MA 113 Calculus I Fall 2009 Exam 4 December 15, 2009

Transcription

1 MA 3 Calculus I Fall 009 Eam December 5, 009 Answer all of the questions - 7 and two of the questions 8-0. Please indicate which problem is not to be graded b crossing through its number in the table below. Additional sheets are available if necessar. No books or notes ma be used. Please, turn off our cell phones and do not wear ear-plugs during the eam. You ma use a calculator, but not one which has smbolic manipulation capabilities. Please: clearl indicate our answer and the reasoning used to arrive at that answer (unsupported answers ma not receive credit), give eact answers, rather than decimal approimations to the answer(unless otherwise stated). Each question is followed b space to write our answer. Please write our solutions neatl in the space below the question. You are not epected to write our solution net to the statement of the question. Name: Ke Section: Last four digits of student identification number: floor[frac[π e ]0 ] Question Score Total Free

2 . Find the following its: sinsin( ) 0 5 ln e 5 e 0 tan sin - sin() cos -cos() H =0 : : both derivatives correct :answer ln+ e e +e e e H = H e = 0e 5 : : both derivatives correct (st application) :algebra : both derivatives correct (nd application) :answer e e H 0 0 tan sec : : both derivatives correct :answer : mention at some point the applicabilit of l Hospital s Rule sinsin( ) ln e 5 e 0 tan /5 Total: 9

3 . Consider the function 3 g ( ) 6 9 on the closed interval [,]. List all of the critical points of g ( ). On what intervals is g ( ) increasing? On what intervals is g( ) concave down? (d) List all - and -coordinates for the absolute maimum. (e) List all - and -coordinates for the absolute minimum. g()=3 9 g() 0 3( )( 3) 0,3 :g() correct : : both critical points The function is increasing on the intervals < < and 3 < < because the g > 0 there. g()=6, so g() is concave down on (,), since g < 0 there. : (-,) (3, ) 3: :reason : nd derivative correct : : answer and reason (d) Evaluate g( ) =, g() = 6 and g() = 6. The absolute maimum occurs at (,6) and at (,6). :(,6) : :(,6) (e) Evaluate g( ) =, g(3) = and g() = 6. The absolute minimum occurs at (, ). : (, ) Total: 3 Critical points: =,3 g() is increasing on (,) U (3,) g() is concave down on (,) (d) The absolute maimum occurs at the point(s): (,6) and (,6) (e) The absolute minimum occurs at the point(s): (, ) 3

4 Consider the curve e. Find the derivative, d, of. d Find the slope of the tangent line to this curve at the point (,0). Find the equation of the tangent line at the point (,0). Epress it in the form = m + b. e = d d e e 0 d d d e d e : use implicit differentiation :d( ) : :product rule and d(e ) :answer 0 d e m 0 d (0) e,0 : evaluate derivative at (,0) : :answer 0=-() : use point (,0) and slope - : :answer d d e e The slope of the tangent line: m= The equation of the tangent line: = + Total: 8

5 . If f 0 ( ) g( ), g(), and g() 3, find f ()., find h ( ). If h ( ) cos e 0 f() = g() 9 0 f() 0 g() g() 9 0 f'() 0 () g() g () h() cos e h'() sin( e )( e ) : correct derivative 3 : : substitute :answer : derivative of outside function 3 : : derivative of inside function :answer f () 3 h( ) sin( e )( e ) Total: 6 5

6 5. Find the following integrals and/or antiderivatives. You must show all of our work to receive full credit. An answer without supporting work will receive no credit. (d) d 3 0 t cos t t sin( ) d 0 9 d. 3 0 d 5 5 t cost cos t t t sin t sin sin() t (d) 0 sin( ) d sin u du u cos u C cos( ) C u 9 d du u u du u 9ln(u) 9 9ln(3) 9ln(9) u Other acceptable answers: +9 ln(9/3); +ln(387089/ ); +ln( ); 3 9 : of 5 antiderivatives correct 3 : : 3 of 5 antiderivatives correct :answer : simplif and antiderivative 3 : : nd antiderivative :answer : a reasonable substitution 3 : : correct antiderivative :answer : reasonable substitution 3 : : correct antiderivative :answer If a change of variables is done incorrectl, take the point from the answer point. d = t cost = sin sin() t sin( ) d= -½ cos( )+C (d) d = + 9ln(9) - 9ln(3) 0 9 Total: 6

7 t 6. Consider the function F( ). t Find all intervals on which F() is increasing. Find all intervals on which F() is concave up. 0 F(). F() is increasing for > 0. 0 for > 0, so F 3 ( ) 3 F''() ( ) ( ) F ()=0 for / 3. F is concave up on, since F () > 0 there. 3 3 : derivative : set derivative > 0 : :answer :reason : F''() : deal with inflection points 5: :find interval :answer :reason F() is increasing on (0, ) or for > 0 F() is concave up on, 3 3 Total: 9 7

8 7. For this problem use the following information about a basketball. 3 Surface Area: A r Volume: V r. 3 A basketball is being inflated and its volume is increasing at the rate of 5 cm 3 /sec. Be sure to remember to include units in our response. Find the rate at which the radius, r, of the ball is changing when the radius is 5 cm. Is the radius increasing or decreasing when the radius is 5 cm? Justif our answer. Find the rate at which the surface area is changing when the radius is 5 cm. (d) Is the surface area increasing or decreasing when the radius is 5 cm? Justif our answer. dv =5 cm 3 /sec dr dv πr 5 dr 5 πr dr 5 cm / sec π5 0π r5 dv : 5 : : find correct epression for dv/ : find correct epression for dr/ :answer dr r5 0 so the radius is increasing when the radius is 5 cm. :answer : :reason ds dr =8πr ds 8π 5 0π ds r5 r5 cm /sec :ds/ : :answer (d) ds r5 0 so the surface area is increasing when the radius is 5 cm. :answer : :reason - point if units are not correct in and. Total: 0 8

9 Work two of the following three problems. Indicate the problems that is not to be graded b crossing through its number on the front of the eam. 8. State the Mean Value Theorem. Use complete sentences. Determine the values for the constants a and b such that the function f defined b 0 f( ) ab 0 3 satisfies all of the hpotheses of the Mean Value Theorem on the interval [0,3]. As usual, show our work to support our answer. Find at least one point in [0,3] where the conclusion of the theorem is satisfied. Let f be a function that satisfies the following hpotheses: () f is continuous on [a,b]; () f is differentiable on (a,b) Then there is a number c(a,b) f f so that f' b a The function must be continuous [0,3] and differentiable on (0,3). Thus: f() f(0) so 0 b =. F must be differentiable (hence continuous) at =, so we must have that the derivative from the left f' () a be the same as the derivative from the right: f' () () 6. Thus a = 6. f(3) f(0). For f to satisf the MVT it must do so in the interval from to 3. f ()= + = /3 gives = 5/3 (which does lie in the necessar interval.). : continuit hpothesis : differentiabilit hpothesis 5: :conclusion :completeness of statement : discuss continuous and differentiable :b 5: : left derivative is a : right derivative is 6 : a=right derivative : find slope of the secant line 5: : = 5/3 : check that answer in correct interval 9 Total: 5

10 9. State both parts of the Fundamental Theorem of Calculus. Use complete sentences. Consider the function f on [, ) defined b f ( ) sin ( u ) du. Eplain wh the function f ( ) is increasing. Find the derivative of the function g( ) sin ( u ) du. 3 Suppose f is continuous on [a,b]. (I) If g() f(t), then g () f(). a (II) f() d F F, where F is b a an antiderivative of f, that is, F = f. (page 387) Since sin ( ) is continuous and differentiable b the FTC I, f ()=sin ( )=(sin( )) 0 for all 0. Thus, f is increasing there. 6: 3:FTC I 3:FTC II :f'() : :reason :reasonftcapplies g() = -f( 3 ) b the properties of the integral. Thus, 3 6 g'() f '( ) (3 ) 3 sin ( ) 3 : recognizing g() = -f( ), somehow 5 : : correct derivative 6 5 : sin ( ) and not sin() Total: 5 0

11 0. A particle moves along the -ais so that its velocit at an time t 0 is given b vt ( ) 3t t miles/minute. The position (t) is 5 for t =. Find the acceleration of the particle at time t = 3. Is the speed of the particle increasing at time t = 3? Give a reason for our answer. Find all values of t at which the particle changes direction. Justif our answer. (d) Find the total distance traveled b the particle from time t = 0 until time t = 3. a(t) = v'(t) a(t) 6t a(3) 6 miles / minute :a(t) : :a(3) At t = 3, v(3) = 0 > 0 and a(3) > 0, so the speed of the particle is increasing. :v(3) 0 : :reason The particle changes direction when the velocit changes sign. v(t) = 0 = 3t t = (3t + )(t - ), so v(t) changes sign at t = and at t = -/3. -/3 is not in the domain, so must be omitted. (d) Between 0 and 3 the velocit changes sign at t =, so the total distance traveled is: 3 v(t) v(t) (3t t ) 0 0 (3t t ) 3 3 (t t t) (t t t) 6 7 miles :v(t)=0 :t = : : check that v(t) changes sign : must deal with ommiting t=-/3 :integral : integrate right parts 6: : correct subintegrals : answer must be positive : units correct in and (d) Total: 5