain, ) Vertical Asymptotes: It is a polynomial, continuous everywhere, and so has no vertical asymptotes. Horizontal Asymptotes: There are no horizontal asymptotes or this since lim ) = and lim ) =. First Derivative Inormation We compute ) = 2 2) ) + ) 2 2) = 2 )[ + )] = 2 )2 ) and set it equal to 0 and solve or to ind critical numbers. The critical points occur where is undeined never here) or zero. The latter happens when = 0, =, = As a result, 2 4
those numbers are the critical numbers. Using sign testing/analysis or, or our chart is 0 ց ր 2 local ց ր local local min ma min, 0) 0, /2) /2, ), ) ) ) ց ր ց ր So is increasing on 0, /2) and, ), and is dereasing on, 0) and /2, ). Moreover, has local mins at = 0 and =, but a local ma at = /2. Here 0) = 0 and ) = 0, and 2 ) = 6 Second Derivative Inormation Meanwhile, is always deined and continuous, and = 2 2 2 + 2 = 0 only at our possible inlection point = 2 ± 2. Using sign testing/analysis or, or our chart is /2 / 2 /2 + / 2 inl. point inl. point, 2 2 ) 2 2, 2 + 2 ) 2 + 2, ) ) ) So is concave down on 2, 2 2 + ) and concave up on, 2 2 ) and 2 2 + ),, with inlection points at = 2 2 ±. You can check that 2 2 ± ) = 2 36. Piece the irst and second derivative inormation together 0 ց ր ր 2 ց ց ր 5 inl. inl. local pt. local pt. local min ma min
Sketch: 6. A conical water tank point acing down) with radius o 3 meters at the top and height o 8 meters is leaking water. At the moment when the water is 2 meters rom the top o the tank water is leaking at a rate o cubic meter per minute. How ast is the water level decreasing at that moment? The cross section with water level drawn in) looks like: Diagram 3 w r at 8 e h r Equation relating the variables: Volume= V = 3 πr2 h Variables Let r = radius o the water level at time t Let h = height o the water level at time t Let V = volume o the water in the tank at time t Find dh dt =? when h = 6 eet 2 rom top) and dv dt = m3 min Etra solvable inormation: Note that r is not mentioned in the problem s ino. But there is a relationship, via similar triangles, between r and h. We must have r 3 = h 8 = r = 3 8 h Ater substituting into our previous equation, we get: 6
V = 3 π 3 8 h ) 2 h = 3 64 πh3 Dierentiate both sides w.r.t. time t. d dt V ) = d ) 3 dt 64 πh3 = dv dt = 3 64 π 3h2 dh dt = dv dt = 9 64 πh2dh dt Substitute Key Moment Inormation now and not beore now!!!): = 9 64 π6)2dh dt Solve or the desired quantity: dh dt = 64 36π 9 = 6 m 8π min Answer the question that was asked: The height is decreasing at a rate o second. 6 8π meters every 7