AP Calculus 2004 AB (Form B) FRQ Solutions

Transcription

1 AP Calculus 4 AB (Form B) FRQ Solutions Louis A. Talman, Ph.D. Emeritus Professor of Mathematics Metropolitan State University of Denver July, 7 Problem. Part a The curve intersects the x-axis at x =, so the desired area is x dx = (x )3/ 3 () = 3 ( )3/ = 8. () 3. Part b The volume generated when the region of Part a is revolved about the horizontal line y = 3 is π [ 9 ( 3 x ) ] dx = π = π [ 6 x + ( x) ] dx (3) [4( x) 3/ + x x ] = 3 π.7. (4)

2 .3 Part c Solving the equation y = x for x in terms of y gives x = y +. Hence, the volume generated by revolving the region about the vertical line x = is π 3 [ (y + ) ] 3 ( dy = π y 4 8y + 8 ) dy () ( ) y 3 = π 6y3 + 8y ( ) 43 = π (6) = 648 π (7) Problem. Part a Because R(t) = t cos(t/) is the rate of change of the number of mosquitos on the island and we have R(6) >, it follows from the continuity of R that the number of mosquitos is increasing throughout some interval centered at t = 6. Note: The statement that the number is increasing at t = 6 is problematic: The standard definition of the term increasing applies only on intervals, and not at an individual point.. Part b R (t) = t cos t t sin t, so (8) R (6).939 <. (9) R (6) <, and R is continuous at t = 6. It follows that R(t) is decreasing near t = 6. Thus, the number of mosquitos is increasing at a decreasing rate near t = 6. (But see the note to Part a, above.)

3 .3 Part c By the Fundamental Theorem of Calculus, the number M(t) of mosquitos at time t is given by M(t) = + Hence (carrying out the integration numerically) M(3) = + To the nearest whole number, this is t R(τ) dτ. () τ τ cos dτ ().4 Part d The maximum number of mosquitos for the period t 3 will occur when t =, or when t = 3, or when R(t) =. The latter condition obtains when t = π/ and when t = π/. Integrating numerically when necessary in (), we find that M() = ; () ( ) π M ; (3) ( ) π M ; (4) M(3) () Thus, the mosquito population peaks at about 39 when t = π/. 3 Problem 3 3. Part a The Midpoint Rule with four subintervals of equal length gives 4 v(t) dt v() ( ) + v() ( ) + v() (3 ) + v(3) (4 3) (6) ( ) = 9 miles. (7) 3

4 The integral gives the distance, in miles, that the plane traveled during the time interval t Part b By Rolle s Theorem, acceleration which is v (t) must be zero at least once in the interval t because v() = v(). Similarly, v (t) must be zero at least once in the interval t 3, because v() = v(3). Thus, acceleration must vanish at least twice in the interval t Part c If the plane s velocity is given by then f(t) = 6 + cos t 7t + 3 sin 4, (8) f (t) = 7t cos 4 4 sin t gives acceleration. At t = 3, this gives acceleration as (9) f (3) = 6 cos sin miles/min ().4769 miles/min. () 3.4 Part d Average velocity over t 4 is 4 4 ( 6 + cos t ) 7t + 3 sin dt = [ 6t + sin t 4 4 ] 7t 4 cos 7 4 = [ 4 + sin 4 ] 4 7 cos 7 [ ] 4 7 () (3).967miles/min. (4) 4

5 4 Problem 4 4. Part a Inflection points are to be found where f changes sign that is, where the slope of f changes from positive to negative or vice versa. Consequently, the function f whose derivative is pictured has inflection points at x = and at x = Part b the function f is decreasing on the interval [, 4] and increasing on the interval [4, ] because f is non-positive, with only isolated zeros, on the first of these intervals and non-negative, with only an isolated zero on the second. The absolute maximum vale of f must fall at one of the points x = or x =. (There can be no absolute maximum for f at any point interior to (, ) because f does not change signs from positive to negative anywhere in that interval.) The (unsigned) area bounded by f and the x-axis on the interval [, 4] is clearly larger than the area between f and the x-axis on the interval [4, ], so whence 4 f (t) dt = f( ) f(4) > f() f(4) = 4 f (t) dt, () f( ) > f(), (6) so the absolute maximum value taken on in the interval [, ] is f( ). 4.3 Part c We are given that g(x) = xf(x), so g () = f() + f () = 6 + ( ) = 4. (7) Also g() = f() =. (8) An equation for the line tangent to the graph at x = is therefore y = + 4(x ), or (9) y = 4x + 4. (3)

6 Problem. Part a See Figure Figure : Problem, Part a. Part b If y = x 4 (y ), then slope can be negative only when the product on the right side of the equation is negative. This is so just when (y ) <, the points in the plane where slope is negative are the points (x, y) for which y <..3 Part c If y = f(x), with f() =, is a solution to the differential equation y = x 4 (y ), then f (x) = x 4 [f(x) ], or (3) f (x) f(x) = x4. (3) As a solution to a differential equation near x =, f must be a continuous function, at least on some interval centered at x =, so we can be sure that (f(x) ) < when x is 6

7 near. Choosing such an x, we integrate both sides of equation (3) from to x: x f (ξ) f(ξ) dξ = x ξ 4 dξ. (33) Making use of the negativity of the denominator on the left side, as well as the fact that f() =, we obtain x ln [ f(ξ)] = ξ x, or (34) From this, it follows that ln [ f(x)] = ln + x. (3) f(x) = e x /, whence (36) ( ) f(x) = e x / (37) 6 Problem 6 6. Part a If n >, then x n dx = xn+ n + = n+ n + n+ n + = n +. (38) 6. Part b If n > and y = x n, then y = nx n, so (39) y x= = n. (4) It follows that the equation of the line tangent to y = x n at (, ) is y = + n(x ). (4) This line crosses the x-axis at x = n, so that the base of the triangle T has length n. The altitude of T is one, so the area of T is n. 7

8 6.3 Part c From what we have seen in Parts a and b, above, the area, A(n) of the region S, as a function of n, is Thus, A(n) = n + n = n n + n. (4) A (n) = (n + n) (n )(4n + ) 4n (n + ) (43) = n n n (n + ). (44) When n >, we see that A (n) = only for n = +, by the Quadratic Formula. Noting that A (n) > for n < + but that A (n) < for + < n, we conclude, by the First Derivative Test, that the maximal area occurs when n = +. 8