Mathematics 36, section 1 -- AP Analysis

Practice Final Exam -- Solutions

December 12, 1997

I.

• A) In each case, the line is the tangent line to the graph at x = 0 -- the line with equation y = f'_i(0)x + 1. Computing the derivatives, you should get:
  1. f₁: y = 3x + 1
  2. f₂: y = 4x + 1
  3. f₃: y = 2x + 1.
• B) For y = f₁(x) + f₃(x), at x = 0, y = 2, and y' = f'₁(0) + f'₃(0) = 5. Answer: y = 5x+2.

II.

• A) f' is negative and decreasing asymptotically from 0 until about x = -4, negative but increasing on (-4,-2.1) (approximately), positive and increasing on (-2.1,0), positive but decreasing on (0,2.1), then negative and decreasing on (-2.1,4), then negative but increasing asymptotically to 0 from x = 4 on.
• B) (0,0) must be a local minimum on y = F(x) where F' = f. There are no other critical points for F.

III. The height function is increasing up to the time t = a that the left hand portion of the "W" is filled up to the level of the middle segment. It is concave down since the walls of the container slope outward -- the water is being poured at a constant volume per unit time so as the water level rises, more water is needed to produce the same increase in level. From time t = a to t = 2a, water will spill over into the right hand portion of the "W" and the water level on the left will not change. Then from that point the water level will be increasing and concave down as before.

IV.

• A) Taylor polynomial:
  p₃(x) = 1 + x/3 - x²/9 + 5x³/81
  Approximation: (1.1)^(1/3) approx = p₃(.1) = 1.03228395 (1.0323, rounded to 4 decimal places).
• B) From the error bound for Taylor polynomials,
  | error | <= max_{0 <= z <= .1} f⁽⁴⁾(z)/4! (.1)⁴
  = (1/24) max_{0 <= z <= .1} 80/(81 (1 + z)^11/3)(.1)⁴
  = (1/24) (80/81)(.1)⁴ (max is at z = 0)
  which is about .000004. The actual error (comparing answer from A with calulator value) is approx.
  |1.032280115 - 1.03228395| = .00000383 = 3.83 x 10^-6.

V.

• A) Half of a circular ring with inner radius a and outer radius a + Delta r, with area approx. Pi a Delta r.
• B) Using a right-hand Riemann sum with Delta r = 1:
  total pop. (in thousands) approx. = Pi(70 + 61.5 x 2 + 58 x 3 + 51 x 4 + 45 x 5 + 29 x 6 + 25 x 7 + 19.5 x 8
  which is approx. 4.087 x 10⁶.

VI.

• A) (Using the substitution u = sin 2x) (1/2) e^{sin 2x} + C.
• B) (Using parts, or number 27 in the table) (x⁸/8) ln(5x) - (x⁸/64) + C.
• C) (Using number 24 in the table, with a = 3, b = 4, c = 1, d = 3: 7 ln|x - 4| - 6 ln|x - 3| |₅⁶ approx. = 2.42.

VII.

• A) int₀^Pi sqrt(x) dx = (2/3) Pi^3/2.
• B) M = int₀^Pi x sin(x/2) dx. Using parts (or the table) int₀^Pi x sin(x/2) dx = -2x cos(x/2) + 4 sin(x/2)|₀^Pi = 4.

VIII.

• A) dT/dt = -k(T - A), k > 0. Separating variables and integrating: T = A + (T(0) - A)e^-kt.
• B) From given information T(0) = 0, A = 350 (temp of oven). So T = 350(1 - e^-kt). At t = 10, T = 50, so 50 = 350(1 - e^-10), or 1/7 = (1 - e^-10k), so k = -ln(6/7)/10 approx. = .0154. When T = 180, 180 = 350(1 - e^(-.0154)t). Solving for t, t = 46.9. So you'll need to wait another 36.9 minutes (get a snack to tide you over!)

IX.

• A) This is a logistic equation, so the general solution has the form P = M/(1 + a e^-Mkt). From the given information, M = 1000 (the leveling off point is the maximum sustainable population, M). At t = 0, P = 100, so 100 = 1000/(1 + a), and a = 9. Then at t = 1, P = 200, so 200 = 1000/(1 + 9e^{-1000 k}), so k = -ln(4/9)/1000.
• The population was growing the fastest at the time when P = 1000/2 = 500 (the rate of growth as a function of P is k P(M - P), which reaches its maximum when P = M/2). Solving 500 = 1000(1 + 9e^ln(4/9)t) for t, t = 2.71 weeks.

X.

• A) The spiral should start at the origin, loop counter-clockwise outward (the distance from the origin as a function of t is t, reach (2 Pi, 0) at t = 2 Pi, then loop around again, reaching (4 Pi, 0) at t = 4 Pi.
• We have x'(t) = -t sin(t) + cos(t), y'(t)= t cos(t) + sin(t) So sqrt((x'(t))² + (y'(t))²) = sqrt(1 + t²). The arclength is the integral of this from t = 0 to t = 4 Pi. Using table entries 30 (+ sign, a = 1) and 29, the value is:
  2 Pi sqrt(1+16 Pi²) + (1/2)ln(4 Pi + sqrt(1+16 Pi²))
• C) This is similar to question IV on Exam 3 -- differentiate x(t), y(t), and show the result is the same as the right-hand sides of the equations.