2004 Free Responses Solutions. Form B

Transcription

1 Free Responses Solutions Form B All questions are available from James Rahn

2 Form B AB Area d 8 B. ( ) π ( ) Volume π d π.7 or.8 or ( ) Volume π 9 y 7. or 68 π

3 Form B AB R(6).79 > so since the rate of change R(t) is positive, the number of mosquitoes in increasing. B. R ' (6)-.9 < so the number of mosquitoes is increasing at a decreasing rate. + Rt ( ) 96. or 96 mosquitoes P Mosquitoes + R( t) D. Rate of mosquitoes P '( ) R( ) P increases from to 7.89 and.69 to, but decreases from.89 to.69. There are two locations for a possible maimum. They are at 7.89 and. P(7.89) 9 mosquitoes P() 96 mosquitoes

4 Form B AB/BC Riemann Summidpoints for v(t) i9. + i7. + i. + i. 9 This Riemann sum is an approimation of the number of miles the plane flew in the first minutes of the flight. B. Since v() v() 7. and v() v()., the MVT for V would guarantee at least two locations where the acceleration will be zero since the average acceleration between and is zero and the average acceleration between and is zero. t 7t f() t 6+ cos + sin f '().7 or.8 miles/min D. f( t).96 miles/min

5 Form B AB/BC points of inflection at and because the f " changes sign at this values. B. Possible locations of minimums:. It is an absolute min and a local min. Possible locations of maimums: - and The function f begins above (,6) since the slope of f is negative or zero to the left of. There is a zero slope on f at. The function falls the right of (,6). Then after (,6) f falls again (ccd) until and changes concavity to ccu. From to f continues to be ccu but f is decreasing. At f stays ccu but f increases to where it obtains the same y value as f(). At the function has a horizontal tangent line. The abs. ma occurs at - since f() is only equal to the f(). Since f() t it was at. the graph of f will end up at the same height at as g ( ) f( ) g'( ) f '( ) + f( ) g'() f '() + f() g '() ( ) + 6 g() f() (6) the tangent line is y ( - ) +

6 AB B. Slope is negative when y-< or y < and does not equal + c ( y ) d d y ln y + c e Ce y y () y + Ce y + C C y e

7 Form B AB6/BC6 n d n+ n + n + The Rate of change of the area is given by B. Tangent line: y n ( - ) + y n - n + -intercept for the tangent line is n Area of Triangle n n n n Area of S d n n n n n ( n)( n + ) n ( n)( n + ) The numerator of the derivative equals zero at ± Let a b + Since the derivative's denominator is always positive, the sign of the derivative is controlled by the numerator. The sign stu would be The sign stu indicates that the maimum area occurs at +

8 Form B BC 7 Velocity d d B. Speed t t t e e ( ) ( d ) t ( e ) ( t 9) ( ) t 8 or 7.6 or 7.66 velocity d 7 Since the point of the path at t is (,) the equation of the tangent line is 7 y ( ) The total distance traveled is found by integrating the length along a path. d + t t ( t 9) ( e e ) or.7 acceleration vector t t,e e t + 9 t d d y, To write the equation of the tangent line we will need the slope of the line or the velocity vector to find the slope. t t velocity vector t + 9,e e t, 7 D. To find the position at time t we can write an equation that uses the starting position plus an integral that represents the distance the particle moves in the -direction. ( ) ( ) () () + t + 9 () + t or 7.9 So the slope of the tangent line is

9 BC Let the Taylor Polynomial be t( ) 7 9( ) ( ) in this problem. From this form f() 7, f ' (), f ''() -9, and f '''() - Using this fact we know 9 f "()! since it is the coefficient of the( ) This yields the value of f "() -8. term B. From the list in part A we know that f ' () since there is not (-) term in the Taylor Polynomial so the coefficient of - must be zero. This leads to being a critical value of f since f ' (). Since f "() -8 this indicates that f is concave down at and there is a critical value at so the function f has a maimum at. C. T () 7 9( ) ( ) No there is no information about the derivative at, on the derivative at. D. Error is () f ( c) ( ) for somec (,)! f ( c) f T for some c! Since f () ( c) 6, it follows that () () () + ( ) (,) 6 f T or f () ( ) () () ( ) f () Thus showing that f() is negative.

10 Form B BC Let g ( ) d B. π d π ln π ln lim b ( b ) lim d lim b lim d lim b b ( b ) ( b ) ( b )( b + ) lim b lim lim ( b + ) So it diverges. b b b