Answers to Some Sample Problems

Transcription

1 Answers to Some Sample Problems. Use rules of differentiation to evaluate the derivatives of the following functions of : cos( 3 ) ln(5 7 sin(3)) e 3 h 3 e i sin( 3 )3 +[ ln ] cos( 3 ) [ln(5) + 7 ln()+lnsin(3)] cos(3) 3 sin(3) ( 8)[5 ] (3 5 +9)[ 5 ] ( 8) 3 e 3 ( 3) + e 3 (3 ) 3 [e ( )] + e [3 ] [5 tan(π) + 7 sec(5)] 5sec (π) π + 7 sec(5)tan(5) 5 [arctan()] +() arcsin( ) q / ( ). For the function f() + 3 +,usethedefinition of the derivative to find f () (no formulas). f( + h) ( + h) h + f( + h) f() ( + h) h ( +h + h 3( +) 3( + h +) )+ ( + h +)( +) h + h 3h + ( + h +)( +) f( + h) f() 3 + h h ( + h +)( +) 3 h ( +) 3. You are given the following information about a function f: Itsvalueat is 7 and its instantaneous rate of change at is. (a) Use this information to find the equation of the tangent line at. Use point-slope m( ) with f( )f() 7 and m f (): 7 ( ), or ( ) + 7 (b) Use part (a) to estimate f(.5). f(.5) tan (.5 ) (c) Suppose that f () < for all. Is our estimate in part (b) an overestimate or underestimate (draw a diagram and eplain). Since f () < the curve is concave down, so its values lie below the tangent lines. Thus, the tangent line values are greater, so are overestimates.. Suppose Q() R()e.IfR() and R () 5, find Q (). Product Rule: Q () R()[e ( )] + R ()e,so Q () R() ( )e + R ()e e +5e e.353.

2 5. The circumference C πr of a circle is measured to be cm, with an error of ±. cm. If the Area A πr is computed using this value of the circumference, what is the maimum error? What will be the relative error? Epress A in terms of C: R C µ C π so A π C π π.notethatda dc C π The tangent-line approimation sas that A C da dc da or A C (here is the error). Thus dc Also, Error in A C (Error in C) π Error in A (±.) ±.3 π Relative Error in A A A ±.3 /(π). or 5 %.. is implicitl defined to be a function of b 3 +. One point on the graph of this function is (, )(5, 3). (a) Find the equation of the tangent line to the graph of as a function of at (, ). We just need the derivative to appl point-slope: 3 d d + d d d d Thus: 3 3 ( 5) or 3 ( 5) + 3 (b) Use our answer to approimate at 5.. Give our answer as precisel as possible. 3 (5. 5) A point mass moves along the -ais in such a wa that its position at time t is (t) e t sin(t). B computing d/dt and d( )/dt, prove that this function (t) satisfies + +. (This is straightforward differentiation, using the product rule, and algebra.) 8. Use logarithmic differentiation to differentiate the functions f() ( +3)9 (5 +) 7 and g() (tan()). + π sin() Take ln of both sides: Now differentiate: Dothesameforg(): ln(f()) 9 ln( +3)+7ln(5 +) ln( + π) ln(sin()) f () f() π cos sin f () ( +3)9 (5 +) 7 µ 8 + π sin() π cos sin ln(g()) ln(tan()) g () g() sec () tan() +ln(tan()) g () (tan()) µ sec () tan() + ln(tan())

3 9. Let f() (a) Find all critical points of this function, and determine which are maima or minima. f () 3 + (3 )( +) set,so /3 and are the critical numbers, which give the points (/3, 3/7) (.33,.85) and (, ) having horizontal tangents. (b) Find all inflection points. f () + set,so /3 and ( /3, /7) (.33, 5.) is an inflection point (f clearl changes sign at /3). (c) Draw a sketch of the curve f() showing the maima, minima and inflection points (d) Give intervals where the function is increasing and where it is decreasing. Function is decreasing between the two critical points; i.e. for <</3; it is increasing elsewhere. (e) Find the absolute maimum and absolute minimum of f() on the interval [, ]. We need to test the critical points and the endpoints: f( ) 3 f( ) f(/3) 3/7 f() Thus, the absolute maimum is and the absolute minimum is 3.. If g() + then g () ( +) () ( +) ( +) and g () ( + )( ) ( +).Usetheseto findallcriticalpointsofg, all inflection points. Does the curve g() have an asmptotes? Draw a sketch of g(). Note that +is alwas positive, so never ; thus,bothg () and its derivative are alwas defined. To find the critical points, set g (). This means (+)( ),so ± are the onl critical points. When <, g () < ; when lies between and, sa, g () > ; when>, g () <. Thus, is a relative minimum and is a relative maimum. (You can also use the second derivative test: g ( ) > and g () <.) To find inflection points, set g () ; ou get and ±. Since + is never, there are no vertical asmptotes. On the other hand: lim + lim so is a horizontal asmptote. Here is a plot: / / +/ lim / ( ) +/ ( ), 3

4 ½ t. Let C be the curve with parametric equations 3t 5 5t 3 +t. (a) Find d in terms of t. d (b) Find all points where C has horizontal or vertical tangents. d d d/dt d/dt 5t 75t + 5(t 5t +) t t 5(t )(t ) t 5(t )(t +)(t )(t +) t Thus, there are horizontal tangents at t ±, t ±, and a single vertical tangent at t. (c) Use our calculator to draw a sketch of C showing all interesting features (see part b); state what window ou used. 3 t 3: Use Newton s method to find the smallest positive solution to cos. We need to find a root of cos. A quick plot shows there is one between and. Newton s method is the formula new old f ( old) f ( old ) old ( old) cos ( old ) ( old )+sin( old ) Starting with the guess.5, repeated iterations of the formula give guesses that stabilize at A closed, rectangular wooden bo with square base and lid is to contain 8 cubic feet. The material for the base and the lid costs twice what the material for the sides costs. What are the dimensions of the least epensive bo

5 that can be made with these specifications? We find the critical point: V s h 8, soh 8/. Cost of a side sh (P ) Cost of top or bottom s (P ) Total Cost C sh P +s P C s(8/s )P +s P 5P/s +s P C 5P/s +8sP set 5/s 8s s 3 s, h 8.. Find the rectangle of maimum area, with a side ling along the -ais, whose top vertices lie on the parabola 8. Width W, Height H 8,soA (8 ) 3. A set / 8/3 p 8/3.33 W 3., H , Area A certain coffee filter is in the shape of a cone of radius cm and height 3 cm. When the depth of the coffee in the filter is 5 cm, coffee is draining out of it at a rate of 5 cc per minute. What is the rate of decrease of the depth of the coffee at that moment? Let h be the coffee level and r the radius of the cone of coffee. From the usual similar triangles computation: h/3 r/, orr h 3.ThevolumeV of the coffee is given b V 3 πr h 3 π h 3 h 3 π h 9 h π 7 h3. dv dt π 9 5 c 3 /m π 9 dh dt dh h dt dh (5) dt,so c/m π 5. Using a subdivision of the interval [, 3] into equal parts, and the evaluation set consisting of the left-hand Z 3 end-points, calculate a Riemann sum that approimates d

6 Z 3 µ d.5 () +.5 (.5) + () +.5 (.5).58 (a) Is Riemann sum an overestimate or underestimate? Use a diagram to eplain. Since the curve is decreasing, the heights are too big, so it is an overestimate. (b) Estimate the integral using subdivisions, but this time use midpoint rectangles Z 3 µ.5 d.5 (.5) +.75 (.75) +.5 (.5) +.75 (.75) 7. Evaluate the following integrals: (a) R d R d ln + C (b) R 5 d 5 R d 5( /( ) 5 + C (c) R 5 sin +8cosd 5 cos +8sin + C (d) R + d arctan + C (e) R 3e + e d 3e e + C Z 5 + (f) d R 5 / + / / d R 5 / + 3/ / d (g) 5 3/ /3+ 5/ /5 / (/3) 3/ +(/5) 5/ / + C Z e d. Letu.Thendu / d or d du. So R e d R e u du e u + C e + C 8. Calculate each of the following integrals b giving a geometrical argument. (a) R 5 5 d This is the area under the top half of the circle + 5,from to 5; i.e. the area of / of the circle of radius 5. R 5 5 d 5π/ (b) R 3 5+3d. This is the area under a trapezoid of width 3, with bases (heights) and 5+9 : R 3 5+3d (8 + ). 9. Suppose that G () g() for all, andg() 7, g(3) 5, G() and G(3) 9. Calculate R 3 g() d. B the Fundamental Theorem of Calculus, R 3 g() d G(3) G() Calculate the definite integrals: (a) Z (3 ) d (3 ln ) (8 ln ) ( ln ) 7 ln.

7 (b) (c) Z π/ Z So R 3sec d3tan π/ 3( ) 3. + d. Letu +.Thendu d or du. Calculate the derivative: + d R du u lnu ln. d. Moreover u and u. (a) f() R et dt. f () e t t e. (b) g() R +t dt. g () +t k t ( ) + ( / ) +.. An object moves along the -ais so that its acceleration at time t is given b a(t) e t t cm/s If we know that its initial (t )velocitis7 cm/s and its initial position is at cm, find its position at time t. Let v (t) be the velocit at time t, and(t) the position at time t. a(t) dv dt et t, sov(t) R e t tdt e t t /+C. Wearegivenv() 7, so 7 v() e /+C +C C v(t) e t t /+. Also, v(t) d dt et t /+,so(t) R e t t /+dt e t t 3 /+t + D. Wearegiven (), so Thus, (t) e t t 3 /+t +3. () e 3 /+ +D D 3. 7