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1998 AP Calculus AB Scoring Guidelines 1. Let R be the region bounded by the x axis, the graph of y = x, and the line x = 4. Find the area of the region R. (b) Find the value of h such that the vertical line x = h divides the region R into two regions of equal area. (c) Find the volume of the solid generated when R is revolved about the x axis. (d) The vertical line x = k divides the region R into two regions such that when these two regions are revolved about the x axis, they generate solids with equal volumes. Find the value of k. y y = x 1: A = 4 x dx 1 R O 1 4 5 x A = 4 x dx = 4 x/ = 16 or 5. (b) h x dx = 8 h/ = 8 h h = 16 or.5 or.519 4 x dx = x dx h/ = 16 h/ h { 1: equation in h 4 (c) V = π ( x) dx = π x 4 = 8π or 5.1 or 5.1 1: limits and constant 1: integrand k (d) π ( x) dx = 4π π k = 4π k π ( 4 x) dx = π ( x) dx k π k = 8π π k { 1: equation in k k = 8 or.88
. Let f be the function given by f(x) = xe x. 1998 Calculus AB Scoring Guidelines Find lim f(x) and lim f(x). x x (b) Find the absolute minimum value of f. Justify that your answer is an absolute minimum. (c) What is the range of f? (d) Consider the family of functions defined by y = bxe bx, where b is a nonzero constant. Show that the absolute minimum value of bxe bx is the same for all nonzero values of b. lim x xex = lim x xex = or DNE { 1: as x 1: or DNE as x (b) f (x) = e x + x e x = e x (1 + x) = if x = 1/ f( 1/) = 1/e or.68 or.67 1/e is an absolute minimum value because: (i) (ii) f (x) < for all x < 1/ and f (x) > for all x > 1/ f (x) + 1/ and x = 1/ is the only critical number 1: solves f (x) = 1: evaluates f at student s critical point /1 if not local minimum from student s derivative 1: justifies absolute minimum value /1 for a local argument /1 without explicit symbolic derivative Note: / if no absolute minimum based on student s derivative (c) Range of f = [ 1/e, ) or [.67, ) or [.68, ) (d) y = be bx + b xe bx = be bx (1 + bx) = if x = 1/b At x = 1/b, y = 1/e y has an absolute minimum value of 1/e for all nonzero b Note: must include the left hand endpoint; exclude the right hand endpoint 1: sets y = be bx (1 + bx) = 1: solves student s y = 1: evaluates y at a critical number and gets a value independent of b Note: / if only considering specific values of b
AB Board Note # 1 Part (d) 1998 AP Calculus AB Scoring Guidelines / Argument with the following three ingredients: 1. The graph of y = bxe bx is a horizontal compression or expansion (with a reflection across the y axis if b < ) of the graph of y = xe x.. The range of y = bxe bx is therefore the same as the range of y = xe x.. Therefore the absolute minimum value of y = bxe bx is the same for all (non zero) values of b. / Analyzing the horizontal compression/expansion of graphs of y = bxe bx for specific values of b.
Velocity (feet per second) v(t) 9 8 7 6 5 4 1 O 1998 Calculus AB Scoring Guidelines 5 1 15 5 5 4 45 5 Time (seconds) t t v(t) (seconds) (feet per second) 5 1 1 15 55 5 7 78 5 81 4 75 45 6 5 7. The graph of the velocity v(t), in ft/sec, of a car traveling on a straight road, for t 5, is shown above. A table of values for v(t), at 5 second intervals of time t, is shown to the right of the graph. During what intervals of time is the acceleration of the car positive? Give a reason for your answer. (b) Find the average acceleration of the car, in ft/sec, over the interval t 5. (c) Find one approximation for the acceleration of the car, in ft/sec, at t = 4. Show the computations you used to arrive at your answer. (d) Approximate 5 v(t) dt with a Riemann sum, using the midpoints of five subintervals of equal length. Using correct units, explain the meaning of this integral. (b) (c) (d) Acceleration is positive on (, 5) and (45, 5) because the velocity v(t) is increasing on [, 5] and [45, 5] Avg. Acc. = v(5) v() 5 or 1.44 ft/sec Difference quotient; e.g. v(45) v(4) 5 v(4) v(5) 5 = = v(45) v(5) = 1 6 75 5 75 81 5 6 81 1 Slope of tangent line, e.g. 5 = 7 5 through (5, 9) and (4, 75): v(t) dt = 7 5 = ft/sec or = 6 5 ft/sec or = 1 1 ft/sec 9 75 = ft/sec 5 4 1[v(5) + v(15) + v(5) + v(5) + v(45)] = 1(1 + + 7 + 81 + 6) = 5 feet This integral is the total distance traveled in feet over the time to 5 seconds. 1: (, 5) 1: (45, 5) 1: reason Note: ignore inclusion of endpoints { 1: method Note: / if first point not earned 1: midpoint Riemann sum 1: meaning of integral
1998 AP Calculus AB Scoring Guidelines 4. Let f be a function with f(1) = 4 such that for all points (x, y) on the graph of f the slope is given by x + 1. y Find the slope of the graph of f at the point where x = 1. (b) Write an equation for the line tangent to the graph of f at x = 1 and use it to approximate f(1.). (c) Find f(x) by solving the separable differential equation dx = x + 1 with the initial y condition f(1) = 4. (d) Use your solution from part (c) to find f(1.). dx = x + 1 y dx x = 1 y = 4 = + 1 4 = 4 8 = 1 (b) y 4 = 1 (x 1) f(1.) 4 1 (1. 1) { 1: equation of tangent line 1: uses equation to approximate f(1.) (c) f(1.).1 + 4 = 4.1 y = (x + 1) dx y = (x + 1) dx y = x + x + C 4 = 1 + 1 + C 14 = C y = x + x + 14 y = x + x + 14 is branch with point (1, 4) f(x) = x + x + 14 1: separates variables 1: antiderivative of term 1: antiderivative of dx term 5 1: uses y = 4 when x = 1 to pick one function out of a family of functions 1: solves for y /1 if solving a linear equation in y /1 if no constant of integration Note: max /5 if no separation of variables Note: max 1/5 [1----] if substitutes value(s) for x, y, or /dx before antidifferentiation (d) f(1.) = 1. + 1. + 14 4.114, from student s solution to the given differential equation in (c)
1998 Calculus AB Scoring Guidelines 5. The temperature outside a house during a 4-hour period is given by ( ) πt F (t) = 8 1 cos, t 4, 1 where F (t) is measured in degrees Fahrenheit and t is measured in hours. Sketch the graph of F on the grid below. (b) Find the average temperature, to the nearest degree Fahrenheit, between t = 6 and t = 14. (c) An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees Fahrenheit. For what values of t was the air conditioner cooling the house? (d) The cost of cooling the house accumulates at the rate of $.5 per hour for each degree the outside temperature exceeds 78 degrees Fahrenheit. What was the total cost, to the nearest cent, to cool the house for this 4 hour period? Degrees Fahrenheit 1 9 8 7 1: bell shaped graph minimum 7 at t =, t = 4 only maximum 9 at t = 1 only 6 (b) Avg. = 1 14 6 (c) 6 1 18 4 14 = 1 8 (697.957795) 6 Time in Hours = 87.16 or 87.161 87 F [ 8 1 cos 1 cos 5. or 5.1 ( πt 1 ) ( πt 1 [ ( )] πt 8 1 cos dt 1 )] 78 18.769 t or 18.77 : integral 1: limits and 1/(14 6) 1: integrand /1 if integral not of the form 1 b a b a F (t) dt { 1: inequality or equation 1: solutions with interval (d) C =.5 18.77 or 18.769 5.1 or 5. ([ ( )] ) πt 8 1 cos 78 dt 1 =.5(11.9741) = 5.96 $5.1 : integral 1: limits and.5 1: integrand /1 if integral not of the form k b a (F (t) 78) dt
1998 AP Calculus AB Scoring Guidelines 6. Consider the curve defined by y + 6x y 1x + 6y = 1. Show that dx = 4x xy x + y + 1. (b) Write an equation of each horizontal tangent line to the curve. (c) The line through the origin with slope 1 is tangent to the curve at point P. Find the x and y coordinates of point P. 6y + 6x + 1xy 4x + 6 dx dx dx = dx (6y + 6x + 6) = 4x 1xy dx = 4x 1xy 6x + 6y + 6 = 4x xy x + y + 1 1: implicit differentiation 1: verifies expression for dx (b) (c) dx = 4x xy = x( y) = x = or y = When x =, y + 6y = 1 ; y =.165 There is no point on the curve with y coordinate of. y =.165 is the equation of the only horizontal tangent line. y = x is equation of the line. ( x) + 6x ( x) 1x + 6( x) = 1 8x 1x 6x 1 = x = 1/, y = 1/ dx = 1 4x xy = x y 1 4x + x = x x 1 4x + 4x + 1 = x = 1/, y = 1/ 4 1: sets dx = 1: solves dx = 1: uses solutions for x to find equations of horizontal tangent lines 1: verifies which solutions for y yield equations of horizontal tangent lines Note: max 1/4 [1---] if /dx = is not of the form g(x, y)/h(x, y) = with solutions for both x and y 1: y = x 1: substitutes y = x into equation of curve 1: solves for x and y 1: sets dx = 1 1: substitutes y = x into dx 1: solves for x and y Note: max / [1-1-] if importing incorrect derivative from part