Mathematical Methods for Actuarial Science I: Exam Questions

MATH1821 Mathematical Methods for Actuarial Science I Examination Questions (17 - 18) Answer All Eight (8) Questions 1. Evaluate the following limits. (a) (3 points) lim x →∞ x 1 / √ x (b) (3 points) lim x → 1 + ( 1 x - 1 - 1 ln x ) (c) (5 points) lim n →∞ [2 n (2 n - 1) · · · ( n + 1)] 1 /n n Solution. (a) Since lim x →∞ ln x 1 / √ x = lim x →∞ ln x √ x = lim x →∞ (ln x ) ′ ( √ x ) ′ = lim x →∞ 1 x 1 2 √ x = lim x →∞ 2 √ x = 0 , lim x →∞ x 1 / √ x = e 0 = 1. (b) Direct verification: lim x → 1 + ( 1 x - 1 - 1 ln x ) = lim x → 1 + ln x - x + 1 ( x - 1) ln x = lim x → 1 + (ln x - x + 1) ′ (( x - 1) ln x ) ′ = lim x → 1 + 1 x - 1 ln x + 1 - 1 x = lim x → 1 + 1 - x x ln x + x - 1 = lim x → 1 + (1 - x ) ′ ( x ln x + x - 1) ′ = lim x → 1 + - 1 ln x + 1 + 1 = - 1 2 (c) Let a n = [2 n (2 n - 1) · · · ( n + 1)] 1 /n n . Then ln a n = ln [ 2 n (2 n - 1) · · · ( n + 1) n n ] 1 /n = 1 n ln 2 n (2 n - 1) · · · ( n + 1) n n = 1 n ln ( n + 1 n · n + 2 n · · · 2 n n ) = 1 n [ ln ( 1 + 1 n ) + ln ( 1 + 2 n ) + · · · + ln ( 1 + n n ) ] is a Riemann sum of ln x over the interval [1 , 2]. Hence lim n →∞ ln a n = ∫ 2 1 ln x dx = [ x (ln x - 1) ] 2 1 = 2 ln 2 - 1 = ln(4 e - 1 ) and lim n →∞ a n = 4 e - 1 . 1

2. (5 points) Show that the equation x 3 - 2 x 2 + cos x = 0 has a unique solution between 0 and 1. Solution. Let f ( x ) = x 3 - 2 x 2 + cos x for all x . Then f is continuous. We have f (0) = cos 0 = 1 > 0 and f (1) = 1 - 2 + cos 1 = - 1 + cos 1 < - 1 + 1 = 0 . By the Intermediate Value Theorem, there is an x ∈ (0 , 1) such that f ( x ) = 0. To show the uniqueness, note that f ′ ( x ) = 3 x 2 - 4 x - sin x = - 3 x (1 - x ) - x - sin x < 0 for all x ∈ (0 , 1). The function f is decreasing and hence one-to-one there. The solution is unique. 2

3. (5 points) For every positive integer n , let I n = ∫ x cos n x dx. Prove the reduction formula I n +2 = n + 1 n + 2 I n + x sin x cos n +1 x n + 2 + cos n +2 x ( n + 2) 2 . Solution. Integrate I n +2 = ∫ x cos n +2 x dx by parts using u = x cos n +1 x and dv = cos x dx . Then du = [cos n +1 x - ( n + 1) x sin x cos n x ] dx and v = sin x. We get I n +2 = x sin x cos n +1 x - ∫ [cos n +1 x - ( n + 1) x sin x cos n x ] sin x dx = x sin x cos n +1 x + ∫ cos n +1 x d (cos x ) + ( n + 1) ∫ x (1 - cos 2 x ) cos n x dx = x sin x cos n +1 x + cos n +2 x n + 2 + ( n + 1) I n - ( n + 1) I n +2 . Therefore, ( n + 2) I n +2 = ( n + 1) I n + x sin x cos n +1 x + cos n +2 x n + 2 . The reduction formula follows by dividing both sides by n + 2. 3

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4. (5 points) Find the area of the surface generated by revolving the curve 9 x 2 = y (3 - y ) 2 for 0 ≤ y ≤ 3 about the y -axis. Solution. The curve is symmetric about the y -axis, we can think of the surface as generated by revolving the curve x = √ y (3 - y ) 3 = y 1 / 2 - 1 3 y 3 / 2 , 0 ≤ y ≤ 3 , about the y -axis. We have dx dy = 1 2 y - 1 / 2 - 1 2 y 1 / 2 and 1 + ( dx dy ) 2 = ( y - 1 / 2 + y 1 / 2 2 ) 2 . Hence the surface area is S = ∫ 3 0 2 πx √ 1 + ( dx dy ) 2 dy = 2 π ∫ 3 0 3 y 1 / 2 - y 3 / 2 3 · y - 1 / 2 + y 1 / 2 2 dy = π 3 ∫ 3 0 (3 + 2 y - y 2 ) dy = π 3 [ 3 y + y 2 - y 3 3 ] 3 0 = 3 π. 4

5. (5 points) A spherical balloon is being inflated so that its volume is increasing at a constant rate of 104 π cm 3 per second. It will burst when its surface area reaches 400 π cm 2 . If initially, the radius of the balloon is 4 cm, when will the balloon burst? (The volume and surface area of a sphere with radius r are V = 4 πr 3 / 3 and S = 4 πr 2 respectively.) Solution. We have 4 πr 2 dr dt = dV dt = 104 π = ⇒ r 2 dr dt = 26 . Hence ∫ r 2 dr = 26 ∫ dt = ⇒ r 3 = 78 t + C. When t = 0, r = 4 so that C = 64 and r 3 = 78 t + 64. Now the balloon will burst when 4 πr 2 = S = 400 π or r = 10. At that instant, 1000 = 78 t + 64, i.e., t = 12. Therefore the balloon will burst 12 seconds later. Alternatively, the balloon will burst when its volume is 4000 π/ 3, which occurs at t = ( 4000 π 3 - 256 π 3 ) ÷ 104 π = 12 (s) . 5

6. (5 points) A right circular cone is inscribed in a sphere of radius 3. What dimensions will give the cone its largest lateral surface area? (The lateral surface area of a cone with height h and base radius r is S = πr √ r 2 + h 2 .) Solution. Let the height and base radius of the cone be h and r respectively. Then (3 - h ) 2 + r 2 = 3 2 when h ≤ 3 and ( h - 3) 2 + r 2 = 3 2 when h > 3. In both cases, h 2 + r 2 = 6 h and r 2 = 6 h - h 2 . The lateral surface area of the cone is S = πr √ r 2 + h 2 = π √ 6 h (6 h - h 2 ) for 0 ≤ h ≤ 6 . We have dS dh = √ 6 π (12 h - 3 h 2 ) 2 √ h (6 h - h 2 ) = 3 √ 6 πh (4 - h ) 2 √ h (6 h - h 2 ) . The only critical point in the interior of the domain of S is h = 4. We have S (4) = 8 √ 3 π and S (0) = S (6) = 0 . Hence the cone will have the largest lateral surface area when its height is h = 4 and base radius r = √ 24 - 16 = 2 √ 2. 6

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7. (a) (3 points) Use the substitution x = 2 sin θ , or otherwise, show that ∫ √ 4 - x 2 dx = 2 sin - 1 x 2 + x √ 4 - x 2 2 + C. (b) (5 points) A 2-meter ladder is leaning against a wall when its base is sliding away from the wall at a speed inversely proportional to the distance from the top of the ladder to the ground. If initially, the ladder is standing vertically and the base is sliding away at 0.25 meter per second, how long will it take for the ladder to lie flat on the ground? Solution. (a) Putting x = 2 sin θ for - π/ 2 ≤ x ≤ π/ 2, then √ 4 - x 2 = 2 | cos θ | = 2 cos θ and dx = 2 cos θ dθ . Hence ∫ √ 4 - x 2 dx = ∫ 2 cos θ · 2 cos θ dθ = 4 ∫ cos 2 θ dθ = 4 ∫ 1 + cos 2 θ 2 dθ = 2 θ + sin 2 θ + C = 2 θ + 2 sin θ cos θ + C = 2 sin - 1 x 2 + 2 · x 2 · √ 4 - x 2 2 + C = 2 sin - 1 x 2 + x √ 4 - x 2 2 + C. (b) Introduce the variables x and y as in the picture below: From the given conditions, dx/dt = k/y for some constant k , and, when t = 0, x = 0 and dx/dt = 1 / 4. Since y = 2 when t = 0, we have k/ 2 = 1 / 4 and so k = 1 / 2. Hence dx dt = k y = 1 2 √ 4 - x 2 . Solve the separable DE: ∫ √ 4 - x 2 dx = 1 2 ∫ dt, 7

we get 2 sin - 1 x 2 + x √ 4 - x 2 2 = t 2 + C. As x = 0 when t = 0, C = 0. Now the ladder will lie flat on the ground when x = 2. At that moment, t = 2[2 sin - 1 (2 / 2) + 2 √ 4 - 2 2 / 2] = 2 π . Therefore the the ladder will lie flat on the ground 2 π seconds later. 8

8. (6 points) Sketch the graph of f ( x ) = ( x + 1) 3 ( x - 1) 2 . Solution. The domain of f is { x : x ̸ = 1 } and it is neither even nor odd. The first two derivatives of f are f ′ ( x ) = 3( x + 1) 2 ( x - 1) 2 - 2( x + 1) 3 ( x - 1) 3 = ( x + 1) 2 (3 x - 3 - 2 x - 2) ( x - 1) 3 = ( x + 1) 2 ( x - 5) ( x - 1) 3 and f ′′ ( x ) = 2( x + 1)( x - 5) ( x - 1) 3 + ( x + 1) 2 ( x - 1) 3 - 3( x + 1) 2 ( x - 5) ( x - 1) 4 = ( x + 1)[2( x - 1)( x - 5) + ( x + 1)( x - 1) - 3( x + 1)( x - 5)] ( x - 1) 4 = ( x + 1)(2 x 2 - 12 x + 5 + x 2 - 1 - 3 x 2 + 12 x + 15) ( x - 1) 4 = 24( x + 1) ( x - 1) 4 . Hence f has two critical points at - 1 and 5. Together with the point 1, where f is not defined, we divide R into 4 subintervals: Interval x < - 1 - 1 < x < 1 1 < x < 5 5 < x Sign of f ′ + + - + Behaviour of f increasing increasing decreasing increasing We see that f does not have an extreme value at x = - 1, but a local minimum at x = 5. When x = 5, y = 27 / 2. The second derivative f ′′ vanishes when x = - 1, together with the point 1, we have Interval x < - 1 - 1 < x < 1 1 < x Sign of f ′ - + + Behaviour of f concave down concave up concave up The point ( - 1 , 0) is a point of inflection. Combining the two tables, we get x < - 1 - 1 < x < 1 1 < x < 5 5 < x increasing increasing decreasing increasing concave down concave up concave up concave up Since f ( x ) = ( x + 5) + 12 x - 4 ( x - 1) 2 , the graph has an oblique asymptote y = x + 5 and a vertical asymptote x = 1 at which lim x → 1 - f ( x ) = lim x → 1 + f ( x ) = ∞ . Moreover, there is another y -intercept at (0 , 1). We sketch the graph below. 9

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