AP Calculus AB AP Calculus BC

AP® Calculus AB AP® Calculus BC

Free-Response Questions and Solutions 1989 – 1997

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For the College Board’s online home for AP professionals, visit AP Central at apcentral.collegeboard.com. 1989 BC1

Let f be a function such that f ′′ (x) = 6x + 8.

(a) Find f (x) if the graph of f is tangent to the line 3xy−=2 at the point (0,−2) .

(b) Find the average value of f (x) on the closed interval [1− ,1].

(a) fx′()=+38x2 x+C f ′()03= C = 3 f ()xx=+3243x+x+d d =−2 fx()=+x3243x+x−2

1 1 (b) ∫ ()xx32++43x−2dx 11−−()−1

1 114 3 =+xx432+x−2x  24 3 2 −1 1143 143 =++−22−−++ 2432 432 =−2 3

x2 Let R be the region enclosed by the graph of y = , the line x = 1, and the x-axis. x 2 +1

(a) Find the area of R .

(b) Find the volume of the solid generated when R is rotated about the y-axis.

⌠ 1 x2 (a) Area =  dx 2 + ⌡0 x 1 ⌠ 1 1 =− 1 2 dx ⌡0 x +1 =−x arctan x 1 0 π =−1 4

1 ⌠ x2 (b) Volume = 2π  xdx 2 + ⌡0 x 1 ⌠ 1 x =−2π  xdx 2 + ⌡0 x 1 1 x2 1 =−2lπ nx2 +1 22 0 =−π ()1ln2

1/2 ⌠ y Volume =−π 1 dy  − ⌡ 0 1 y 1/2 =+π ()2lyyn−1 0 =−π ()1ln2

Consider the function f defined by f (x) = ex cos x with domain [0,2π ] .

(a) Find the absolute maximum and minimum values of f (x).

(b) Find the intervals on which f is increasing.

(a) fx′()=−exxsin x+ecos x =−exx []cos sin x ππ5 fx′()==0 when sin xcos x, x=, 44

() x fx 0 1 π 2 π e /4 4 2 π 5 2 5/π 4 − e 4 2 2π 2π e

ππ2 Max: ee25;Min: − /4 2

(b) f ′()x + − + π 5π 2π 0 4 4

ππ5  Increasing on 0, , , 2π  44 

Point of inflection at x = π

Consider the curve given by the parametric equations x =−2tt323 and y=t3−12t

dy (a) In terms of t , find . dx

(b) Write an equation for the line tangent to the curve at the point where t =−1.

(c) Find the x- and y-coordinates for each critical point on the curve and identify each point as having a vertical or horizontal tangent.

dy (a) =−31t 2 2 dt dx =−66tt2 dt 22−−()tt+−22() dy ===31t 2t 4 dx 66t 22−−t 2t 2t 2t ()t −1

(b) xy=−5, =11 dy 3 =− dx 4 3 yx−=11 −()+5 4 or 329 yx=− + 44 43yx+=29

−−2 ()28,16 horizontal

00(),0 vertical 11()−−,11vertical 2 ()4,−16 horizontal

At any time t ≥ 0 , the velocity of a particle traveling along the x-axis is given by the dx differential equation −10x = 60e4 t . dt

(a) Find the general solution x(t) for the position of the particle.

(b) If the position of the particle at time t = 0 is x =−8, find the particular solution x(t) for the position of the particle.

−∫ 10dt (a) Integrating Factor: ee= −10t

d −− ()xe 10tt= 60e4 e 10t dt xee−−10tt=−10 6 +C x()te=−10 41tt+Ce0

()= 10t xth Ce = 4t xAp e 41Ae44tt−=0Ae 60e4t A =−10 xt()=−Ce10tt10e4

(b) −=81CC−0;=2 x()te=−2110tt0e4

dx (c) =20ee10tt−40 4 dt 20ee10tt−=40 4 0 1 t = ln 2 6

dx −−10()10ee41tt+2 0=60e4t dt 0 +−100ee41tt20 0=60e4t 1 t = ln 2 6

Let f be a function that is everywhere differentiable and that has the following properties.

f (x) + f (h) (i) f (x + h) = for all real numbers h and x . f (−x) + f (−h) (ii) f (x) > 0 for all real numbers x .

(iii) f ′ (0) =−1.

(a) Find the value of f (0) .

1 (b) Show that f (−x) = for all real numbers x . f (x)

(d) Use the definition of the derivative to find f ′()xf in terms of (x).

(a) Let xh==0 ff()00+ () ff()00=+( 0)= =1 ff()00+ ()

(b) Let h = 0 fx()+ f(0) fx()+=0 fx()= fx()−+f(−0) 1 Use ff()0 ==1 and solve for (x) f ()−x

f ()−+xf(0) Note that fx(0−+)= is the reciprocal of f(x). fx()+ f(0)

fx()+ f()h (c) fx()+=h 11+ fx() f()h fx()+ f()h = f ()xf()h fh()+ f()x = fx()f()h

f ()xh+−f(x) (d) fx′()= lim h→0 h f ()xf()h− f()x = lim h→0 h fh()−1 = fx()lim h→0 h ==f ()xf′(0) −f()x

A particle starts at time t = 0 and moves along the x-axis so that its position at any time t ≥ 0 is given by x(t) = (t −1) 3 (2t − 3).

(a) Find the velocity of the particle at any time t ≥ 0.

(b) For what values of t is the velocity of the particle less than zero?

1990 BC1 Solution

(a) vt()= x′()t =−31()tt23(2−3)+2()t−1 =−()tt182 (−11)

(b) vt()<−0 when (t 1)2 (8t−11)<0 Therefore 8tt−<11 0 and ≠1 11 or tt<≠ and 1 8 11 Since tt≥≤0, answer is 0 <, except t=1 8

1990 BC2

− Let R be the region in the xy-plane between the graphs of y = ex and y = e x from xx==0 to 2.

(a) Find the volume of the solid generated when R is revolved about the x-axis.

(b) Find the volume of the solid generated when R is revolved about the y-axis. 1990 BC2 Solution

2 (a) Ve=−π ∫ ()22xxe− dx 0 2 1122xx− =+π ee 220

11− 11 =+π ee44−+ 22 22

π − =+ee44−2 2

2 (b) Vx=−2π ∫ exxe− dx 0  2 =+π xx−−−xx+ 2 xe()e ∫ e e dx 0 2 =+2π xe()xxe−−−(ex−ex) 0 22−−22 =+22π ()ee−(e−e)−0−()1−1 − =+23π ee22

1990 BC3

Let f (x) = 12 − x 2 for x ≥ 0 and f (x) ≥ 0.

(a) The line tangent to the graph of f at the point (,kf(k)) intercepts the x-axis at x = 4. What is the value of k ?

(b) An isosceles triangle whose base is the interval from (0,0) to (,c 0) has its vertex on the graph of f . For what value of c does the triangle have maximum area? Justify your answer. 1990 BC3 Solution

(a) fx()=−12 x2 ; f′()x=−2x

slope of tangent line at ()kf,2()k =− k

line through ()4,0 &()kf, (k) has slope

fk()− 0 12 − k 2 = −− kk44 12 − k 2 so −=2kk⇒2 −8k+12 =0 k − 4

kk==2 or 6 but f()6 =−24 so 6 is not in the domain. k = 2

11cc2 (b) Ac=⋅f=c12 − 22 24 3 c  =−6c on 0,4 3 8 dA 33c22c =−6;6− =0 when c=4. dc 88

Candidate test First derivative

cA A′ + − 00 0 4 43 416 Max

43 0

second derivative 2 dA =− < = 2 3 0 so c 4 gives a relative max. dc c=4 c = 4 is the only critical value in the domain interval, therefore maximum 1990 BC4

Let R be the region inside the graph of the polar curve r = 2 and outside the graph of the polar curve r = 2(1 − sinθ).

(a) Sketch the two polar curves in the xy-plane provided below and shade the region R .

5 4

3 2

1 x −5 −4 −3 −2 −1 1 2 3 4 5 −1 −2

−3 −4

−5

(b) Find the area of R . 1990 BC4 Solution y y (a)

R xx −5 5 −5 O 5

− −5 5

π 1 2 =−⌠ 2 ()()−θθ (b) Ad⌡ 221sin 2 0 π =−22∫ ()sinθθsin2 dθ 0 ππ =−4s∫∫inθθdd()1−cos2θ θ 00 π π 1 =−4cosθθ−− sin2θ 0  2 0 =−41()− +4(1)−[]π −0 =−8 π 1990 BC5

1 Let f be the function defined by f (x) = . x −1

(a) Write the first four terms and the general term of the Taylor series expansion of f (x) about x = 2.

(b) Use the result from part (a) to find the first four terms and the general term of the series expansion about x = 2 for ln x −1 . 3 (c) Use the series in part (b) to compute a number that differs from ln by less than 2 0.05. Justify your answer. 1990 BC5 Solution

(a) Taylor approach Geometric Approach

f ()21= 11= − f ′()22=−(−1)2 =−1 xx−+11()−2 23 n n − f ′′ ()2 =−11uu+ −u+""+()− u+ f ′′ ()22=−(21)3 =2; =1 2! where ux=−2 − f ′′′ ()2 f ′′′ ()26=− (2−1)4 =−6; =−1 3! 1 23 nn Therefore =−1 ()xx−2 +()−2 −()x−2 +""+(−1)()x−2 + x −1

(b) Antidifferentiates series in (a): nn+1 112314()−−12(x ) ln xC−1 =+x−()x−+2 ()x−2 −()x−+2 ""+ + 234 n +1 0l=−n21⇒C =−2

1 4 Note: If C ≠ 0 , “first 4 terms” need not include −()x −2 4

35 1112311 (c) ln =−ln 1 =−+−" 22 22232 11 1 =−+ −" 2824 1111 since <−, =0.375 is sufficient. 24 20 2 8

Justification: Since series is alternating, with terms convergent to 0 and decreasing in absolute value, the truncation error is less than the first omitted term. n+1 =<111 <5 RCn +  , where 2 Alternate Justification: ()C −1 n 1 n +12 2

< 11 n +12n+1 1 <≥ when n 2 20 1990 BC6

Let f and g be continuous functions with the following properties. (i) gx( ) =−A f(x) where A is a constant 23 (ii) ∫∫f ()xdx= g()xdx 12 3 (iii) ∫ f ()xdx=−3A 2

3 (a) Find ∫ f ()xdx in terms of A . 1

(b) Find the average value of g(x) in terms of A , over the interval [1, 3] .

1 (c) Find the value of k if ∫ f (1xd+)x=kA. 0

1990 BC6 Solution

323 (a) ∫∫fx()dx=+fx()dx ∫fx()dx 112 33 =+∫∫g ()x dx f ()x dx 22 33 =−∫∫()Af()xdx+f()xdx 22 33 =−A ∫∫f()xdx+ f()xdx 22 = A

1133 (b) Average value ==∫∫gx()dx ()A−f()x dx 2211 =−1 3 2Af∫ ()xdx 2 1 11 =−[]2AA=A 22

12 (c) kA=+∫∫fx()1 dx=fx()dx 01 3 = ∫ gx()dx 2 =+AA3 =4A Therefore k = 4

1990 BC5

1 Let f be the function defined by f (x) = . x −1

(a) Write the first four terms and the general term of the Taylor series expansion of f (x) about x = 2.

(a) Taylor approach Geometric Approach

f ()21= 11= − f ′()22=−(−1)2 =−1 xx−+11()−2 23 n n − f ′′ ()2 =−11uu+ −u+""+()− u+ f ′′ ()22=−(21)3 =2; =1 2! where ux=−2 − f ′′′ ()2 f ′′′ ()26=− (2−1)4 =−6; =−1 3!

1 23 nn Therefore =−1 ()xx−2 +()−2 −()x−2 +""+(−1)()x−2 + x −1

(b) Antidifferentiates series in (a):

nn+1 112314()−−12(x ) ln xC−1 =+x−()x−+2 ()x−2 −()x−+2 ""+ + 234 n +1 0l=−n21⇒C =−2

1 4 Note: If C ≠ 0 , “first 4 terms” need not include −()x −2 4

35 1112311 (c) ln =−ln 1 =−+−" 22 22232 11 1 =−+ −" 2824 1111 since <−, =0.375 is sufficient. 24 20 2 8

Justification: Since series is alternating, with terms convergent to 0 and decreasing in absolute value, the truncation error is less than the first omitted term. + 111n 1 5 RC=<, where 2 < Alternate Justification: n n+1  ()C −1 n +12 2

< 11 n +12n+1

Let f and g be continuous functions with the following properties.

(i) gx( ) =−A f(x) where A is a constant

23 (ii) ∫∫f ()xdx= g()xdx 12

3 (iii) ∫ f ()xdx=−3A 2

3 (a) Find ∫ f ()xdx in terms of A . 1

(b) Find the average value of g(x) in terms of A , over the interval [1, 3] .

1 (c) Find the value of k if ∫ f (1xd+)x=kA. 0

323 (a) ∫∫fx()dx=+fx()dx ∫fx()dx 112 33 =+∫∫g ()x dx f ()x dx 22 33 =−∫∫()Af()xdx+f()xdx 22 33 =−A ∫∫f()xdx+ f()xdx 22 = A

1133 (b) Average value ==∫∫gx()dx ()A−f()x dx 2211 =−1 3 2Af∫ ()xdx 2 1 11 =−[]2AA=A 22

12 (c) kA=+∫∫fx()1 dx=fx()dx 01 3 = ∫ gx()dx 2 =+AA3 =4A Therefore k = 4

A particle moves on the x-axis so that its velocity at any time t ≥ 0 is given by v(t) = 12t2 − 36t +15. At t =1, the particle is at the origin.

(a) Find the position x(t) of the particle at any time t ≥ 0.

(b) Find all values of t for which the particle is at rest.

(d) Find the total distance traveled by the particle from t = 0 to t = 2.

1991 BC1 Solution

(a) x()tt=−41328t+15t+C 0==x()1 41−81+5+C Therefore C =−1 xt()=−41t328t +15t−1

(b) 01==v()t2t2 −36t+15 32()tt−−1(2 5)=0 15 t = , 22

dv (c) =−24t 36 dt dv 3 ==0 when t dt 2

v()01= 5

3 v=−12 2

v()29=−

Maximum velocity is 15

1/2 2 (d) Total distance = ∫∫vt()dt− vt()dt 01/2 11 = xx−−()02x()−x 22 55 = −−()11−−1− =17 22

1991 BC2

− Let f be the function defined by f (x) = xe1 x for all real numbers x .

(a) Find each interval on which f is increasing.

(b) Find the range of f .

(d) Using the results found in parts (a), (b), and (c), sketch the graph of f in the xy- plane provided below. (Indicate all intercepts.)

x −3 −2 −1 1 2 3

−1

−2

−3

1991 BC2 Solution

(a) f ′()xx=−e11−−xx(11)+e=(−x)e1−x f increases on (−∞,1]

(b) ff()11==;lim(x) −∞ x→−∞ Range: (−∞,1]

(d) y

x −1 1 2 3

−1

−2

1991 BC3

Let R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of y = sin x and y = cos x , as shown in the figure above.

(a) Find the area of R .

(b) Find the volume of the solid generated when R is revolved about the x-axis.

(c) Find the volume of the solid whose base is R and whose cross sections cut by planes perpendicular to the x-axis are squares.

1991 BC3 Solution

π /4 (a) Area =∫ cos x −sin xdx 0 π /4 =+()sin xxcos 0 22 =+−()01+ 22 =−21

π /4 (b) Vx=−π ∫ cos22sin xdx 0 π /4 = π ∫ cos 2xdx 0 π π /4 = sin 2x 2 0 ππ =−()10= 22

π /4 (c) Vx=−∫ ()cos sin x2 dx 0 π /4 =−∫ 12sincx osxdx 0 π /4 =− ()xxsin2 0

π 1 =− −()00− 42 π =− 1 42

1991 BC4

2x Let F(x) = t2 + t dt . 1

(a) Find F ′ (x) .

(b) Find the domain of F .

(d) Find the length of the curve y = F(x) for 1 ≤ x ≤ 2 .

1991 BC4 Solution

(a) Fx′()=+24x2 2x

(b) tt2 +≥0; therefore ()t≥0 or (t≤−1) Since 1≥≥0, want 2x 0 Therefore x ≥ 0

1 (c) lim Fx()==F0 x→ 1 2 2

2 2 (d) LF=+⌠ 1 ()′()xdx ⌡1 2 =+∫ 116x2 +8xdx 1 2 =+∫ 41xdx 1 2 =+27xx2 = 1

1991 BC5

4 Let f be the function given by f (t) = and G be the function given by 1 + t2 x G(x) = f (t) dt . 0

(a) Find the first four nonzero terms and the general term for the power series expansion of f (t) about t = 0.

(b) Find the first four nonzero terms and the general term for the power series expansion of G(x) about x = 0.

(c) Find the interval of convergence of the power series in part (b). (Your solution must include an analysis that justifies your answer.)

1991 BC5 Solution

4 (a) f ()ta==, geometric with 4, r=−t2 1+ t 2 ft()=−44t24+4t−4t6+""+(−1)n 4t2n +

x ⌠ 4 x (b) Gx()== dt ∫ ()44−t24+4t−4t6+" dt 2 0 ⌡ 0 1+ t x ()−14n t 21n+ =−44435+−7++ + 4tt t t"" 357 2n +1 0 n + 444 ()−14x21n =−4xxxx35+ −7+""+ + 357 2n +1

n+1 + ()−14x23n 21nn++21 ⋅=x2 23nn++()−14n x21n+ 23 21n + lim xx22=<; x21 for −1

1991 BC6

dy A certain rumor spreads through a community at the rate = 2y(1 − y) , where y is the dt proportion of the population that has heard the rumor at time t .

(a) What proportion of the population has heard the rumor when it is spreading the fastest?

(b) If at time t = 0 ten percent of the people have heard the rumor, find y as a function of t .

1991 BC6 Solution

(a) 21yy()−=2y−2y2 is largest when 2−4y=0 1 so proportion is y = 2

1 (b) dy = 2dt yy()1−

⌠ 1  dy = ∫ 2dt ⌡ yy()1−

⌠ 11  +=dy ∫ 2dt ⌡ − yy1 ln yy−−ln ()1 =2t+C

y ln =+2tC 1− y y = ke2t 1− y

1 yk()00=⇒.1 = 9 e2t y = 9 + e2t

1 1 (c) 2 = e2t 1 1− 9 2 1 1 = e2t 9 1 t ==ln 9 ln 3 2

1992 AB4/BC1

Consider the curve defined by the equation y + cosy = x +1 for 0 ≤ y ≤ 2π .

dy (a) Find in terms of y . dx

(b) Write an equation for each vertical tangent to the curve.

d 2y (c) Find in terms of y . dx2

1992 AB4/BC1 Solution

dy dy (a) −=sin y 1 dx dx dy ()1s−=iny 1 dx

dy = 1 − dx 1siny

dy (b) undefined when sin y =1 dx π y = 2 π +=01x + 2 π x =−1 2

1 d  2 − (c) dy= 1siny 2 dx dx  −− dy cos y = dx 2 ()1s− iny

1 cos y  1s− iny = ()− 2 1siny = cos y ()1s− iny 3

1992 AB5/BC2

− Let f be the function given by f (x) = e x , and let g be the function given by g(x) = kx , where k is the nonzero constant such that the graph of f is tangent to the graph of g .

(a) Find the x-coordinate of the point of tangency and the value of k .

(b) Let R be the region enclosed by the y-axis and the graphs of f and g . Using the results found in part (a), determine the area of R .

(c) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated by revolving the region R , given in part (b), about the x-axis.

1992 AB5/BC2 Solution

y (a) − f ′′()xe=− x ; g()x=k

−=ek−x −x ek= x

x =−1 and ke=−

f ()xe= −x

gx()= kx

00 (b) ∫∫()ee−−xx−−()xdx= (e+ex)dx −−11 0 2 − ex =−e x + 2 −1 e =−()10+ −−+e 2 e =−1 2

0 ⌠ −x 2 2 (c) π  (()ee−−()x)dx ⌡ −1

1992 BC3

At time t , 0 ≤ t ≤ 2π , the position of a particle moving along a path in the xy-plane is given by the parametric equations x = et sint and y = et cost .

π (a) Find the slope of the path of the particle at time t = . 2 (b) Find the speed of the particle when t =1.

1992 BC3 Solution

dx (a) =+etttsin ecost dt dy =−etttcos esin t dt t ()− dy ==dy / dt etcos sin t dx dx /sdt et ()int + cost π π dy e /2 ()01− at t ==, =−1 21dx eπ /2 ()+ 0

22 (b) speed =+()ettsin tetcos +(ettcos −etsin t) when t =1 speed is ()eesin1++cos1 22(ecos1−esin1)=e2

1992 BC4

2x − x2 for x ≤1, Let f be a function defined by f (x) = x2 + kx + p for x >1.

(a) For what values of k and p will f be continuous and differentiable at x =1 ?

(b) For the values of k and p found in part (a), on what interval or intervals is f increasing?

(c) Using the values of k and p found in part (a), find all points of inflection of the graph of f . Support your conclusion.

1992 BC4 Solution

(a) For continuity at x = 1, ()−=22()=()++ lim−+2x xf1 lim xkxp xx→→11 Therefore 1 =+1 kp+ Since f is continuous at x = 1 and is piecewise polynomial, left and right derivatives exist. ′′ f−+()1 ==0 and fk()1 2 +

For differentiability at x = 1, 0 =+2 k . Therefore kp=−2, =2

(b) f ′()xx=−22,x≤1 22−>x 0 x <1 fx′()=−22x ,x>1 22x −>0 x >1 Since f increases on each of (−∞,1) and (1, ∞) and is continuous at x = 1, f is increasing on ()−∞,∞ .

(c) fx′′ ()=−2, x<1 fx′′ ()=>2, x1 Since fx′′ ()<−0 on (∞,1) and fx′′ ()>∞0 on (1, ) and f ()1 is defined, ()1, f ()1 = ()1,1 is a point of inflection.

1992 BC5

The length of a solid cylindrical cord of elastic material is 32 inches. A circular cross 1 section of the cord has radius inch. 2

(a) What is the volume, in cubic inches, of the cord?

(b) The cord is stretched lengthwise at a constant rate of 18 inches per minute. Assuming that the cord maintains a cylindrical shape and a constant volume, at what rate is the radius of the cord changing one minute after the stretching begins? Indicate units of measure.

(c) A force of 2x pounds is required to stretch the cord x inches beyond its natural length of 32 inches. How much work is done during the first minute of stretching described in part (b)? Indicate units of measure.

1992 BC5 Solution

1 2 (a) Vr==ππ2h⋅32 =8π 2

dV dr dh (b) 02==ππrh +r 2 ; dt dt dt at th==1, 50 and so 8ππ=⋅r 2 50,

2 so r = 5 22dr 2 Therefore 0 =+2ππ()50 ()18 55dt 

dr 72 =+π 40 dt 25 dr 9 =− in/min dt 125

or 8 Vr==8,ππ2h so r= h

−1 dr 182 −8dh  Therefore =⋅  ⋅  dt 2 h h2 dt  at th==1, 50 so

−1 dr 182 −8 =⋅ ()18 dt 2 50 2500  9 =− in/min 125

18 18 (c) Work ==∫ 2xdx x22=18 0 0 = 324 in-pounds = 27 foot-pounds

1992 BC6

∞ 1 ≥ Consider the series p , where p 0. n= 2 n ln(n)

(a) Show that the series converges for p > 1.

(b) Determine whether the series converges or diverges for p = 1. Show your analysis.

1992 BC6 Solution

11 (a) 0 << for ln ()nn>1, for ≥3 nnppln () n 1 by p-series test, ∑ converges if p > 1 n p ∞ 1 and by direct comparison, ∑ p converges. n=2 nnln ()

1 ∞ (b) Let fx()= , so series is ∑ f()n xxln n=2 ∞ ⌠ 1 b  dx ==lim ln ln x lim[ln(ln(b)) −ln(ln 2)] =∞ bb→∞ →∞ ⌡ 2 xxln 2 Since f (x) monotonically decreases to 0, the integral test shows ∞ 1 ∑ diverges. n=2 nnln

11 (c) >>0 for p <1, nnp ln nln n

∞ 1 ≤< so by direct comparison, ∑ p diverges for 01p n=2 nnln

1993 AB3/BC1

Consider the curve y 2 = 4 + x and chord AB joining the points AB( −4,0) and (0,2) on the curve.

(a) Find the x- and y-coordinates of the point on the curve where the tangent line is parallel to chord AB .

(b) Find the area of the region R enclosed by the curve and the chord AB .

1993 AB3/BC1 Solution

21 (a) slope of AB == 42 B0(),2

A ()−4, 0

dy 111 Method 1: yx=+4; = ; = dx 24++xx24 2 so xy=−3, =1

dy 1 Method 2: 2yy==1; so 2 1 and y=1, x=−3 dx 2

(b)

0 0 ⌠ 123/2 1 Method 1:  42+−x xd+ x=()4+x−x2 −2x  ⌡ −4 234−4

213/2 64 =−()44(−+8)=−4= 333

3 2 2 y Method 2: ∫ ()24yy−−()22−4dy=y− 0  3 0 84 =−4 = 33

0 16 Method 3: ∫ 4 +=xdx ; Area of triangle =4 −4 3 16 4 Area of region =−4 = 33

1993 AB3/BC1 Solution, continued

(c)

0 2 ⌠ 2 1 Method 1: π  ()42+−x xd+ x  2 ⌡ −4  0 ⌠ 1 =+π  42x −xx2 ++4dx  ⌡ −4 4 16 8π =−π 8 =≈8.378 33

2 π 2 8 Method 2: ∫ 22π yy()−−4()y−4dy= 0 3

0 0 2 2 ππ⌠ +=+x Method 3: ⌡ ()44xdx x −4 2 −4 =−01ππ()−6+8=8

112 6π Volume of cone ==π ()2 ()4 33 16ππ8 Volume =−8π = 33

1993 AB4/BC3

Let f be the function defined by f (x) = ln(2 + sin x) for π ≤ x ≤ 2π.

(a) Find the absolute maximum value and the absolute minimum value of f . Show the analysis that leads to your conclusion.

(b) Find the x-coordinate of each inflection point on the graph of f . Justify your answer.

1993 AB4/BC3 Solution

1 (a) fx′()= cos x; 2s+ inx 3π In []ππ, 2 , cos xx==0 when ; 2

x f ()x

π ln ()2 = 0.693 2π ln ()2 3π ()= ln 1 0 2

absolute maximum value is ln 2

absolute minimum value is 0

()−+sin x (2 sin xx)−cos cos x (b) fx′′ ()= ()2s+ inx 2 −−2sin x 1 = ; ()2s+ inx 2 1 fx′′ ()==0 when sin x− 2 71ππ1 x = , 66

sign of f ′′ −−+

concavity down up down π π π π 7 11 2 6 6

7π11π x = and since concavity changes as indicated at these points 66

1993 BC2

2 The position of a particle at any time t ≥ 0 is given by x(t) = t2 − 3 and y(t) = t3 . 3

(a) Find the magnitude of the velocity vector at t = 5.

(b) Find the total distance traveled by the particle from t = 0 to t = 5 .

dy (c) Find as a function of x . dx

1993 BC 2 Solution

(a) x′′()tt==22y()tt2 xy′′()51==0 ()550 v()5 =+102250 =2600 =≈10 26 50.990

5 (b) ∫ 44tt24+ dt 0 5 =+∫ 21tt2 dt 0 5 2 3/2 =+()1 t 2 3 0 2 =−()263/2 1 ≈87.716 3

dy yt′() 2t 2 (c) ===t ′ dx x ()t 2t xt=−223; t=x+3

tx=+3 dy =+x 3 dx

xt=−2 3; t=x+3

223/2 yt==3;3y()x+ 33 dy =+x 3 dx

1993 BC4

Consider the polar curve r = 2sin(3θ) for 0 ≤ θ ≤ π .

(a) In the xy-plane provided below, sketch the curve.

x −2 −1 1 2 O

−1

−2

(b) Find the area of the region inside the curve.

π (c) Find the slope of the curve at the point where θ = . 4

1993 BC4 Solution y

(a) 2

x −2 −1 O 1 2

−1

−2

π ππ 112  (b) Ad==∫∫4sin 3θθ ()1−cos6θdθ=θ−sin6θ =π 00  260 3 π /3 or ∫ 4sin2 3θθd ==" π 2 0 6 π /6 or ∫ 4sin2 3θθd ==" π 2 0

dx =−2sin 3θθsin +6cos3θcosθ dθ dy =+2sin 3θθcos 6cos3θsinθ dθ π dy dx At θ ==, −2 and =−4, so 4 ddθθ − dy ==21 − dx 42 or 2 ()xy22+=62x2y−y3 dy dy dy 22()xy22++x2y =6x2+12xy−6y2 dx dx dx π At θ ==, xy1 and =1 so 4 dy dy dy 42+=2 6 +12−6 dx dx dx Copyright © 2003 by College Entrance Examination Board. All rights reserved. dy dy 1 Available at apcentral.c88+=ollege12boa⇒rd.com= dx dx 2 1993 BC5

x Let f be the function given by f (x) = e 2 .

(a) Write the first four nonzero terms and the general term for the Taylor series expansion of f (x) about x = 0.

(b) Use the result from part (a) to write the first three nonzero terms and the general term x e 2 −1 of the series expansion about x = 0 for g(x) = . x ∞ ′ n = 1 (c) For the function g in part (b), find g (2) and use it to show that + . n=1 4(n 1)! 4

1993 BC 5 Solution

xx23 xn (a) exx =+1 + + +""+ + 2! 3! n! 23 n x ()xx/2 ()/2 ()x/2 ex /2 =+1 + + +""+ + 22! 3! n! xx23x xn =+1 + + +""+ + 22232!23! 2n n!

23 n xx++x++x+ (b) x /2 "" e −1 22232!23! 2n n! = xx

1 xx21xn− =+ + +""+ + 22232!23! 2n n!

− ()nx−1 n 2 (c) ′()=+12x ++ + gx 23""n 22! 23! 2n! − 1 x ()nx−1 n 2 =+ +""+ + 824 2n n! − 12⋅2 ()n −12n 2 g′()2 =++""+ + 2223! 23! 2n n! 11 n −1 =+ +""+ + 812 4n! ∞ = ∑ n ()+ n=1 41n !

ex /2 −1 Also, gx()= x 1 xe⋅−xx/2 ()11()e/2 − gx′()= 2 x2 1 21⋅−ee()− 1 g′()2 ==2 44

1993 BC6

Let f be a function that is differentiable throughout its domain and that has the following properties. f (x) + f (y) (i) f (x + y) = for all real numbers x, y, and x+ y in the domain of f 1− f (x) f (y) (ii) lim f (h) = 0 h→0 f (h) (iii) lim = 1 h→0 h

(a) Show that f (0) = 0 .

(b) Use the definition of the derivative to show that f ′()x=+1 [f(x)]2 . Indicate clearly where properties (i), (ii), and (iii) are used.

1993 BC6 Solution

(a) Method 1: f is continuous at 0, so ff()0l==im(x) 0 x→0

ff()00+ () Method 2: ff()00=+( 0)= 10− ff() ()0

2 ff()01( −=()0 ) 2f()0

2 ff()01(−−()0) =0 f ()00=

fx()+−h f(x) (b) fx′()= lim → h 0 h fx()+ f()h − fx() 1− fx()f()h = lim [By (i)] h→0 h

2 fh() 1+fx() =⋅lim  h→0 hf1− ()xf()h  2 1+ fx() =⋅1 [By (iii) & (ii)] −⋅() 10fx 2 =+1 fx()

dy (c) Method 1: Let yf==()x; 1+y2 dx

dy = dx 1+ y2 tan−1 yx=+C

=+() yxtan C fC()00=⇒ =0o[]r C=nπ ,n∈Z

fx()==tan xor f()x tan (x+nπ )

1993 BC6 Solution, continued

Method 2: Guess that fx()= tan x

+=()2 +22==′() 11f xxtansecxfx f ()0t==an()00

Since the solution to the D.E. is unique f ()x= tan x is the solution.

1994 AB 2-BC 1

Let R be the region enclosed by the graphs of ye=x , y=x, and the lines x = 0 and x = 4.

(a) Find the area of R .

(b) Find the volume of the solid generated when R is revolved about the x-axis.

(c) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the y-axis.

1994 AB 2- BC 1

4 (a) Area =−∫ exx dx 0 4 x2 =−ex 2 0 16 =−ee40−()−0 2 R =−e4 9 ≈ 45.598 x 4 OR Using geometry (area of triangle)

4 1 ∫ edx x−⋅4⋅4 0 2

or Using y-axis

14 e4 ∫∫ydy+−y ln ydy+∫4 −ln ydy 01 4

4 2 (b) Ve=−π ∫ ()x ()x2 dx x 0 4 = π ∫ ex22x − dx 0 4 1 x3 = π e2x − 23 0 1641 = π ee80− − − 0 232  1 131 = π e8 − 26 ≈1468.646π ≈ 4613.886

1994 AB 2- BC 1 (continued)

[or] Using geometry (Volume of the cone)

4 2 1 ππ∫ ()edx x−⋅442 ⋅ 0 3 4 1 π =−π e2x ⋅64  230 1164π =−π e8 − 223

Using y- axis

14 e4 2lπ ∫∫yy⋅+dy y()y−nydy+∫y(4−lny)dy 01 4

4 (c) Vx= 2π ∫ ()ex − xdx y 0

14 e4 V =+π ∫∫y22dy y −()ln y 22dy +∫16 −()ln y dy y 01 4

1994 AB 5-BC 2

A circle is inscribed in a square as shown in the figure above. The circumference of the circle is increasing at a constant rate of 6 inches per second. As the circle expands, the square expands to maintain the condition of tangency. (Note: A circle with radius r has circumference C= 2π r and area Ar= π 2 )

(a) Find the rate at which the perimeter of the square is increasing. Indicate units of measure.

(b) At the instant when the area of the circle is 25π square inches, find the rate of increase in the area enclosed between the circle and the square. Indicate units of measure.

1994 AB 5-BC 2

(a) PR= 8 dP dR = 8 dt dt R dC dR 62==π dt dt dR 32dP 4 ==; inches/second dt ππdt ≈ 7.639 inches/second

(b) Area =−4RR22π d ()Area dR dR =−82RRπ dt dt dt dR =−()42π R dt

Area of circle ==25ππR2 R = 5 d ()Area 120 =−30 inches2/second dt π 30 =−()4 π inches2/second π

≈ 8.197 inches2/second

1994 BC 3

A particle moves along the graph of y = cos x so that the x-coordinate of acceleration is always 2. At time t = 0, the particle is at the point (π ,−1) and the velocity vector of the particle is ()0,0 .

(a) Find the x- and y-coordinates of the position of the particle in terms of t.

(b) Find the speed of the particle when its position is ()4,cos4 .

1994 BC 3

(a) xt′′ ()=⇒22x′ ()t=t+C x′′()00=⇒Cx=0;(t)=2t x()tt=+2 k,0x( )=π =k xt()=+t2 π yt()=+cos()t2 π

(b) dy =−2sttin()2 +π dt

dx 22dy  st()=+ dt dt 

2 =+()22tt2 ()−sin()t2 +π

=+44tt22sin2()t2+π when xt=+4, 22ππ=4; t=4 − s =−44()ππ+44()−sin2 4 ≈ 2.324

1994 BC 4

Let f (x) = 6 − x 2 . For 0 < w < 6 , let A(w) be the area of the triangle formed by the coordinate axes and the line tangent to the graph of f at the point (w,6− w 2 ).

(a) Find A(1) .

(b) For what value of w is A(w) a minimum?

1994 BC 4

(a) fx()=−6;x2 f′()x=−2x f ′()12=− yx−=52−(−1)or y=−2x+7 7 xyint: int:7 2 17 49 A()17==() 22 4

(b) f ′()ww=−2;y−()6−w2 =−2w()x−w 6 + w2 xy int: int: 6 + w2 2w

2 ()6 + w2 Aw()= 4w 2 42ww()()6+−22()2w4()6+w Aw′()= 16w2 Aw′()=+0 when ()6 w22(3w−6)=0 w = 2

′ − + A w

0 2 6

1994 BC 5

− 2 Let f be the function given by f (x) = e 2x .

(a) Find the first four nonzero terms and the general term of the power series for f (x) about x = 0.

(b) Find the interval of convergence of the power series for f (x) about x = 0. Show the analysis that leads to your conclusion.

(c) Let g be the function given by the sum of the first four nonzero terms of the power series for f (x) about x = 0. Show that f (x) − g(x) < 0.02 for − 0.6 ≤ x ≤ 0.6 .

1994 BC 5

uu23 un (a) euu =+1 + + +""+ + 23! n! 46 n nn2 − 2 48xx ()−12x ex22x =−12 + − +""+ + 23! n!

(b) The series for eu converges for −∞ < u < ∞

2 So the series for e−2x converges for −∞ < −2x2 < ∞

And, thus, for −∞ < x < ∞

n+1 + ()+ a ()−12n 1 x21n n! lim n+1 =⋅lim nn→∞ an→∞ ()+1! ()− n nn2 n 12x 2 ==lim x2 0 <1 n→∞ n +1

2 So the series for e−2x converges for −∞ < x < ∞

16xx8132 6 (c) fx()−=g()x − +" 4! 5!

This is an alternative series for each x, since the powers of x are even.

a +1 2 Also, n =

8 16x8 16()0.6 Thus fx()−≤g()x ≤ 4! 4! =<0.011" 0.02

1994 BC 6

Let f and g be functions that are differentiable for all real numbers x and that have the following properties.

(i) f ′ (x) = f (x) − g(x)

(ii) g ′ (x) = g(x) − f (x)

(iii) f (0) = 5

(iv) g(0) =1

(a) Prove that f (x) + g(x) = 6 for all x .

(b) Find f (x) and g(x). Show your work.

1994 BC 6

(a) f′′()x+=g()x f()x−gx()+gx()−f()x=0 so fg+ is constant. fg()00+=() 6, so f(x)+g(x)=6

(b) fx()=−6g()x so g′()x=−gx() 62+gx()=gx()−6 dy dy =−26yd; =x dx 26y − 1 ln 2yx−=6 +C 2

ln 2yx−=6 2 +K 26ye−=2xK+

2x 26yA−=e xy=⇒01= so −4=A

24ye=− 2x +6 ye=−322x =g()x

2x fx()=−63g()x=+2e

1995 AB4/BC2

Note: Figure not drawn to scale.

2 The shaded regions RR12 and shown above are enclosed by the graphs of fx() x and gx() 2x .

(a) Find the x- and y-coordinates of the three points of intersection of the graphs of f and g .

(b) Without using absolute value, set up an expression involving one or more integrals that gives the total area enclosed by the graphs of f and g . Do not evaluate.

that gives the volume of the solid generated by revolving the region R1 about the liney 5 . Do not evaluate.

(a) 2,4 4,16 0.767,0.588 or (–0.766, 0.588)

24 (b) 22xxxdx22 x dx ³³0.767 2

0.588 4 16 2 y dy ylnyy dy ln y dy ³³0 0.588 ln 2 ³ 4 ln 2

2 22 (c) S 552xdx2 x ³0.767

As shown in the figure above, water is draining from a conical tank with height 12 feet and diameter 8 feet into a cylindrical tank that has a base with area 400S square feet. The depth h , in feet, of the water in the conical tank is changing at the rate of (h 12) 1 feet per minute. (The volume V of a cone with radius r and height h is Vrh S 2 .) 3

(a) Write an expression for the volume of water in the conical tank as a function of h .

(b) At what rate is the volume of water in the conical tank changing when h 3? Indicate units of measure.

(c) Let y be the depth, in feet, of the water in the cylindrical tank. At what rate is y changing when h 3? Indicate units of measure.

dVS h2 dh (b) dt9 dt S h2 h 12 9S 9 V is decreasing at 9S ft/min3

Two particles move in the xy-plane. For time t t 0 , the position of particle A is given by 3t xt 2 and yt (2)2 , and the position of particle B is given by x 4 and 2 3t y 2 . 2

(a) Find the velocity vector for each particle at time t 3 .

(b) Set up an integral expression that gives the distance traveled by particle A from t = 0 to t = 3. Do not evaluate.

(c) Determine the exact time at which the particles collide; that is, when the particles are at the same point at the same time. Justify your answer.

(d) In the viewing window provided below, sketch the paths of particles A and B from t 0 until they collide. Indicate the direction of each particle along its path.

7 7

5 Viewing Window [7,7][5,5]u

3 2 (b) distance 12 2tdt 4 ³0

3t (c) Set tt24;4 2 When t 4, the y-coordinates for A and B are also equal. Particles collide at (2,4) when t 4.

(d) 5

7 7

5 Viewing Window [7,7][5,5]u

Let f be a function that has derivatives of all orders for all real numbers. Assumeff (1) 3, ccc(1) 2, f (1) 2 , and f ccc (1) 4 .

(a) Write the second-degree Taylor polynomial for f about x 1 and use it to approximatef (0.7) .

(b) Write the third-degree Taylor polynomial for f about x 1 and use it to approximatef (1.2) .

(c) Write the second-degree Taylor polynomial for f c , the derivative of f , about x 1 and use it to approximate f c(1.2) .

2 2 (a) Tx2 32 x 1 x 1 2 f 0.7| 3 0.6 0.09 3.69

234 (b) Tx3 32 x 1 x 1 x 1 6 2 f 1.2| 3 0.4 0.04 0.008 2.645 3

c 2 (c) Tx3 22 x 1 2 x 1 f c 1.2| 2 0.4 0.08 1.52

Let fx( ) x22 , gx( ) cos x , and hx ( ) x cos x. From the graphs of f and g shown above in Figure 1 and Figure 2, one might think the graph of h should look like the graph in Figure 3.

(a) Sketch the actual graph of h in the viewing window provided below.

6 6

6 Viewing Window [ 6, 6] x [ 6, 40]

(b) Usehxcc ( ) to explain why the graph of h does not look like the graph in Figure 3.

(c) Prove that the graph of yx 2 cos( kx ) has either no points of inflection or infinitely many points of inflection, depending on the value of the constant k .

(a) 40

6 6

6 (b) hxccc 2sin; x xhx 2cos x 2cos0!x for all x , so graph must be concave up everywhere

(c) yxcc 2cos k2 kx If ky2 dt2, cc 0 for all x , so no inflection points. If ky2 ! 2, cc changes sign and is periodic, so changes sign infinitely many times. Hence there are infinitely many inflection points.

Graph of f

Let f be a function whose domain is the closed interval >@0,5 . The graph of f is shown above. x 3 Let hx() 2 f () tdt. ³ 0

(a) Find the domain of h .

(b) Findhc (2) .

x (a) 035dd 2 d64x d

(c) hc is positive, then negative, so minimum is at an endpoint 0 hftdt 60 ³0 5 hftdt40 ³0 since the area below the axis is greater than the area above the axis therefore minimum at x 4

The rate of consumption of cola in the United States is given by St ( ) Cekt , where S is measured in billions of gallons per year and t is measured in years from the beginning of 1980.

(a) The consumption rate doubles every 5 years and the consumption rate at the beginning of 1980 was 6 billion gallons per year. Find C and k .

(b) Find the average rate of consumption of cola over the 10-year time period beginning January 1, 1983. Indicate units of measure.

7 (c) Use the trapezoidal rule with four equal subdivisions to estimate St() dt. ³ 5

7 (d) Using correct units, explain the meaning of St() dt in terms of cola ³ 5 consumption.

(a) St Cekt SC 06 6 Se 512126 5k 2 e5k ln 2 k 0.138 or 0.139 5

ln 2 1 13 ()t (b) Average rate 6edt5 13 3 ³3 3 ªºee2.6ln 2 0.6ln 2 billion gal/yr ln 2 ¬¼ (19.680 billion gal/yr)

7 1 (c) S t dt ªº S 5 2 S 5.5 2 S 6 2 S 6.5 S 7 ³5 4 ¬¼

(d) This gives the total consumption, in billions of gallons, during the years 1985 and 1986.

This problem deals with functions defined by fx ( ) x b sin x, where b is a positive constant and Sd 2x d 2 S.

(a) Sketch the graphs of two of these functions, yx sin x and yx 3sin x.

y y 6 6

x x 6 6 6 6

6 6

(b) Find the x-coordinates of all points, Sd 2x d 2 S, where the line yxb is tangent to the graph of fx ( ) x b sin x.

(d) For all values of b ! 0 , show that all inflection points of the graph of f lie on the line yx .

(a) y y

6 6

x x 6 6 6 6

6 6

(b) ybxc 11 cos bxcos 0 cosx 0

yxbxb sin x bb sin x 1sin x

3SS x or 22

(d)fxcc bsin x 0 sinx 0 fx xb0 x at x-coordinates of any inflection points

2 Consider the graph of the function h given by hx() ex for 0 d x f .

(a) Let R be the unbounded region in the first quadrant below the graph of h . Find the volume of the solid generated when R is revolved about the y-axis.

(b) LetAw ( ) be the area of the shaded rectangle shown in the figure above. Show that Aw() has its maximum value when w is the x-coordinate of the point of inflection of the graph of h .

f 2 (a) Volume 2S xex dx ³0

1 2 1 Volume SSS lnydy lim ln ydy ³³0 ao0 a

(b) Maximum: 2 Aw wew , 22 Awc eww2 we2 2 eww 122 . Awc 0 when w 1 , ! 2 Awc 0 when w 1 , 2 Awc 0 when w 1 . !2 Therefore, max occurs when w 1 2

Inflection: 22 hx exx,2, hc x xe 22 hxcc 222 exx x xe 2 212.exx 2 hxcc 0 when x 1 , 2 hxcc 0 when x 1 , 2 hxcc 0 when x 1 . !!2 Therefore, inflection point when x 1 . 2 Therefore, the maximum value of A (w ) and the inflection point of hx() occur when x and w are 1 . 2

xx23 x xn The Maclaurin series for fx ( ) is given by 1 "" 2! 3! 4! (n 1)! (a) Find ffc(0) and(17) (0) .

(b) For what values of x does the given series converge? Show your reasoning.

(d) Write gx() in terms of a familiar function without using series. Then, write fx() in terms of the same familiar function.

f n 0 1 (a) a n nn!1! 1 fac 0 1 2

(b) xn1 n 2! x lim lim 0 1 nnofxn of n 2 n 1! Converges for all x , by ratio test

xx2 n (d) exx 1 "" 2!n ! egxxfxx 1 ex 1 ° ifx z 0 fx ® x ° ¯1 if x 0

An oil storage tank has the shape as shown above, obtained by revolving the curve 9 yx 4 from xx 0 to 5 about the y-axis, where x and y are measured in feet. 625 Oil weighing 50 pounds per cubic foot flowed into an initially empty tank at a constant rate of 8 cubic feet per minute. When the depth of oil reached 6 feet, the flow stopped.

(a) Let h be the depth, in feet, of oil in the tank. How fast was the depth of oil in the tank increasing when h 4 ? Indicate units of measure.

(b) Find, to the nearest foot-pound, the amount of work required to empty the tank by pumping all of the oil back to the top of the tank.

h ´ 25 (a) Vydy S µ ¶ 0 3 dV25S dh h dt3 dt 25S dh at h 4, 8 4 3 dt dh 12 ft/min dt 25S

6 ´ §·25S (b) Wyydy 50µ 9 ¨¸ ¶ 0 ©¹3 6 §·25S ´ §·13 Wyydy 50¨¸ 9 22 ¨¸3 µ ©¹¶ 0 ©¹ 6 §·25S §· 235 2 Wy 50¨¸ 9 22y ¨¸33 5 ©¹©¹0 W 69,257.691 ft-lbs to the nearest foot-pound 69,258 ft-lbs

The figure above shows a spotlight shining on point Px(,) y on the shoreline of Crescent Island. The spotlight is located at the origin and is rotating. The portion of the shoreline on which the spotlight shines is in the shape of the parabola yx 2 from the point 1,1 to the point 5,25 . Let T be the angle between the beam of light and the positive x- axis.

(a) For what values of T between 0 and 2S does the spotlight shine on the shoreline?

(b) Find the x- and y-coordinates of point P in terms of tanT .

(c) If the spotlight is rotating at the rate of one revolution per minute, how fast is the point P traveling along the shoreline at the instant it is at the point 3,9 ?

1 S (a) tanTT or 0.785 1114 25 tanTT tan1 5 or 1.373 225 S Therefore, ddT tan1 5 4

yx2 (b) tanT x xx Therefore,x tanT yx 22 tan T

dT (c) 2S dt dx dTT dy d sec22TT ; 2 tan sec T dt dt dt dt dx At 3,9 : 10 2SS 20 dt dy 2 3 10 2SS 120 dt

22 Speed 20SS 120 20S 37 or 382.191

The graph of the function f consists of a semicircle and two line segments as shown x above. Let g be the function given by gx() ftdt () . ³ 0

(a) Find g (3) .

(b) Find all the values of x on the open interval 2,5 at which g has a relative maximum. Justify your answer.

(d) Find the x-coordinate of each point of inflection of the graph of g on the open interval 2,5 . Justify your answer.

3 (a) gftdt3() ³0 111 SS22 422

(b) gx( ) has a relative maximum at x 2 becauseg' x f ( x ) changes from the positive to negative atx 2

(d) graph of g has points of inflection with x-coordinates x = 0 and x = 3

becausegxcc fx c ( ) changes from the positive to negative atxx 0 and from negative to positive at 3 or becausegxc fx changes from increasing to decreasing at x 0 and from decreasing to increasing atx 3

Letvt ( ) be the velocity, in feet per second, of a skydiver at time t seconds, t t 0 . After dv her parachute opens, her velocity satisfies the differential equation 232v , with dt initial condition v (0) 50 .

(a) Use separation of variables to find an expression for v in terms of t , where t is measured in seconds.

(b) Terminal velocity is defined as limvt ( ) . Find the terminal velocity of the skydiver tof to the nearest foot per second.

dv (a) 2322vv 16 dt dv 2dt v 16 lnvtA 16 2 veee16 22tA A t vCe16 2t 50 16 Ce0 ; C 34 ve 342t 16

(b) limvt lim 34 e2t 16 16 ttof of

During the time period from tt 0 to 6 seconds, a particle moves along the path given by xt ( ) 3 cos(S t ) and yt ( ) 5 sin(S t ) .

(a) Find the position of the particle when t 2.5.

(b) On the axes provided below, sketch the graph of the path of the particle from t = 0 to t = 6 . Indicate the direction of the particle along its path.

(d) Find the velocity vector for the particle at any time t .

(e) Write and evaluate an integral expression, in terms of sine and cosine, that gives the distance the particle travels from t = 1.25 to t = 1.75 .

(a) x 2.5 3cos 2.5S 0 y 2.5 5sin 2.5S 5

(b) y

(d) xtcc() 3SS sin t yt () 5 S cos S t JG vt 3sinSSSS t ,5cos t

1.75 (e) distance 9SSSS22 sinttdt 25 2 cos 2 ³1.25 5.392

Let Px()73(4)5(4)2(4)6(4) x x 234 x x be the fourth-degree Taylor polynomial for the function f about 4. Assume f has derivatives of all orders for all real numbers.

(a) Find ff(4) andccc (4).

(b) Write the second-degree Taylor polynomial for f c about 4 and use it to approximate f c(4.3) .

x (c) Write the fourth-degree Taylor polynomial for gx() ftdt () about 4. ³ 4

(d) Can f (3) be determined from the information given? Justify your answer.

(a) fP 447 f ccc 4 2, f ccc 4 12 3!

23 (b) Px3 73 x 4 5 x 4 2 x 4 2 Px3c 310464 x x 2 f c 4.3| 3 10 0.3 6 0.3 0.54

x (c) Pgx43 ,() Ptdt ³4 x 23 ªº7345424 tt tdt ³4 ¬¼

351234 74 xxxx 4 4 4 232

(d) No. The information given provides values for ff 4,cccccc 4,ff 4, 4 and f 4 4 only.

Let R be the region enclosed by the graphs of yx ln 2 1 and yx cos .

(a) Find the area of R .

(b) Write an expression involving one or more integrals that gives the length of the boundary of the region R . Do not evaluate.

(c) The base of a solid is the region R . Each cross section of the solid perpendicular to the x-axis is an equilateral triangle. Write an expression involving one or more integrals that gives the volume of the solid. Do not evaluate.

y (a) ln xx2 1 cos x r0.91586 let B 0.91586

R B area ªº cosxx ln2 1 dx ³B ¬¼ x 1.168

(c)

3 base 2

base

13ªº area of cross section ªºcosxx ln22 1 cos xx ln 1 ¬¼u «» 22¬¼

B 3 2 volume ´ ªºcosxx ln2 1 dx µ ¬¼ ¶ B 4

Let xky 2 2 , where k ! 0 . (a) Show that for all k ! 0 , the point 4, 2 is on the graph of xky 2 2 . k (b) Show that for all k ! 0 , the tangent line to the graph of xky 2 2 at the point 4, 2 passes through the origin. k

(c) Let R be the region in the first quadrant bounded by the x-axis, the graph of xky 2 2 , and the line x 4 . Write an integral expression for the area of the region R and show that this area decreases as k increases.

44 R 2 4 x

(b) xky 2 2 dy 12 ky dx dy 1

dx yk 2/ 22k

ª «the tangent line is « « 21 yx 4 « k 22k « « 1 « yx which contains (0,0) ¬ 22k or ª 2/k 1 «slope of the line through 0,0 and 4, 2 /k is « 4 22k ¬«which is the same as the slope of the tangent line

2/k (c) Akydy 4 2 2 ³0 or 1 4 Axdx 2 k ³2

42 Ak 0.5 3 dA 22 kk1.5 0 for all ! 0 dk 3

thus the area decreases as k increases