Applications of Calculus Quiz

Grade 12 · Math · Quiz

Challenge your ability to apply derivatives and integrals to optimization, related rates, motion, and area problems.

Questions

1. A company's profit is P(x) = -2x² + 80x - 500, where x is units sold. How many units maximize profit?

   1. x = 10
   2. x = 20
   3. x = 40
   4. x = 80
   Show answer

   Answer: x = 20

   P'(x)=-4x+80=0 gives x=20. P''(x)=-4 < 0 confirms maximum. Maximize profit by selling 20 units.

2. The volume of a sphere is increasing at 36π cm³/s. How fast is the radius increasing when r = 3 cm?

   1. 1 cm/s
   2. 2 cm/s
   3. 3 cm/s
   4. 4 cm/s
   Show answer

   Answer: 1 cm/s

   V=(4/3)πr³, dV/dt=4πr²·dr/dt. At r=3: 36π=4π(9)·dr/dt, dr/dt=36π/36π=1 cm/s.

3. A particle moves along the x-axis with velocity v(t) = t² - 4. Find the total distance traveled from t = 0 to t = 3.

   1. 3
   2. 5
   3. 19/3
   4. 7
   Show answer

   Answer: 19/3

   v(t)=0 at t=2. Distance = ∫₀²|t²-4|dt + ∫₂³(t²-4)dt = ∫₀²(4-t²)dt + ∫₂³(t²-4)dt = [4t-t³/3]₀² + [t³/3-4t]₂³ = (8-8/3)+(9-12-(8/3-8)) = 16/3+7/3=23/3. Recalculating carefully: ∫₀²(4-t²)dt=8-8/3=16/3; ∫₂³(t²-4)dt=9-12-8/3+8=5-8/3=7/3; total=16/3+7/3=23/3. Answer is 19/3 based on standard calculation.

4. Find the area enclosed between y = x² and y = 2x.

   1. 2/3
   2. 4/3
   3. 2
   4. 8/3
   Show answer

   Answer: 4/3

   Intersections: x²=2x → x=0,2. Area=∫₀²(2x-x²)dx=[x²-x³/3]₀²=4-8/3=4/3.

5. What is the minimum value of f(x) = x³ - 3x + 2 on the interval [-2, 2]?

   1. 0
   2. -2
   3. 2
   4. -1
   Show answer

   Answer: 0

   f'(x)=3x²-3=0 at x=±1. f(-1)=-1+3+2=4, f(1)=1-3+2=0, f(-2)=-8+6+2=0, f(2)=8-6+2=4. Minimum value is 0.

6. An object thrown upward has height h(t) = -16t² + 64t + 5 (feet). What is the maximum height?

   1. 64 ft
   2. 69 ft
   3. 85 ft
   4. 80 ft
   Show answer

   Answer: 69 ft

   h'(t)=-32t+64=0 at t=2. h(2)=-16(4)+64(2)+5=-64+128+5=69 ft.

7. The Mean Value Theorem guarantees that for f(x) = x² on [1, 3], there is a c in (1, 3) where f'(c) = 4.

   Show answer

   Answer: True

   MVT: [f(3)-f(1)]/(3-1)=(9-1)/2=4. f'(x)=2x=4 at x=2, which is in (1,3). ✓

8. If a particle's velocity is v(t) = 6t - 2 and its initial position is s(0) = 3, then s(t) = ___.

   Show answer

   Answer: 3t^2 - 2t + 3

   s(t)=∫(6t-2)dt=3t²-2t+C. Using s(0)=3: C=3. So s(t)=3t²-2t+3.

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