Mark Howell Gonzaga High School, Washington, D.C.

Transcription

1 Be Prepared for the Calculus Exam Mark Howell Gonzaga High School, Washington, D.C. Martha Montgomery Fremont City Schools, Fremont, Ohio Practice exam contributors: Benita Albert Oak Ridge High School, Oak Ridge, Tennessee Thomas Dick Oregon State University Joe Milliet St. Mark's School of Texas, Dallas, Texas Skylight Publishing Andover, Massachusetts

2 Copyright 5-4 by Skylight Publishing Chapter. Annotated Solutions to Past Free-Response Questions This material is provided to you as a supplement to the book Be Prepared for the AP Calculus Exam. You are not authorized to publish or distribute it in any form without our permission. However, you may print out one copy of this chapter for personal use and for face-to-face teaching for each copy of the Be Prepared book that you own or receive from your school. Skylight Publishing 9 Bartlet Street, Suite 7 Andover, MA 8 web: sales@skylit.com support@skylit.com

3 4 AB AP Calculus Free-Response Solutions and Notes Question AB- (a) The average rate of change is A A.97 pounds per day. (b) A ( 5).64 pounds per day. On day 5, the amount of grass clippings remaining in the bin is decreasing at the rate of.64 pounds per day. (c) The average amount of grass clippings is A() t dt Solving At () =.7565 gives t.45 days. (d) A.7898 and A Lt = A + A ( t ). Solving pounds. Lt = gives t 5.54 days.. Store this result in a calculator variable, then use the stored value to solve the equation.. Again, store these results in calculator variables (keep greater accuracy to assure that the final answer is correct), then use the stored values in the formula for Lt. ()

4 4 FREE-RESPONSE SOLUTIONS ~ 4 AB Question AB- (a) Solving f ( x ) = 4 gives x = and x =... ( 4 ) ( ) V = π + f x + dx. (b) The area of each isosceles right triangle is ( f ( x) ). is ( f ( x) ) dx k (c).. ( 4 f x ) dx = ( 4 f ( x) ) dx. k 4. The volume of the solid. This is the exact value for the x-coordinate of the point. In this case, it s just as easy to enter. in subsequent calculations. f x dx f x dx k.. Or: ( 4 ) = ( 4 ).

5 FREE-RESPONSE SOLUTIONS ~ 4 AB 5 Question AB- (a) Using the areas of two triangles, g = = 9. = = 5 4 f t dt (b) The graph of g is increasing where g ( x) = f ( x). The graph of g is concave down where g ( x) f ( x) for < x <. (c) 5x g ( x) 5 g( x) h ( x) = 5x h 5 ( ) = <. Both of these conditions hold for 5< x < and =. (d) p ( x) f ( x x) ( x ) 5 g 5 g 5 f 5 g = = = =. The slope at x = is equal to f p = = = 6.. You can leave it at that to avoid arithmetic mistakes.. Or, using the areas of three triangles, g = 4+ 4 = 9.

6 6 FREE-RESPONSE SOLUTIONS ~ 4 AB Question AB-4 (a) The average acceleration is the average rate of change of velocity, which is v( 8) v = = meters per minute per minute (b) Since va ( t ) is differentiable, it is also continuous. Since is between v A ( 5) = 4 and A ( 8) va on the interval [5, 8] guarantees that va ( t ) = for some t between 5 and 8. v =, the Intermediate Value Theorem applied to () t (c) The position at t = is + v () A t dt. Using a trapezoidal sum approximation, this is approximately ( ) ( ) + ( 5) + (5 ) + (8 5) + ( 8) meters. (d) Let x() t and y() t be the positions of trains A and B, respectively, and z() t be the dz dx dy distance between the trains. Then z = x + y z = x + y dt dt dt dz dx dy z = x + y. At t =, xt = and yt = 4, so zt = 5. dt dt dt dx dy = va () =. = vb () t = 5t + 6t+ 5, so dt t= dt dy = = 5. Therefore, at t =, dt t= dz dt = + dz = meters per dt t= 5 minute.. = 5, so the train is approximately 5 meters west of Origin Station.. = 6

7 FREE-RESPONSE SOLUTIONS ~ 4 AB 7 Question AB-5 (a) f has only one relative minimum on [,], at x =, because this is the only number on [,] f x changes sign from negative to positive. where (b) Since f is a twice-differentiable function, the Mean Value Theorem applies to f x on the interval [,], such that f ( c). Thus, there is a c in the open interval () f ( ) f = =. f ( x) (c) h ( x) = h f ( x) f = = =. f 7 4 (d) An antiderivative for f g( x) g x is f g( x ). So f ( g( x) ) g ( x) dx f ( g( x) ) f ( g ) f ( g ) = = = f() f( ) = 8 = 6.. Alternatively, since f ( ) = f () =, you can refer to Rolle s Theorem.

8 8 FREE-RESPONSE SOLUTIONS ~ 4 AB Question AB-6 (a) (b) At (, ), the slope is ( )cos() =. An equation of the tangent line is y = x. For x =., this gives y =.4. dy x dx y = dy cos( x) dx ln y sin ( x) C y = = + f () =, we get ln = C C = ln (c) Separating variables, we get cos. Substituting the initial condition ln y = sin x ln ln y = sin x + ln ( x) ( x) sin + ln sin y = e = e. Two possible solutions satisfy this equation, sin( x) e y = and condition y () =. Therefore, sin( x) e y = +, but only the first one of them satisfies the initial sin( x) e f( x) =.. The statement f (.) =.4 is incorrect and could result in lost points.. The domain of this solution is all real numbers.

9 4 BC AP Calculus Free-Response Solutions and Notes Question BC- See AB Question. Question BC- (a) π The curves intersect at θ = and θ = π. The area of R consists of the area in the first quadrant plus the area of the quarter circle in the second quadrant: π Area ( ) 9π = sin d θ θ = dx d dθ = dθ. Therefore, dx.66. dθ π = (b) x rcos θ ( sin ( θ ) cos( θ )) θ 6 (c) (d) The distance between the curves as a function of θ is the difference between their respective values of r for a given θ, that is, ( sin( θ )). The rate of change d π of that distance is ( ( sin( θ ))). At θ =, the rate of change is. dθ dr dr dθ π =. At θ =, this is = 6. dt dθ dt 6 d dθ π when θ =. ( ( sin θ )) sin( θ) 4cos( θ). Or write d = =, which is equal to dθ 9

10 FREE-RESPONSE SOLUTIONS ~ 4 BC Question BC- See AB Question. Question BC-4 See AB Question 4. Question BC-5 x (a) Area = x xe x dx = e + x = e + xe x dx. x (b) Volume = π ( + ) ( + ). (c) x dy x x Given y = xe, = x e + e. The vertical line dx x = intersects x y = xe at y = e and intersects y = x at y =, so the length of the vertical segment is e +. The length of the linear segment with slope can be evaluated using the Pythagorean Theorem (or the distance formula): length = + = 5. x x Perimeter = ( + ) e x e e dx.. = e +.

11 FREE-RESPONSE SOLUTIONS ~ 4 BC Question BC-6 (a) ( x ) n+ n+ n+ a n n = = n n a n+ n+ n ( x ) ( x ) n lim ( x ) = x < x < n n +. R =. n 8 n The n n+ (b) The series for f is ( x ) ( x ) ( x ) ( x ) n n series for f is ( x ) ( x ) ( x ) + n (c) The first term of the series for f is and the common ratio is ( x ) series converges to f ( x) f ( x) = dx= ln x + C x Since < x <, = = + x ( x ). f = x is positive, so f ( x) ln ( x ), so the C = f ( x) ln x =. =.