REVIEW SHEET 1 SOLUTIONS ( ) ( ) ( ) x 2 ( ) t + 2. t x +1. ( x 2 + x +1 + x 2 # x ) 2 +1 x ( 1 +1 x +1 x #1 x ) = 2 2 = 1

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1 REVIEW SHEET SOLUTIONS Limit Concepts and Problems e sin t + t t e sin t + t t e cos t + + t t + + t + t t t t + + a b c - d DNE e f NO. a removable discontinuity at - b infinite discontinuity at Derivative Concepts and Problems. def d d def + +

2 7 + def cos + ln + sin + + ln + tan + sec e tan e tan sec e tan e + e sin sin + e 6 t tan t tan + t + t. a f g g f g b fg f g + g f c f g f g g f g w u v v u v Related Rates/Implicit Differentiation. + y + y + y d d + y y at,y,-, d

3 y + y + 9y y + y + 6y d d y d 9 6y y so at,y,, d d d y 9 6y y. m Given dv.m r r meters 6 m meters V volume m min, Find d wen m Formulas: by similar triangles: Differentiating wit respect to t: r 6 r and volume, V r 7 dv 7 d d dv 7. 7 Wen te water is meters ig, te water level is dropping at.7 meters per minute.. S v ds v v + dv v dv so ds. ft y ft v + v +.7m min +..7 y ft Given d Find ft sec wen feet. Formula: by Pytagorean Teorem, + y d so differentiation wit respect to time we obtain + y y Also, by te Pytagorean Teorem wen, y Te top of te ladder is dropping at feet per second wen te base is feet from te wall. d ft s. 6. L D W

4 L lengt cm D diagonal cm W wi cm A area cm P perimeter cm Given dl cm sec and dw Formulas: cm sec da, Find, dp dd, and A LW da dl W + L dw P L + W dp dl + dw D L + W dd L + + L + W + wen L cm and W cm + cm sec + cm sec dl + + W L + W dw cm sec 7. ft y Formulas: A + y and B d Given tat 8 ft/sec. d Find wen y feet. tan From B we ave by differentiation wit respect to t tat; d sec d y d d d y 8 So te angle is decreasing at te rate of. rad/sec wen feet of kite string as been let out.. rad/sec BASIC MAX/MIN. Find/test critical points, end points. f + f. Solving for te critical points,. Testing we ave f +. For te endpoints we ave: f +., and f +. So te absolute maimum occurs at and te absolute minimum occurs at.. Find/test te critical points and end points for f 9 on [-,]. We ave f 9 wen. Testing we ave f 9 9. For te endpoints we ave f and f So te eact minimum value is y and te eact maimum value is y.

5 . For f, te derivative is positive so te minimum occurs at te left endpoint and te ma at te rigt endpoint: Minimum Maimum 7 9. Let te base ave dimensions by square and te eigt be y. Since te volume is fied we ave te equation: V y Te total material used in te open bo is te AREA of te base plus te sides of te bo. We let A area, ten tis total is: A y + We wis to FIND and y, WHEN A is MINIMUM. We solve for y and plug in to. y V A V + V +. We find te critical points by solving A : A V V V. We can see tat te second derivative will be positive, so te function is concave up, so tis must be were te minimum occurs. To find y we use, V y V V. So te dimensions are V V V. Te spere at rigt as radius cm. Te cylinder as radius r and eigt. By Pytagorean teorem we ave: r +. Te volume of te cylinder is V r. We use and substitute for r in giving us. V. To find te maimum, we find te critical points: V Solving for, we obtain. Noting te second derivative is negative for positive, tis must be a maimum. To find r we use so r. So te can wit maimum volume will ave radius r cm and eigt cm. r cm 6 We will let te wi and eigt of te printing region ave dimensions w and inces. Ten w square inces. Te total paper used will be given by A w We solve for w and substitute into to obtain A We now find te critical points for A by solving: A and find in. in in Using again we find w in. Now te OVERALL dimensions of te poster ave te margins included, so te eigt will be 8 inces, te wi 9 inces.

6 Curve Sketcing Problem, PAGE : Given te grap of f, te grap of te derivative would ave to pass below te -ais near te origin. Tat is to say tat te derivative would be negative because te function is decreasing near te origin. Grap C is te only one tat meets tis criterion. Also notice tat te grap of f flattens out at its etremities, terefore its derivative would be close to, like C or D. Te grap of f reaces its ma and min eactly were graps B and C ave zeros or roots. Multiple Coice Questions PAGE 7-8. Δ for all subintervals;,,, f - f [ ] + [ ] + [ ] + + i i. Δ for all subintervals;.,.,., f +. f [. + ] + [. + ] + [. + ] 9 i i answer works probably e was printed incorrectly 8 Note no multiple coice. Δ for all subintervals;,,,,, f + f [ + ] + [ + ] + [ + ] + [ + ] + [ + ] i i Δ. for all subintervals;,., f f.[] +.[.] + 9 i i Antiderivatives, te Fundamental Teorem, applications. f sec tan + + sin + - e. f sec + - cos + arcsin - e.. a d + c b d 6

7 + c d d sin d cos e cos sin d sin + cos + +. vt t, < t < displacement t / t - t 7/ units t dist. traveled t - + t - - / t - t + / t - t t t - 7/ / - - 7/. 7/. units. at t +, < t <, v t t vt a t t + + t + c, v c, so, vt + t + units/time t t st v t + t + + t + t, s - s units n t 6. E Distance traveled t t Note misprint of time sould be first seconds 7. F 8. G + A sin A sin cos sin cos sin cos 9. F f. G A A + +