Calculus BC Section II PART A A GRAPHING CALCULATOR IS REQUIRED FOR SOME PROBLEMS OR PARTS OF PROBLEMS

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1 Calculus BC Section II PART A A GRAPHING CALCULATOR IS REQUIRED FOR SOME PROBLEMS OR PARTS OF PROBLEMS. An isosceles triangle, whose base is the interval from (0, 0) to (c, 0), has its verte on the graph of f. For what value of c oes the triangle have maimum area? y To maimize the Area of a triangle you nee an area formula: A bh. The base of the triangle is c units long, an since the triangle is isosceles an the verte is on the curve, it's height is given by plugging in c into the equation of the parabola. Thus, A c c A6c c 8 (0, 0) (c, 0) : When fining a maimum or minimum on an interval be sure to check the enpoints when the interval is CLOSED since the enpoints of this interval woul not create a triangle, the interval is open an therefore you o not nee to check the enpoints. To fin the maimum, we take the erivative to get A' 6 c. 8 To fin the critical numbers, first set the erivative equal to 0 an solve for c. 06 c 8 c 6 8 c 6 c 4 Since A ' is a polynomial, the erivative is never unefine. Since c cannot be more than twice the istance from the origin to the - intercept of the parabola, we know 0 c, an we can ignore c = 4. Therefore, the triangle has a maimum area when c = 4. Justify your answer: Using the first erivative test we can look at a sign chart or the graph of the erivative. Looking at a sign chart of A ' with the enpoints Looking at a graph of A ', we see a parabola opening an critical numbers, we have ownwar with -intercepts of 4 an c Since A ' changes from + to at c = 4, then there is a maimum when c = 4. Using the secon erivative test plug c = 4 into the secon erivative. Fining the secon erivative gives us A'' c. 4 Since A'' 4 0, c = 4 gives a maimum value of A.

2 . Given the following table of values at = an =, fin the inicate erivatives in parts a l. f f ' g 5 7 f g a) f f ' When =, f f ' g) g f b) f g g ) f f g f ' When =, f g f ' 7 f c) g g f ' f g g f ' f When =, 9 g f g f ' f f g f ' 9 When =, f 4 e) f g f ' g When =, f ' g 5 f) f g f ' g When =, f ' g f f ' When =, f f ' 5 h) g g g When =, g 7 i) f g 46 k) f l) f f ' f ' g When =, f ' g 6 6 j) g g When =, g f f ' When =, f f ' f ' When =,

3 for. Let f be a function efine by f. k p for a) For what values of k an p will f be continuous an ifferentiable at =? To be continuous at =, lim f lim f f. Evaluating the left an right limits we get lim f lim f k p k p k p0 To be ifferentiable at =, we take the erivative of f as a piecewise function to get for f ' k for To be ifferentiable at =, the erivative from the left of must equal the erivative from the right of. Thus, k 0k k Substituting this into the equation from the first part we get p =. b) For the values of k an p foun in part a, on what interval or intervals is f increasing? for Using the values of p an k from part a, we have f '. The function f is increasing when for f ' is positive. Accoring to the erivative function above (or looking at the graph below), when <, f ' > 0, an when >, f ' > 0. Thus f is increasing on the interval,,. Graph of c) Using the values of k an p foun in part a, fin all points of inflection of the graph of f. Support your conclusion. Points of inflection are foun using the secon erivative. So, changes sign at =, there is a point of inflection at =. for f ''. Since the secon erivative for

4 4. Consier the curve efine by y y 7. a) Write an epression for the slope of the curve at any point (, y). Use implicit ifferentiation to fin the erivative of the curve. Remember, taking the erivative of the y term involves the prouct rule. y y y y 0 y y y y y y y y y y b) Determine whether the lines tangent to the curve at the intercepts of the curve are parallel. Show the analysis that leas to your conclusion. The intercepts of the curve are where the y values equal zero. Plugging y = 0 into the original equation we have 00 7 Thus, we have two points,0 an,0 7. To show they are parallel, we must show that the slopes at these points are the same. Plugging y = 0 into the equation for y above, we get y 0 0 y y Since the slope is equal to whenever y = 0, the tangent lines at the intercepts of the curve are parallel. c) Fin the points on the curve where the lines tangent to the curve are vertical. The tangent lines are vertical when y is unefine. This occurs when 0 y. Solving this for, we get = y. Plugging this into the original equation we get y y yy 7. 4y y y 7 y 7 y 9 y Substituting these values back into = y, we have two points where the tangent line is vertical ( 6, ) an (6, ).

5 5. The figure below shows the graph of f ', the erivative of a function f. The omain of the function is the set of all such that. y Note: This is the graph of the erivative of f, not the graph of f. a) For what values of, < <, oes f have a relative minimum? A relative maimum? Justify your answer. f has a relative minimum when f ' changes from to +. Since we are given the graph of f ', this happens at = 0. f has a relative maimum when f ' changes from + to. Since we are given the graph of f ', this happens at =. b) For what values of is the graph of f concave up? Justify your answer. f is concave up when f '' is positive. Since we are given the graph of f ', we are looking for when the slopes of f ' are positive. This occurs on the intervals (, ) an (, ). c) Use the information foun in parts a an b an the fact that f ( ) = 0 to sketch a possible graph of f on the aes provie below. y Your graph may be ifferent, but must show the following: Starting at (, 0) your curve must be Increasing on (, ) an (0, ) an (, ) Decreasing on (, 0) Horizontal Tangents at =, 0, an Concave Up on (, ) an (, ) Concave Down on (, ) an (, )

6 6. Fin the value of c that satisfies the Mean Value Theorem for f 5 over [ 4, ]. The MVT says there is at least one point between 4 an (we'll call it c) where the erivative (instantaneous rate of change) at c equals the slope of the line that connects the enpoints of the interval (average rate of change). Thus we have f 4 f f ' c 4 08 c 4c5 6 c 4c c c Solving this equation using the quaratic formula (or graphing an fining the -intercepts) we get c.097 an c.4. Since both of these values are in the interval, then we have two values of c that satisfy the MVT over [ 4, ]. 7. An object moves along the -ais with velocity vt t t 5t where t 0,. a) When is the object stoppe? Justify your answer. The object is stoppe when v (t) = 0. Use your calculator to fin where the graph of velocity crosses the -ais in the interval [0, ]. This occurs at t an t b) When is the object moving right? Justify your answer. The object is moving right when the velocity is positive. This occurs on (0, ) an (, ) c) What is the acceleration at time t =.? Show all your work. at t 4t 5. Thus, a ) When is the object speeing up? Justify your answer. The object is speeing up when the velocity an acceleration have the same sign. Looking at the graphs of both velocity an acceleration (or using a sign chart for both) we see that on ( , ) an (, ) the velocity an acceleration have the same sign. e) The object was at a position of +4 when t = 0. Where will it be at t =? Show all your work. 4 5 Since position is the antierivative of velocity, the position function is t t t t t C. 4 Since (0) = 4, we get 4 = C. Thus 4 5 t t t t t 4. 4 To fin where the particle will be when t =, we fin

7 8. Given the graph of f consisting of two line segments an a semicircle f () as shown an that g f. a) Fin g 0, g, an 4 g area from = to = 0 is going to be the opposite sign of the g f f 0 area from = 0 to =. f no area accumulate g 0 4 area of half a circle minus (it s below the -ais) f g or.07 the area of the triangle. b) When oes g have a minimum? Justify your response. Since g f, then f. Thus, g () has a relative minimum when f to +. Since this oes not occur, there is no relative minimum for g () between = 0 an = 4. c) When oes g () have a point of inflection? Justify your response. changes from Net, check the enpoints. To be a relative minimum at an enpoint, the value of g () at the enpoint must be smaller than the values close to the enpoint. Since the slope to the right of the left enpoint is greater than 0, = 0 is a minimum. Since the slope to the left of the right enpoint is less than 0, = 4 is a minimum. A point of inflection occurs when ' f ' the ' f ' changes sign at = an =. changes sign. Looking at the slopes of the graph of f () we see that ) Sketch g. Label the vertical ais. The slope of g is 0 at = an =. On (0, ) g is increasing/concave own On (, ) g is increasing/concave up On (, ) g is increasing/concave own On (, 4) g is ecreasing/concave own g () 4 These points shoul be on your graph g ()

8 9. [00 AP Calculus AB Free Response Question #5] A container has the shape of an open right circular cone, as shown below. 0 cm h r 0 cm The height of the container is 0 cm an the iameter of the opening is 0 cm. Water in the container is evaporating so that its epth h is changing at the constant rate of cm/hr. 0 (Note: The volume of a cone of height h an raius r is given by V rh.) a) Fin the volume V of water in the container when h = 5 cm. Inicate units of measure. Using similar triangles (as shown below) when h = 5, r =, so r b) Fin the rate of change of the volume of water in the container, with respect to time, when h = 5 cm. Inicate units of measure. Since 5 r, r h. Substituting this into the formula for V gives us V h h h. 0 h V h Taking the erivative with respect to time we get h. t 4 t V 5 When h = 5, 5 = cm. t hr c) Show that the rate of change of the volume of water in the container ue to evaporation is irectly proportional to the epose surface area of the water. What is the constant of proportionality? To show the rate of change of the volume of water in the container ue to evaporation is irectly proportional to the V epose surface area of the water, we nee to show that k r, where k is the constant of proportionality. t V h h V Since h, an h = r an, we have r r area. t 4 t t 0 t The constant of proportionality is. 0

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