IB Practice - Calculus - Differentiation Applications (V2 Legacy)

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1 IB Math High Level Year - Calc Practice: Differentiation Applications IB Practice - Calculus - Differentiation Applications (V Legacy). A particle moves along a straight line. When it is a distance s from a fied point, where s >, the (s + ) velocity v is given by v. Find the acceleration when s. (s ) (Total 4 marks). Give eact answers in this part of the question. The temperature g (t) at time t of a given point of a heated iron rod is given by ln t g (t), where t > 0. t (a) Find the interval where g (t) > 0. (4) (b) Find the interval where g (t) > 0 and the interval where g (t) < 0. (5) (c) Find the value of t where the graph of g (t) has a point of infleion. () (d) Let t* be a value of t for which g (t*) 0 and g (t*) < 0. Find t*. () (e) Find the point where the normal to the graph of g (t) at the point (t*, g (t*)) meets the t-ais. () (Total 8 marks). The acceleration, a(t) m s, of a fast train during the first 80 seconds of motion is given by a(t) t + 0 where t is the time in seconds. If the train starts from rest at t 0, find the distance travelled by the train in the first minute. (Total 4 marks) n k, > 0 4. Consider the function f k (), where k 0, 0 (a) Find the derivative of f k (), > 0. (b) Find the interval over which f 0 () is increasing. The graph of the function f k () is shown below. y () () 0 A Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 7

2 IB Math High Level Year - Calc Practice: Differentiation Applications (c) (i) Show that the stationary point of f k () is at e k. (ii) One -intercept is at (0, 0). Find the coordinates of the other -intercept. (4) (d) Find the area enclosed by the curve and the -ais. (5) (e) Find the equation of the tangent to the curve at A. () (f) Show that the area of the triangular region created by the tangent and the coordinate aes is twice the area enclosed by the curve and the -ais. () (g) Show that the -intercepts of f k () for consecutive values of k form a geometric sequence. () (Total 0 marks) 5. The velocity, v, of an object, at a time t, is given by v, where t is in seconds and v is in m s. Find the distance travelled between t 0 and t a. Working: Answer:... (Total marks) 6. Find the coordinates of the point which is nearest to the origin on the line L: λ, y λ, z. Working: Answer:... ke t (Total marks) 7. A rectangle is drawn so that its lower vertices are on the -ais and its upper vertices are on the curve y sin, where 0 n. (a) Write down an epression for the area of the rectangle. (b) Find the maimum area of the rectangle. Working: Answers: (a)... (b)... (Total marks) 8. Let f :! e sin. (a) Find f (). There is a point of infleion on the graph of f, for 0 < <. (b) Write down, but do not solve, an equation in terms of, that would allow you to find the value of at this point of infleion. Working: Answers: (a)... (b)... (Total marks) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 7

3 IB Math High Level Year - Calc Practice: Differentiation Applications 9. The diagram shows the graph of y f (). y y f ( ) Indicate, and label clearly, on the graph (a) the points where y f () has minimum points; (b) the points where y f () has maimum points; (c) the points where y f () has points of infleion. Working: (Total marks) 0. Let f () ln 5, 0. 5 < <, a, b; (a, b are values of for which f () is not defined). (a) (i) Sketch the graph of f (), indicating on your sketch the number of zeros of f (). Show also the position of any asymptotes. () (ii) Find all the zeros of f (), (that is, solve f () 0). () (b) Find the eact values of a and b. () (c) Find f (), and indicate clearly where f () is not defined. (d) Find the eact value of the -coordinate of the local maimum of f (), for 0 < <.5. (You may assume that there is no point of infleion.) (e) Write down the definite integral that represents the area of the region enclosed by f () and the -ais. (Do not evaluate the integral.) (). (a) Sketch and label the curves y for, and y ln for 0 <. () (b) Find the -coordinate of P, the point of intersection of the two curves. (c) (d) If the tangents to the curves at P meet the y-ais at Q and R, calculate the area of the triangle PQR. Prove that the two tangents at the points where a, a > 0, on each curve are always perpendicular. (4) (Total 4 marks) () () () (6) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 4 of 7

4 IB Math High Level Year - Calc Practice: Differentiation Applications. (a) Let y a + bsin, where 0 < a < b. b + a sin (i) ( b + a )cos Show that. ( b + asin ) (4) (ii) Find the maimum and minimum values of y. (4) (iii) Show that the graph of y a + bsin, 0 < a < b cannot have a vertical asymptote. b + a sin () (b) For the graph of y 4 + 5sin for 0 π, 5 + 4sin (i) write down the y-intercept; (ii) find the -intercepts m and n, (where m < n) correct to four significant figures; (iii) sketch the graph. (5) (c) The area enclosed by the graph of y 4 + 5sin and the -ais from 0 to n is denoted 5 + 4sin by A. Write down, but do not evaluate, an epression for the area A. () (Total 7 marks). The diagram shows a sketch of the graph of y f () for a b. y y f ( ) a b On the grid below, which has the same scale on the -ais, draw a sketch of the graph of y f () for a b, given that f (0) 0 and f () 0 for all. On your graph you should clearly indicate any minimum or maimum points, or points of infleion. y a b (Total marks) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 5 of 7

5 IB Math High Level Year - Calc Practice: Differentiation Applications 4. An astronaut on the moon throws a ball vertically upwards. The height, s metres, of the ball, after t seconds, is given by the equation s 40t + 0.5at, where a is a constant. If the ball reaches its maimum height when t 5, find the value of a. Working: Answer:... (Total marks) 5. Let f ( ) ( ),.4.4 (a) Sketch the graph of f (). (An eact scale diagram is not required.) On your graph indicate the approimate position of (i) each zero; (ii) each maimum point; (iii) each minimum point. (4) (b) (i) Find f (), clearly stating its domain. (ii) Find the -coordinates of the maimum and minimum points of f (), for < <. (7) (c) Find the -coordinate of the point of infleion of f (), where > 0, giving your answer correct to four decimal places. () (Total marks) 6. Consider the tangent to the curve y (a) Find the equation of this tangent at the point where. (b) Find the coordinates of the point where this tangent meets the curve again. (Total marks) 7. A point P(, ) lies on the curve y. Calculate the minimum distance from the point A, to the point P. Working: Answer:... (Total marks) 8. Let θ be the angle between the unit vectors a and b, where 0 < θ < π. Epress a b in terms of sin θ. (Total marks) 9. A particle is moving along a straight line so that t seconds after passing through a fied point O on the line, its velocity v (t) m s is given by π v ( t) t sin t. (a) Find the values of t for which v(t) 0, given that 0 t 6. () (b) (i) Write down a mathematical epression for the total distance travelled by the particle in the first si seconds after passing through O. (ii) Find this distance. (4) (Total 7 marks) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 6 of 7

6 IB Math High Level Year - Calc Practice: Differentiation Applications 0. Consider the function f ( ), where (a) (b) Show that the derivative +. Sketch the function f (), showing clearly the local maimum of the function and its horizontal asymptote. You may use the fact that n lim 0. (c) Find the Taylor epansion of f () about e, up to the second degree term, and show that this polynomial has the same maimum value as f () itself. (5) (Total marks). A particle is projected along a straight line path. After t seconds, its velocity v metres per second is given by v. + t (a) Find the distance travelled in the first second. (b) Find an epression for the acceleration at time t.. A curve has equation y + y. Find the equation of the tangent to this curve at the point (, ).. A rectangle is drawn so that its lower vertices are on the -ais and its upper vertices are on the curve e y. The area of this rectangle is denoted by A. (a) Write down an epression for A in terms of. (b) Find the maimum value of A. 4. The diagram below shows the graph of y f (), 0 4. y n f ( ) f ( ). () (5) 0 4 On the aes below, sketch the graph of y y 0 f ( t), marking clearly the points of infleion. 0 4 Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 7 of 7

7 IB Math High Level Year - Calc Practice: Differentiation Applications 5. A particle moves in a straight line with velocity v, in metres per second, at time t seconds, given by v(t) 6t 6t, t 0 Calculate the total distance travelled by the particle in the first two seconds of motion. Working: Answer: Find the -coordinate of the point of infleion on the graph of y e,. Working: Answer: Air is pumped into a spherical ball which epands at a rate of 8 cm per second (8 cm s ). Find the eact rate of increase of the radius of the ball when the radius is cm. Working: Answer: The point B(a, b) is on the curve f () such that B is the point which is closest to A(6, 0). Calculate the value of a. Working: Answer: (a) On the same aes sketch the graphs of the functions, f () and g (), where f () 4 ( ), for 4, g () ln ( + ), for 5. () (b) (i) Write down the equation of any vertical asymptotes. (ii) State the -intercept and y-intercept of g (). () (c) Find the values of for which f () g (). () (d) Let A be the region where f () g () and 0. (i) On your graph shade the region A. (ii) Write down an integral that represents the area of A. (iii) Evaluate this integral. (4) (e) In the region A find the maimum vertical distance between f () and g (). () (Total 4 marks) 0. A curve has equation y 8. Find the equation of the normal to the curve at the point (, ). Working: Answer:.... A particle moves in a straight line. Its velocity v m s after t seconds is given by v e sin t. Find the total distance travelled in the time interval [0, π]. t Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 8 of 7

8 IB Math High Level Year - Calc Practice: Differentiation Applications. The diagram below shows the graph of y f (). y On the aes below, sketch the graph of y f (). y. The function f is defined by f (), for > 0. (a) (i) Show that ln f () (ii) Obtain an epression for f (), simplifying your answer as far as possible. (b) (i) Find the eact value of satisfying the equation f () 0 (ii) Show that this value gives a maimum value for f (). (c) Find the -coordinates of the two points of infleion on the graph of f. () (Total marks) 4. An airplane is flying at a constant speed at a constant altitude of km in a straight line that will take it directly over an observer at ground level. At a given instant the observer notes that the angle θ is π radians and is increasing at radians per second. Find the speed, in kilometres per hour, at which 60 the airplane is moving towards the observer. Airplane km (5) (4) Observer Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 9 of 7

9 IB Math High Level Year - Calc Practice: Differentiation Applications 5. The following diagram shows an isosceles triangle ABC with AB 0 cm and AC BC. The verte C is moving in a direction perpendicular to (AB) with speed cm per second. C A B Calculate the rate of increase of the angle CÂB at the moment the triangle is equilateral. Working: Answer: Find the equation of the normal to the curve + y 9y 0 at the point (, 4). Working: Answer: A closed cylindrical can has a volume of 500 cm. The height of the can is h cm and the radius of the base is r cm. (a) Find an epression for the total surface area A of the can, in terms of r. (b) Given that there is a minimum value of A for r > 0, find this value of r. Working: Answers: (a)... (b) The displacement s metres of a moving bo B from a fied point O at time t seconds is given by s 50t 0t (a) Find the velocity of B in m s. (b) Find its maimum displacement from O The line L is given by the parametric equations λ, y λ, z. Find the coordinates of the point on L which is nearest to the origin Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 0 of 7

10 IB Math High Level Year - Calc Practice: Differentiation Applications 40. ln The function f is defined on the domain by f (). (a) (i) Show, by considering the first and second derivatives of f, that there is one maimum point on the graph of f. (ii) State the eact coordinates of this point. (iii) The graph of f has a point of infleion at P. Find the -coordinate of P. Let R be the region enclosed by the graph of f, the -ais and the line 5. (c) Find the eact value of the area of R. (d) The region R is rotated through an angle π about the -ais. Find the volume of the solid of revolution generated. () (Total marks) 4. The normal to the curve y k + ln, for 0, k, at the point where, has equation + y b, where b. Find the eact value of k. 4. The function f is defined by f () e p ( + ), here p. (a) (i) Show that f () e p (p( + ) + ). (ii) Let f (n) () denote the result of differentiating f () with respect to, n times. Use mathematical induction to prove that f (n) () p n e p (p( + ) + n), n +. (7) (b) When p, there is a minimum point and a point of infleion on the graph of f. Find the eact value of the -coordinate of (i) the minimum point; (ii) the point of infleion. (4) (c) Let p. Let R be the region enclosed by the curve, the -ais and the lines and. Find the area of R. () (Total marks) 4. The diagram shows a trapezium OABC in which OA is parallel to CB. O is the centre of a circle radius r cm. A, B and C are on its circumference. Angle OĈB θ. (9) () (6) O r A C B Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 7

11 IB Math High Level Year - Calc Practice: Differentiation Applications Let T denote the area of the trapezium OABC. (a) Show that T r (sin θ + sin θ). (4) For a fied value of r, the value of T varies as the value of θ varies. (b) Show that T takes its maimum value when θ satisfies the equation 4 cos θ + cos θ 0, and verify that this value of T is a maimum. (5) (c) Given that the perimeter of the trapezium is 75 cm, find the maimum value of T. (6) (Total 5 marks) 44. The curve y + 4 has a local maimum point at P and a local minimum point at Q. Determine the equation of the straight line passing through P and Q, in the form a + by + c 0, where a, b, c. 45. The following diagram shows the points A and B on the circumference of a circle, centre O, and radius 4 cm, where AÔB θ. Points A and B are moving on the circumference so that θ is increasing at a constant rate. O A B Given that the rate of change of the length of the minor arc AB is numerically equal to the rate of change of the area of the shaded segment, find the acute value of θ. 46. A man PF is standing on horizontal ground at F at a distance from the bottom of a vertical wall GE. He looks at the picture AB on the wall. The angle BPA is θ. E B A P D F Let DA a, DB b, where angle Pˆ DE is a right angle. Find the value of for which tan θ is a maimum, giving your answer in terms of a and b. Justify that this value of does give a maimum value of tan θ. (Total 9 marks) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 7 G

12 IB Math High Level Year - Calc Practice: Differentiation Applications 47. Let f () + 4. Find the values of m for which the line y m + is a tangent to the graph of f Particle A moves in a straight line, starting from O A, such that its velocity in metres per second for 0 t 9 is given by v A t + t +. Particle B moves in a straight line, starting from O B, such that its velocity in metres per second for 0 t 9 is given by v B e 0.t. (a) Find the maimum value of v A, justifying that it is a maimum. (5) (b) Find the acceleration of B when t 4. The displacements of A and B from O A and O B respectively, at time t are s A metres and s B metres. When t 0, s A 0, and s B 5. (c) Find an epression for s A and for s B, giving your answers in terms of t. (d) (i) Sketch the curves of s A and s B on the same diagram. (ii) Find the values of t at which s A s B. 49. Let f be the function defined for > by f () ln ( + ). (a) Find f (). (b) Find the equation of the normal to the curve y f () at the point where. Give your answer in the form y a + b where a, b () (7) (8) (Total marks) 50. The radius and height of a cylinder are both equal to cm. The curved surface area of the cylinder is increasing at a constant rate of 0 cm /sec. When, find the rate of change of (a) the radius of the cylinder; (b) the volume of the cylinder Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 7

13 IB Math High Level Year - Calc Practice: Differentiation Applications ln 5. The function f is defined by f (),. (a) Find f () and f (), simplifying your answers. (b) (i) Find the eact value of the -coordinate of the maimum point and justify that this is a maimum. (ii) Solve f () 0, and show that at this value of, there is a point of infleion on the graph of f. (iii) Sketch the graph of f, indicating the maimum point and the point of infleion. The region enclosed by the -ais, the graph of f and the line is denoted by R. (c) Find the volume of the solid of revolution obtained when R is rotated through 60 about the - ais. (d) Show that the area of R is (4 ln ) The volume of a solid is given by (6) () () (6) 4 V πr + πr h. At the time when the radius is cm, the volume is π cm, the radius is changing at a rate of cm/min and the volume is changing at a rate of 04π cm /min. Find the rate of change of the height at this time A particle moves in a straight line. At time t seconds, its displacement from a fied point O is s metres, and its velocity, v metres per second, is given by v t 4t +, t 0. When t 0, s. Find the value of t when the particle is at O Car A is travelling on a straight east-west road in a westerly direction at 60 km h. Car B is travelling on a straight north-south road in a northerly direction at 70 km h. The roads intersect at the point O. When Car A is km east of O, and Car B is y km south of O, the distance between the cars is z km. Find the rate of change of z when Car A is 0.8 km east of O and Car B is 0.6 km south of O. Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 4 of 7

14 IB Math High Level Year - Calc Practice: Differentiation Applications 55. A television screen, BC, of height one metre, is built into a wall. The bottom of the television screen at B is one metre above an observer s eye level. The angles of elevation ( Aˆ OC, Aˆ OB ) from the observer s eye at O to the top and bottom of the television screen are α and β radians respectively. The horizontal distance from the observer s eye to the wall containing the television screen is metres. The observer s angle of vision ( Bˆ OC) is θ radians, as shown below. (a) (i) Show that θ arctan arctan. (ii) Hence, or otherwise, find the eact value of for which θ is a maimum and justify that this value of gives the maimum value of θ. (iii) Find the maimum value of θ. (b) Find where the observer should stand so that the angle of vision is Given that y find (a) d y ; e (7) (5) (Total marks) (b) the eact values of the -coordinates of the points of infleion on the graph of y, justifying that they are points of infleion. () 57. (a) A curve is defined by the implicit equation y + 6 y 6. 0 Show that the tangent at the point A with coordinates, has gradient. (6) (b) The line cuts the curve at point A, with coordinates,, and at point B. a Find, in the form r c s b d (i) the equation of the tangent at A; (ii) the equation of the normal at B. (0) (c) Find the acute angle between the tangent at A and the normal at B. (4) (Total 0 marks) e () Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 5 of 7

15 IB Math High Level Year - Calc Practice: Differentiation Applications 58. The acceleration in m s of a particle moving in a straight line at time t seconds, t > 0, is given by the formula a. When t, the velocity is 8 m s. (+ t) (a) Find the velocity when t. (b) Find the limit of the velocity as t. (c) Find the eact distance travelled between t and t. (6) () (4) (Total marks) 59. If f (), > 0, (a) find the -coordinate of the point P where f () 0; () (b) determine whether P is a maimum or minimum point. () (Total 5 marks) 60. The function f is defined by f () e. It can be shown that f (n) () ( n + n n ) e for all n +, where f (n) () represents the n th derivative of f (). (a) By considering f (n) () for n and n, show that there is one minimum point P on the graph of f, and find the coordinates of P. (7) (b) Show that f has a point of infleion Q at. (5) (c) Determine the intervals on the domain of f where f is (i) concave up; (ii) concave down. () (d) Sketch f, clearly showing any intercepts, asymptotes and the points P and Q. (4) (e) Use mathematical induction to prove that f (n) () ( n + n n ) e for all n +, where f (n) () represents the n th derivative of f (). (9) (Total 7 marks) 6. A gourmet chef is renowned for her spherical shaped soufflé. Once it is put in the oven, its volume increases at a rate proportional to its radius. (a) Show that the radius r cm of the soufflé, at time t minutes after it has been put in the oven, dr k satisfies the differential equation, where k is a constant. r (b) Given that the radius of the soufflé is 8 cm when it goes in the oven, and cm when it s cooked 0 minutes later, find, to the nearest cm, its radius after 5 minutes in the oven. (8) (Total marks) 6. A normal to the graph of y arctan ( ), for > 0, has equation y + c, where c. Find the value of c. (5) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 6 of 7

16 IB Math High Level Year - Calc Practice: Differentiation Applications 6. André wants to get from point A located in the sea to point Y located on a straight stretch of beach. P is the point on the beach nearest to A such that AP km and PY km. He does this by swimming in a straight line to a point Q located on the beach and then running to Y. When André swims he covers km in 5 5 minutes. When he runs he covers km in 5 minutes. (a) If PQ km, 0, find an epression for the time T minutes taken by André to reach point Y. (b) Show that dt d T (c) (i) Solve 0. (ii) Use the value of found in part (c) (i) to determine the time, T minutes, taken for André to reach point Y. (iii) d T 0 5 Show that ( + 4) and hence show that the time found in part (c) (ii) is a minimum. () (Total 8 marks) 64. A family of cubic functions is defined as f k () k k +, k +. (a) Epress in terms of k (i) f k () and f k (); (ii) the coordinates of the points of infleion P k on the graphs of f k. (6) (b) Show that all P k lie on a straight line and state its equation. () (c) Show that for all values of k, the tangents to the graphs of f k at P k are parallel, and find the equation of the tangent lines. (5) (Total marks) 65. Consider the curve with equation f () e for < 0. Find the coordinates of the point of infleion and justify that it is a point of infleion. (4) () (Total 7 marks) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 7 of 7

17 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme. Given v IB Practice - Calculus - Differentiation Applications (V Legacy) - MarkScheme (s + ) (s ) dv dv ds dv then acceleration a v ds ds (M) (s ) (s + ) (s + ) therefore a (s ) (s ) (M) 7(s + ) a (s ) (M) 56 therefore when s, a 7 (A) (C4) ln t ln t. (a) g (t). So g (t). (M)(A) t t Hence, g (t) > 0 when > ln t or ln t < or t < e. (M) Since the domain of g (t) is {t:t > 0}, g (t) > 0 when 0 < t < e. (A) 4 (b) ln t t t ( ln t) Since g (t), g (t) t 4t (M) t[8 ln t] 4t (A) Hence g (t) > 0 when 8 ln t < 0 ie t > e 8/. Similarly, g (t) < 0 when 0 < t < e 8/. (A) 5 (c) g (t) 0 when t 0 or 8 ln t. e Since, the domain of g is {t:t > 0}, g (t) 0 when t e 8/. (M) (M) Since g (t) > 0 when t > e 8/ and g (t) < 0 when t < e 8/, (M) 8 4 /, e is the point of infleion. The required value of t is e 8/. (A) 8 / Note: Award (A) for evaluating t as e 8/. [4] (d) g (t) 0 when ln t or t e. (M) Also g (e e [8 ln e ] ) < 0 (M) 6 5 4e e Hence t* e (A) (e) At (t*, g (t*)) the tangent is horizontal. (M) So the normal at the point (t*, g (t*)) is the line t t*. (M) Thus, it meets the t ais at the point t t* e and hence the point is (e, 0). (A). a(t) t + 0 v(t) t + t + c (M) 40 v 0 when t 0, and so c 0 Thus, v(t) t + t t(t 80). (A) [8] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 6

18 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme Since v(t) 0 for 0 t 80, the distance travelled (M) m. (A) (C4) 4. (a) fk () ln k f () ln + k (M)(A) (b) f () ln + 0 k f0 () increases where f () > 0 (M) 0 ln > > (A) e (c) (i) Stationary points happen where f () 0 k ln k (M) e k (A) (ii) intercepts are where fk () 0 ln k 0 (ln k) 0 (M) 0 or ln k e k (e k, 0) (A) 4 (d) k k e e Area ln k ( k ln ) 0 0 (M) Integrate ln by parts. ln ln (M) ln + C 4 (A) Area ln 4 k e 0 k 60 0 v( t) 0 t + t 60 0 [4] (A) 5 4 Note: Given ln k fk () 0 when 0. (e) Gradient of the tangent at A(e k, 0), m is f (e k ) ln e k + k (M) (f) e k Therefore, an equation of the tangent is y e k. (A) The tangent forms a right angle triangle with the coordinate aes. The perpendicular sides are each of length e k. k k k Area of the triangle e e e k k e e ie The area of the triangle is twice the area 4 k (M) (A) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 6

19 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme enclosed by the curve and the -ais. (AG) (g) Since the -intercepts are of the form k e k, for k (M) then k+ e k+l and e (A) k Therefore, the -intercepts 0,,...k,... form a geometric sequence with 0 and a common ratio of e. (R) 5. Total distance k e (M) [e ] (M) k(e a/ ) metres (or equivalent eg k( e a/ )) (A) (C) Note: Award (C) if k is omitted in the final answer. 6. Let s be the distance from the origin to a point on the line, then 7. k + t / a k 0 s ( λ) + ( λ) a t / 0λ 4λ + 9 d( s ) 0λ 4 dλ (M) d( s ) 7 For minimum put 0, λ dλ 0 (A) The point is,,. 0 0 (A) (C) d( s ) Note: At this point > 0, but this is not required. dλ OR The position vector for the point nearest to the origin is perpendicular to the direction of the line. At that point: λ λ. 0 0 (M) Therefore, 0λ Therefore, λ 0 (A) Therefore, the point is,,. 0 0 (A) (C).0 [0] [] [] π (a) Area (π ) sin. (M)(A) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 6

20 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme (b) Maimum Area. units (A) (C) 8. (a) Given f () e sin (b) Then f () cos e sin f () cos e sin sin e sin (C) (A) (C) e sin (cos sin ) (M) For the point of infleion, put f () 0 e sin (cos sin ) 0 (or equivalent) (A) (C) Note: Award (C) if the candidate only writes f () 0 9. y y f () [] [] min inf 0 ma min 4 0 inf 0 (a) Minimum points (A) (C) (b) Maimum point (A) (C) (c) Points of infleion (A) (C) Note: There is no scale on the question paper. For eaminer reference the scale has been added here and the numerical answers are minima at and, maimum at 0 and points of infleion at.79 and.. 0. (a) (i) y ln 5.5 [] (b) asymptote asymptote (G) Note: Award (G) for correct shape, including three zeros, and (G) for both asymptotes (ii) f () 0 for 0.599,.5,.5 (G)(G)(G) 5 f () is undefined for ( 5 ) 0 (M) ( ) 0.5 Therefore, 0 or / (A) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 4 of 6

21 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme (c) f () (M)(A) or 5 4 (d) (e). (a) f () is undefined at 0 and / (A) For the -coordinate of the local maimum of f (), where 0 < <.5 put f () 0 (R) (M) 6 5 (A) The required area is.5 A f ( ) (A) Note: Award (A) for each correct limit. y 4 y R Q O y ln P Note: Award (C) for y, (C) for y ln. (C) (b) + ln 0 when Therefore, the -coordinate of P is (G) (c) The tangent at P to y has equation y , (G) and the tangent at P to y ln has equation y (G) Thus, the area of triangle PQR ( )(0.548). (M) 0.0 ( sf) (A) OR y (M) Therefore, the tangent at (p, p ) has equation p y p. (C) y ln (M) Therefore, the tangent at (p, p ) has equation + py p + p. (C) Thus, Q (0, p ) and R (0, p + ). Thus, the area of the triangle PQR (p + )p (M) 0.0 ( sf) (A) 6 (d) y when a, a (C) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 5 of 6

22 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme y ln when a, a (a > 0) (C) Now, (a) for all a > 0. a (M) Therefore, the tangents to the curve at a on each curve are always perpendicular. (R)(AG) 4. (a) (i) a + bsin y, 0 < a < b b + a sin ( b + a sin )( b cos ) ( a + b sin )( a cos ) ( b + a sin ) (M)(C) b cos + absin cos a cos absin cos (M)(C) ( b + a sin ) ( ) b a cos ( b + a sin ) (AG) 4 (ii) 0 cos 0 since b a 0. π This gives (+πk, k ) (M)(C) π a + b When, y, b + a π a b and when, y b a. Therefore, maimum y and minimum y. (A) 4 (iii) A vertical asymptote at the point eists if and only if b + a sin 0. (R) b Then, since 0 < a < b, sin <, which is impossible. a (R) Therefore, no vertical asymptote eists. (AG) (b) (i) y-intercept 0.8 (A) (ii) 4 For -intercepts, sin 4.069, (A) (iii) y [4] 0 π/ π m π/ n π (C) 5 (c) Area 5sin sin sin sin (M)(C) OR sin Area d sin (M)(C) [7] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 6 of 6

23 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme. y y f ( ) a b y maimum infleion infleion a minimum b Note: Award (A) for the sketch (A) for maimum and minimum points (A) for points of infleion (A)(A)(A) 4. Given s 40t + 0.5at ds, then the maimum height is reached when 0 (M) at (M) 40 a.6 (units not required) 5 (A) (C) 5. (a) f () ( ).5 [] [] ma ma zero zero 0.5 min 0.5 min zero.5 (A4) 4 Notes: Award (A) for the shape, including the two cusps (sharp points) at ±. (i) Award (A) for the zeros at ± and 0. (ii) Award (A) for the maimum at and the minimum at. (iii) Award (A) for the maimum at appro. 0.65, and the minimum at appro There are no asymptotes. Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 7 of 6

24 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme (b) (i) Let f () The candidates are not required to draw a scale. 4 Then f () ( ) + ( ) (M)(A) f () ( f () 7 ( ) (or equivalent) 7 f () (or equivalent) The domain is.4.4, ± (accept.4 < <.4, ±) (A) (ii) For the maimum or minimum points let f () 0 ie (7 ) 0 or use the graph. (M) Therefore, the -coordinate of the maimum point is (or 0.655) and (A) 7 the -coordinate of the minimum point is (or 0.655). (A) 7 7 Notes: Candidates may do this using a GDC, in that case award (M)(G). (c) The -coordinate of the point of infleion is ±.9 (G) OR 4(7 9) f (), ± (M) 4 9 ( ) For the points of infleion let f () 0 and use the graph, 9 ie.9. (A) 7 Note: Candidates may do this by plotting f () and finding the -coordinate of the minimum point. There are other possible methods. 6. METHOD (a) The equation of the tangent is y 4 8. (G) (C) (b) The point where the tangent meets the curve again is (, 0). (G) (C) METHOD (a) y 4 and at. (M) Therefore, the tangent equation is y 4 8. (A) (C) (b) 7. METHOD This tangent meets the curve when which gives ( + ) ( + ) 0. The required point of intersection is (, 0). (A) (C) Let S AP ( ) + ( + ). (M) The graph of S is as follows: ( ) ( 4 ) + ( ) ) [] [] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 8 of 6

25 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme S 4 / 4 0 The minimum value of S is (G) Therefore the minimum distance ( sf) (A) OR The minimum point is (0.68,.6) (G) The minimum distance is.6 ( sf) (G) (C) METHOD Let S AP ( ) + ( + ). (M) ds ( ) + 4( + ) 4( + + ) Solving + 0 gives 0.68 Therefore the minimum distance.6 ( sf) (A) (C) 8. METHOD b ( 0.68 ) + ( ) (G) [] a b a a b + ()() cosθ (M) ( cosθ ) (A) 4sin θ sin θ. (A) (C) METHOD O a b A M ab B AB ab Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 9 of 6

26 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme In ΔOAM, AM OA sin θ. (M)(A) Therefore, a b sin θ. (A) (C) 9. (a) π v(t) t sin t 0 when t 0, t or t 6 (C)(C)(C) (b) (i) π 6 π The required distance, d t sin t t sin t 0 (M) (ii) d (G) m. (A) OR (i) 6 π The required distance, d. t sin t 0 (M) (ii) d.5 m. (G)(A) OR (i) π 6 π The required distance, d t sin t t sin t 0 (M) (ii) 9 π π 6 π π d t sin t t sin t π 0 (C) 6 9 π π π π π π sin t t cos t sin t t cos t π 0 [from (i)] (C) 9 π (sin π π cos π sin π + π cos π + sin π π cos π) 6 m (.5 m). π (A) 4 8 Note: Award (A) for ± ( ± 5.7) which is obtained by integrating v π [] from 0 to (a) The derivative can be found by logarithmic differentiation. Let y f (). (b) y ln y ln y ln ln + y (M) (M)(M) ln y y ln that is, f () f () (AG) This function is defined for positive and real numbers only. To find the eact value of the local maimum: y 0 ln e (M) y e e To find the horizontal asymptote: (A) [7] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 0 of 6

27 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme ln lim ( y ) lim ln y lim 0 lim y (M)(A) y ( ) (e,e e y (A) 5 (c) By Taylor s theorem we have f (e) P() f (e) + f (e)( e) + ( e) (A) ln ln f () f () + f ( ) (M) Also, f (e) 0, and f (e) 0 + f (e) e e e e (M)(A) e e hence P() e e e e ( e) which is a parabola with verte at e and P(e) e e f (e) (R)(AG) 5 + t. (a) Distance (M)(A) 0 arctan t (A) (A) (C4) dv t (b) Acceleration (M) ( + t ) (A) (C). y + y + 4y + 0 (M)(A) d ( y + 4y) or y (A) y + At (,), (A) Equation of tangent is y l( ) or + y (A) (C6) e e. (a) A (M)(A) (C) tan [] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 6

28 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme (b) d A ( ) e (A) da 0 when (A) e Ama (or 0.858) (A) (C4) 4. denotes pts of infleion y v(t) 6t 6t 6t (t ) METHOD Note: Award (A) for the shape, and (A) for each point of infleion. Distance travelled 6t 6t d + 6t 6t d (M)(M) METHOD ( ) t ( ) t (G)(G) 6 m. (G) (C6) ( ) t ( ) t Distance travelled 6t 6t d + 6t 6t d (M)(M) 0 0 [t t ] + [t t ] (A)(A) ( ) + (7) () 6 m. (A) (C6) Note: Award (G)(or (A)) if the units are missing. 6. METHOD y e e + e (M)(A) d y e + e (M)(A) e ( + ) (A) Therefore the -coordinate of the point of infleion is (A) (C6) METHOD Sketching y f '() vt () (A)(A)(A) (C6) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 6

29 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme y f () 0 (G4) f '() has a minimum when (G) Thus, f () has point of infleion when (G) 7. dv 8 (cms ), V πr dv > 4πr dr (M)(A) dv dv dr dr dv dv > dr dr (M) dr When r, 8 (4π ) (M)(A) π 8. y 4 (cm s ) (do not accept 0.59) (A) (C6) y B( a, b) 0 A(6, 0) b a AB (a 6) + a 4 (M)(A) Minimum value of ( 6) + 4 occurs at. (G) > a. (G) (C6) 9. (a) (A)(A) 6 4 A g () f () Note: Award (A) for showing the basic shape of f (). Award (A) for showing both the vertical asymptote and the basic shape of g (). Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 6

30 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme (b) (i) is the vertical asymptote. (A) (ii) -intercept: 4.9 ( e ) (G) y-intercept: y 0.90 ( ln ) (G) (c) f () g ().4 or.05 (G)(G) (d) (i) See graph (e) (ii) Area of A 4 (ln ( + ) ) (M)(A) (iii) Area of A 0.6 (G) 4 y f () g () y 5 + ln( + ) Maimum occurs when 0 0. METHOD (M) (A) y 4.6 (A) OR Vertical distance is the difference f () g (). (M) Maimum of f () g () occurs at (G) The maimum value is 4.6. (G) y + y 0 (M)(A) At (, ), (M) Gradient of normal (A) Equation of normal is y ( ) (A) (C6) METHOD (A) y (A) (M)(A) 4 5 when (A) Gradient of normal ( ( ) ) Equation of normal is y ( ) (A) (C6) (A) [4] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 4 of 6

31 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme. Distance travelled sin t +. First integral 0.60 Second integral (±)0. Distance travelled 0.85 y π e t π t e sin td 0 π t (G) (G) (G) (C6) C A B (C6) Notes: Award (A) for zero at A, (A) for correct (concave) shape to left of dotted line, (A) for correct (concave) descent to right of dotted line, (A) for zero at B, (A) for maimum at C, (A) for asymptotic to -ais as. Please note that the first and fourth (A) marks are given for the candidate s graph hitting the -ais at A and B. No marks are given for the eact shape of this graph at A and B since it is not possible to deduce from the given graph whether or not the gradient of y is continuous at these points.. (a) (i) f () (M)(A) (AG) (ii) f "() (M)(A) (A) 5 Note: Award the second (A) for some form of simplification, eg accept. (b) (i) ln 0 giving (M)(A) (c) Note: Award (M)(A0) for.89. (ii) With this value of, f () < 0 (M)(A) Therefore, a maimum. (AG) 4 Points of infleion satisfy f (0) 0, ie (ln ) 4 ln ln ± ln ln ln ln ( ln) 4ln + [ ] [ ln] venumber 8( ln) ( ln) ( ln 4) ln + ln (M) (A) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 5 of 6

32 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme ± ln ( 0.845, 4.9) (A) OR 0.845, 4.9 (M)(G)(G) 4. tan θ sec dθ θ (M) π when θ, and sec θ 4 sec θ dθ (A)(A) (M) (4) 60 km s 5 40 km h (A) The airplane is moving towards him at 40 km h (A) (C6) Note: Award (C5) if the answer is given as 40 km h. 5. Let h height of triangle and θ CÂB. (A) Then, h 5 tan θ h d 5sec θ θ (A) (M)(A) π Put θ. dθ 5 4 (A) d θ 8 rad per sec Accept per second or 5.7 per second 0 π (A)(A)(C6) Note: Award (A) for the correct value, and (A) for the correct units y 9y 0 Differentiating w.r.t. + y 9y 9 0 Note: Award (A)for + y, and (A) for 9y 9. 9y y 9 EITHER at point (, 4) gradient 0.8. Gradient of normal.5 OR (A) (A) (A)(A) (A) [] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 6 of 6

33 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme y + 9 Gradient of normal 9y (A) at point (, 4), gradient is.5 (A) THEN Equation of normal is given by y 4.5( ) or y (A) (C6) 7. (a) V 500 cm πr 500 h 500 h πr (A) Now S πr + πrh S πr r (M)(A) (C) (b) ds 000 4πr dr r (A) d S for min S we need 0 dr 50 r (or r 4.0) π (A) (A) 8. (a) s 50t 0t v ds (M) 50 0t A (N) (b) Displacement is ma when v 0, M ie when t 5. A Substituting t 5, s (M) s 06.5 m A (N) 4 9. EITHER Let s be the distance from the origin to a point on the line, then s (l λ) + ( λ) + 4 (M) 0λ 4λ + 9 A d( s ) 0λ 4 A dλ For minimum ( s ) 0, λ 7 dλ 0 A OR The position vector for the point nearest to the origin is perpendicular to the direction of the line. At that point: λ λ 0 0 (M)A Therefore, 0λ 7 0 A Therefore, λ 7 0 A THEN, y 0 0 (A)(A) d (C) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 7 of 6

34 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme The point is,,. N 0 0 ln 40. (a) (i) Attempting to use quotient rule f () (M) ln f () A ( ln ) f () 4 M ln f () A Stationary point where f () 0, M ie ln, (so e) A f (e) < 0 so maimum. RAG N0 (ii) Eact coordinates e, y e AA N 9 (b) Solving f (0) 0 M n (A) e (4.48) A N (c) 5 ln Area d A EITHER Finding the integral by substitution/inspection u ln, du (M) (ln ) d u u u MA 5 ( ) (n ) Area (n 5) (ln) A Area (ln 5) (.0) A N 6 OR Finding the integral I by parts (M) u ln, dv du, v ln I uv udv (n ) ln (n ) I M I (ln ) (n ) I A 5 (n ) area (n 5) (n ) A Area (ln 5) (.0) A N 6 Note: Award N for.0 with no working. ( ) Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 8 of 6

35 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme (d) Using V y (M) 5 π n A.8 A N 4. k + (M)(A) When, gradient of normal k 4 + k 4 (A) (A) (M)(A) (C6) p p 4. (a) (i) f () pe ( + ) + e (A) 4. p e p( + ) + (ii) The result is true for n since LHS e p p ( + ) + (AG) and RHS e p p p p( + ) + e p( + ) +. (M) ( k) k p Assume true for n k: f ( ) p e p( + ) + k (M) (M)(A) k p p e p( + ) + k+ (A) Therefore, true for n k true for n k+ and the proposition is proved by induction. (R) 7 (b) (i) f ( ) e ( + ) + 0 (M) (c) (ii) area f ( )d + f ( ) OR b π a area f ( ) 8.08 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( k ) ( k) k p k p f + ( ) f ( ) p pe p( + ) + k + p e p ( ) ( ) + + ( ) f + + ( ) e ( ) f() e ( + ) EITHER 8.08 (A) N (M) (M) (A) N (M) (A) N 4 (A) N [] [] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 9 of 6

36 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme O r A C N B (a) h rsinθ CB CN rcos h Using T ( r+ CB) θ (M) (A)(A) 44. r T (sinθ + sinθcos θ) r (sinθ + sin θ) (A) (AG) N0 4 (b) dt r (cosθ cos θ) 0 (for ma) dθ cosθ + (cos θ ) 4cos θ + cosθ 0 (M) (M)(AG) cosθ 0.59 ( θ 0.959) (A) d T r ( sinθ 4sin θ) (M) dθ d T θ r < 0 dθ there is a maimum (when θ ) (R) 5 (c) θ In triangle AOB: AB r sin (M)(A) θ Perimeter OABC r+ rcosθ + rsin 75 (M) When θ 0.959, r 8.5 cm (A) r 8.5 Area OABC (sinθ + sin θ) (sin sin.87) 96 cm (M) (A) N 6 y + 4 (M) at 0, ( )( + ) 0 (M) 7, ; y 5, So P(, 5) and Q 7, (A)(A) [5] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 0 of 6

37 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme 45. y + 5 Equation of (PQ) is (M) y y y y 9 0 (A) (C6) O A B Arc length, s 4θ d s dθ 4 A da dθ 8( cosθ) Segment area, When numerically equal, dθ 4 8 cosθ ( cosθ) ( 4) ( θ sin θ) 8( θ sin θ) dθ (A) (A) (A) (A) (M) 46. cosθ π θ Accept 60 ( ) E (A) (C6) B A P D F G Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 6

38 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme Let BPˆD α and APˆ D β then θ α β tan α tan β tan θ tan (α β) + tan α tan β at maimum (ab ) 0, b at ab (M) (A) (M) (A) a (A) (M) (A) (A) since < 0 at this value is a maimum. (R) 47. METHOD d, d Line and graph intersect when + 4 m + ie ( + m) + 0 y m + tangent to graph b a ( b a) ab + ab + tan θ + ab b a + ab ( + m) + 0 has equal roots i.e b 4ac 0 (A) ( + m) 6 0 A m 5, m 7 A N METHOD f () 6 A + 4 m + Substitute m 6 into above ( ) ( )( ) ( b a) ( tan θ) d d ( tan θ) d d ( ) ( b a)( ab ) ( + ab) ab M (A) (M) (M) M (A) ± A m5, m 7 A N ( ) ( ) ( ) ( )( + ab 4 ab + ab ) ( ) b a 4 + ab ( b a) ( ) [ ] ab 4ab+ 4 + ab ( b a)( 6ab) ( + ab) ( tan θ) ( b a)( 4ab ab ) 8a b b a ab ( ) a b ab [9] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 6

39 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme 48. (a) maimum when (or any correct method) (M) t t, maimum 6 (m s ) A N (b) Using acceleration M (A) (A) when t 4, a (m s ) A N (c) using (M) (d) d v A s A d v 0 maimum v A dv e 5 0.t v or s s A t + t + t + c 6 when t 0, s A 0 c 0 t t sb e 5e + d when t 0, s B 5 d 0 s t + t + t A 6 0.t s B 5e (i) s A B v B R A M A (M) A A N N AAAA Note: For each curve, award A for the s intercept and A for the correct shape. (ii) t.95 and t 7.8 AA 49. (a) ( ) f MA N (b) Hence when, gradient of tangent (A) t [] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page of 6

40 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme gradient of normal is 7 y ln 7 ( ) 7 4 y + + ln 7 (accept y ) 50. (a) A π (M) d A d 4π d 0 d t 4π π (b) V π A 5 A 5. (a) Using quotient rule ln f ( ) A 6 ln 4 A N 4 4 ( ln ) f ( ) MA ln 5 A N (b) (i) For a maimum, f () 0 giving (M) ln EITHER (A) M A A A M A MA maimum AG N0 OR for ( ) d V d π 5 π 4π e f e 7 7 < 0 5 e e <, f ( ) > 0 N4 N Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 4 of 6

41 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme (ii) (iii) for e >, f ( ) < 0 MA maimum AG N0 7 f ( 0 ) 0 ln ( ) M e 7 (.79) f (.5) 0.8 f () 0.04 Note: Accept any two sensible values either side of.79. Change of sign point of infleion A A A R Note: Award A for shape, A for marking values which locate the maimum point and point of infleion correctly. AA (c) Using V πy (M) ln π (A) A N (d) ln Area A A Using integrating by parts (M) A ln + ln 8 4 AA ln ln A ( 4 ln ) 8 AG N π π ( ) + π ( ) h (M) 8π 6π + 9πh h 5 (cm) A A Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 5 of 6

42 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme 4 V π r +πr h Attempt to differentiate with respect to time. V dr 4πr + dr dh π rh +πr M A A dh 04π 4π ( ) ( ) + π ( )( 5)( ) + π ( ) dh 04π 7π + 60π + 9π dh 7π 9π dh 8 (cm/min) A N 5. METHOD Attempting to integrate the RHS Note: The A above must include C. Using s (0) gives C ( ) A A (M) Solving 0 for t (M) t.8 (sec) A N METHOD A AA Solving for T (M) T.8 (sec) A N 54. z + y (or equivalent) (M) Attempting to differentiate implicitly with respect to t Rate is 90 (km h ) A N0 55. (a) (i) ( t 4t ) s + s t t + t + C t t + t T 0 ( t 4t + ) ds ( ) 0 z (,initially) dz z + y dz ( ) ( ) s t t + t A M A A Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 6 of 6

43 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme Attempting to epress α and β in terms of arctan (M) α arctan and β arctan AA θ α β A θ arctan arctan AG N0 (ii) METHOD Attempting to differentiate either arctan or arctan. d arctan d arctan + + M AA A dθ Simplifying, ( + 4)( + ) A d θ For a maimum 0 (R) + 4 ( + ) (or equivalent) (as > 0 ) A N EITHER Justifying the maimum using appropriately chosen -values M eg when, and when, Correct gradients calculated for chosen values. A dθ dθ eg, and, 0 0 Showing that dθ > 0 for < and dθ < 0 for > R is a maimum. AG N0 Note: Award MAR for a clearly labelled OR dθ sketch of either θ or against. A Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 7 of 6

44 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme Attempting to find a second derivative. 4 d θ 4 4 ( + ) ( + ) 4 + d θ When ( 0.57). 9 Since < 0 for, R then is a maimum. AG N0 METHOD Given that 0 < θ < tanθ is a maimum., θ will be a maimum when Using to get tanθ in terms of. (M) M A R A MA For a maimum 0 (R) 0 A (as > 0) A N EITHER Justifying the maimum using appropriately chosen -values M eg when, Since > 0 for < and < 0 for > then is a maimum. AG N0 Note: Award MAR for a clearly labelled d sketch of either tanθ or ( tan θ) against. OR Attempting to find a second derivative. M d ( ) ( 6) tan θ + When Since ( 4) ( ) ( ) + 4 d θ π tan α tan β tan θ + tan α tan β tan θ + + d ( ) ( + ) ( ) tan θ + d ( ) tanθ d 9 d tanθ ( ) ( + ) d ( tan θ) and when, ( tan θ) d ( ) ( ) d then is a maimum. AG N0 d tanθ ( ) 8 ( tan θ) ( 0.77) ( tan θ) < 0for, 8 A R A R Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 8 of 6

45 IB Math High Level Year - Calc Practice: Differentiation Applications - MarkScheme (iii) METHOD arctan Substituting into θ arctan (M) θ arctan arctan ( 0.40 radians ). A N METHOD Substituting into θ arctan + (M) 56. (a) (b) θ arctan 4 ( 0.40 radians ). A N (b) METHOD π Attempting to solve arctan arctan 5 for M 0.649,.08 (m) AA N4 METHOD Attempting to solve tan5 (or equivalent) for + M 0.649,.08 (m) AA N4 y e e y e d y when 0, ( < 0). d y when ±, e ( > 0). Hence the points are points of infleion. 57. (a) y + 6 y 6 d + 4 ( + ) 0 e ± e + y + 6y 0 Note: Award A for each correctly differentiated term A MA M A R MAAAA A AG N0 + s 0 (b) (i) The equation of the tangent at A is r (M)AAN Note: Award MA0A for omitting r or for writing L. (ii) Substituting y + 6 y 6 M [] Macintosh HD:Users:Shared:Dropbo:Desert:HL:7Calculus:HL.CalcDiffAppsPractice.doc on 09/8/06 at 9:4 AM Page 9 of 6