Linear Approximation Practice Problems with Answers

These linear approximation practice problems start with building a linearization, use it to estimate roots, sines, logs and powers, then move to differentials, measurement error and over or under estimates. Try each one first. The answers come right after the list, then the full solutions.

The linearization of f at x = a is its tangent line:

For x near a, f(x) \approx L(x). For the reasoning and more examples, read the linear approximation guide.

Problems

Linearization

1. Find the linearization of f(x) = \sqrt{x} at a = 4.
2. Use problem 1 to estimate \sqrt{4.1}.
3. Estimate \sqrt[3]{8.2}.

Estimating values

4. Estimate \sin(0.1).
5. Estimate \ln(1.05).
6. Estimate e^{0.02}.
7. Estimate (1.01)^{10}.
8. Estimate \cos 62^\circ.

Differentials and error

9. For y = x^3, find dy and \Delta y when x = 2 and dx = 0.1.
10. The radius of a ball is measured as 10 cm with a possible error of 0.05 cm. Estimate the possible error in the computed volume.

Using the result

11. Is the estimate in problem 2 too big or too small? Explain.
12. A function has f(3) = 5 and f'(3) = -2. Estimate f(3.2).

Answers

Step-by-step solutions

1. f(4) = 2 and f'(x) = \frac{1}{2\sqrt{x}}, so f'(4) = \frac{1}{4}:

2. Plug x = 4.1 into the line:

3. Use f(x) = \sqrt[3]{x} at a = 8, since \sqrt[3]{8} = 2. The derivative is f'(x) = \frac{1}{3}x^{-2/3}, so f'(8) = \frac{1}{3}\cdot\frac{1}{4} = \frac{1}{12}:

4. Use f(x) = \sin x at a = 0. f(0) = 0 and f'(0) = \cos 0 = 1, so L(x) = x:

This is the small-angle approximation \sin x \approx x, which only works in radians.

5. Use f(x) = \ln x at a = 1. f(1) = 0 and f'(1) = 1, so L(x) = x - 1:

6. Use f(x) = e^x at a = 0. f(0) = 1 and f'(0) = 1, so L(x) = 1 + x:

7. Use f(x) = x^{10} at a = 1. f(1) = 1 and f'(x) = 10x^9, so f'(1) = 10:

8. Work in radians. Use a = 60^\circ = \frac{\pi}{3}, where \cos\frac{\pi}{3} = \frac{1}{2} and the slope is -\sin\frac{\pi}{3} = -\frac{\sqrt{3}}{2}. The step is 2^\circ = \frac{2\pi}{180} = \frac{\pi}{90}:

9. The differential is dy = f'(x)\,dx = 3x^2\,dx:

The actual change is \Delta y = (2.1)^3 - 2^3 = 9.261 - 8 = 1.261. The differential is a close but simpler estimate.

10. The volume is V = \frac{4}{3}\pi r^3, so dV = 4\pi r^2\,dr:

The volume is \frac{4000\pi}{3} \approx 4189 cm³, so the relative error is \frac{20\pi}{4000\pi/3} = 0.015, or 1.5\%. That is three times the 0.5\% error in the radius, because the radius is cubed.

11. For f(x) = \sqrt{x}, f''(x) = -\frac{1}{4}x^{-3/2} < 0, so the graph is concave down. A concave down curve lies below its tangent lines, so the estimate 2.025 is a little too big. The actual value 2.02485 confirms it.

12. Use the linearization at a = 3:

Common mistakes on these problems

• Choosing a far away a. To estimate \sqrt{4.1}, use a = 4, not a = 1. The closer a is, the better the estimate.
• Using degrees. In problem 8 the step must be in radians, \frac{\pi}{90}, not 2.
• Mixing up dy and \Delta y. dy is the change along the tangent line, and \Delta y is the real change along the curve.
• Plugging in x - a wrong. In problem 3 the step is 8.2 - 8 = 0.2, not 8.2.

Keep practicing

These 12 problems use fixed numbers. For unlimited practice with new numbers and instant feedback, try the linear approximation and MVT lesson. The Mean Value Theorem is the exact version of the same idea: the average slope over an interval equals the tangent slope at some point inside it. When you want more accuracy than a line can give, the next step is Taylor polynomials.