Module 4 Practice Problems: Applications of Derivatives I
Instructions: Work through each problem on paper first, then reveal the solution.
Problem 1
A balloon rises at 3 m/s from a point 100 m away from an observer. How fast is the distance from observer to balloon increasing when the balloon is 100 m high?
Solution
Let y = height, d = distance. d² = 100² + y². Differentiate: 2d(dd/dt) = 2y(dy/dt). When y = 100: d = 100√2. dd/dt = (100)(3)/(100√2) = 3√2/2 ≈ 2.12 m/s.
Problem 2
Oil spills in a circle. The radius grows at 0.5 m/min. How fast is the area growing when r = 20 m?
Solution
dA/dt = 2πr(dr/dt) = 2π(20)(0.5) = 20π m²/min.
Problem 3
Use linearization of f(x) = √x at a = 25 to approximate √26.
Solution
f(25) = 5, f'(25) = 1/10. L(26) = 5 + 0.1(1) = 5.1. (Actual: 5.0990...)
Problem 4
A sphere's radius is measured as 8 cm with possible error ±0.05 cm. Estimate the maximum error in the surface area (S = 4πr²).
Solution
dS = 8πr dr = 8π(8)(0.05) = 3.2π ≈ 10.05 cm².
Problem 5
Find the absolute max and min of f(x) = x³ − 6x² + 9x + 2 on [0, 4].
Solution
f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3). Critical points: x = 1, 3. f(0) = 2, f(1) = 6, f(3) = 2, f(4) = 6. Abs max = 6 at x = 1 and x = 4. Abs min = 2 at x = 0 and x = 3.
Problem 6
Find the critical points of f(x) = x4 − 4x³.
Solution
f'(x) = 4x³ − 12x² = 4x²(x − 3). Critical points: x = 0 and x = 3.
Problem 7
Verify the MVT for f(x) = x² − x on [0, 3] and find c.
Solution
f is polynomial: continuous and differentiable. Average rate: [f(3) − f(0)]/3 = 6/3 = 2. f'(x) = 2x − 1 = 2 gives x = 3/2, which is in (0, 3).
Problem 8
A 15-ft ladder: the base slides at 1 ft/s. How fast is the angle between the ladder and the ground changing when the base is 9 ft from the wall?
Solution
cos θ = x/15. −sin θ (dθ/dt) = (1/15)(dx/dt). When x = 9: sin θ = 12/15 = 4/5. −(4/5)(dθ/dt) = 1/15. dθ/dt = −1/12 rad/s.
Problem 9
Use linearization to approximate (1.02)5.
Solution
f(x) = x5 at a = 1. f(1) = 1, f'(1) = 5. L(1.02) = 1 + 5(0.02) = 1.1. (Actual: 1.10408...)
Problem 10
Show that x³ + 3x + 1 = 0 has exactly one real root. (Hint: use MVT consequences.)
Solution
f(x) = x³ + 3x + 1. f'(x) = 3x² + 3 > 0 for all x, so f is strictly increasing. A strictly increasing function crosses the x-axis at most once. Since f(−1) = −3 < 0 and f(0) = 1 > 0, by the IVT there is at least one root. Therefore there is exactly one real root.