2011 Form B Solution. Jim Rahn

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1 Form B Solution By Jim Rahn

2 Form B AB 6 6 S'( t) dt 7.8 mm 6 S '( t) dt.86 mm or.864 mm c) S '(7) dv d( r h) dh dh r r dt dt dt dt dr since r is constant, dt dv dh r mm dt dt day d) D()=M () - S () = < D(6) = M (6) - S (6) = > Since the value of D at is negative and the value of D at 6 is positive and D is continuous because M and S are continuous, there must exists a value of t, <t<6, where D(t)= by the Intermediate Value Theorem. Rahn Modified on May,

3 Form B AB 6 5 r(5) 75 5 lim rt ( ) 75 t 5 lim rt ( ) ( e ) 75 t 5 solim r( t) does not exist t 5 Therefore, since the limit does not exist at 5 the function is not continuous at rtdt () rtdt () 5 ( ) liters per hour r '() 5.litersper hour per hour c). This is is the rate at which the drainage rate is increasing at time t= hours 5 A d) r t dt r t dt 5 ( ) ( ) 9 Rahn Modified on May,

4 Form B AB 4 6 f ( x) dx g( x) dx x 6 x 6x x y ; x 6 y (6 )( ) Volume y y y dy c) Slope of g is -. Therefore we are looking for a location on f where the slope is +. 4 f '( x) x x, y Rahn Modified on May,

5 Form B AB4 f (x)= at x = 4 f (x)> when x >4 so f is increasing when x>4 f (x)< when x<4 so f is decreasing when x<4 So there is a relative maximum at x = 4. f x x x x x ( x) "( ) (4 )( 4 ) ( )( ) 4 since f (x)< when x<6 f is concave down when < x <6. c) (4 )( ) x x dx x x dx (4 ) 4x f( x) x c f() 4() () c f x x x ( ) c Rahn Modified on May,

6 Form B AB5 v() v()... at ( ). meters/ sec 6 6 vt ( ) dt total distance in meters traveled in the first 6 seconds v( t) dt meters c) Since B is twice differentiable, this means that v is differentiable and continuous. S(4)=9 S(6)= 49 B(6) B(4) 49 9 Average velocity Therefore, by the Mean Value Theorem for v, there must be a value of 4 t 6 where B (t)=v(t)=. d) ( Lt ( )) ( Bt ( )) ( Lt ( ))( L'( t)) ( Bt ( ))( B'( t)) BtB () '() t L'( t) Lt () Using the right triangle: sides are 9--5 (9)(.5) L '(4).5 meters/second (5) Rahn Modified on May,

7 Form B AB6 4 x gx ( ) cos dx 4 4 x gxdx ( ) cos dx ( )( ) (4 )( ) x sin 4 6 f (x) x f '( x) g'( x) sin 4 Critical values occur when f is not defined or f is zero. Since g (x) does not exist at x =, then f does not exist at x =. For x g (x)= so x g'( x) sin x () sin no solution x 4 For g (x)=-½ x g'( x) sin x ( ) sin x sin x So there are critical values at x = and x = π c) h'( x) g( x) h' g g( ) Rahn Modified on May,

8 Form B BC a Area r( ) d 47.5 x r( )cos sin cos We need to know where. Using the graphing calculator in function mode we can determine that this at θ=.7 The y coordinate would be y r( )sin ( sin )(sin ) ((.7) sin(.7))(sin(.7)) 6.7 c) y r( )sin ( sin )(sin ) dy d( sin )(sin ) d dt d dt Using the calculator to find the dy/dθ at θ = π/ gives Therefore: dy d( sin )(sin ) d dt d dt.4954().89 This means that when θ = π/, the y coordinate is decreasing at the rate of -.89 units of length per unit of time. Rahn Modified on May,

9 Form B BC4 Average value of f on [,5] is f( x) dx ( ) f( x) dx 7 7 f( x) dx (7) 7 c) Because g (x)=f(x), the graph of g is decreasing where f(x) is less than zero. This is on the interval x 5. The graph of g is concave up where the graph of f is increasing. This occurs on the interval <x<8. So g is both concave up and decreasing on the interval < x < 5. d) Arc length of h= ( '( )) h x dx x f ' dx x let u ; du dx x, u, x, u f ' u du This states that the arc length of h is twice the arc length of f on the interval x. The arc length of f is +8 or 9 so the arc length of h from x is 58. Rahn Modified on May,

10 Form B BC6 6 9 n x x x n x ln( x ) x ( ) 4 n ( ) 4 n n At x = : The series is. Using the alternating series test we see that this converges. ( ) 4 n At x = -: The series is. This series diverges because it is the negative of the harmonic series. x The interval of convergence is. 6t 9t t C) f '( t) t t t t t f '( t ) t t t t x 4 gx ( ) t t dt 4 g() t t dt t t 5 5 d) The alternating series error does not exceed the absolute value of the next omitted term. The next term is 7 6 t t dt Rahn Modified on May,

2018 FREE RESPONSE QUESTION TENTATIVE SOLUTIONS

2018 FREE RESPONSE QUESTION TENTATIVE SOLUTIONS 8 FREE RESPONSE QUESTION TENTATIVE SOLUTIONS L. MARIZZA A BAILEY Problem. People enter a line for an escalator at a rate modeled by the function, r given by { 44( t r(t) = ) ( t )7 ) t t > where r(t) is