Calculus 1 Quiz: Linear Approximation

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Linear ApproximationQuestion 1 of 10

The volume of a spherical balloon is increasing. A linear approximation is used to estimate the increase in volume as the radius increases from $r=5$ cm to $r=5.1$ cm. Which of the following is the best estimate for the increase in volume? (The volume of a sphere is $V = \frac{4}{3}\pi r^3$ .)

10\pi

^3

100\pi

^3

4\pi

^3

\frac{530\pi}{3}

^3

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Calculus 1 Quiz

Calculus 1 Quiz: Linear Approximation

Practice Linear Approximation in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Linear Approximation, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$10\pi$ cm $^3$ (correct answer)
$100\pi$ cm $^3$
$4\pi$ cm $^3$
$\frac{530\pi}{3}$ cm $^3$

Explanation: The increase in volume

\Delta V

can be approximated by the differential

dV

. The formula is

dV = V'(r)dr

. First, we find the derivative of the volume function

V(r) = \frac{4}{3}\pi r^3

, which is

V'(r) = 4\pi r^2

. We are using a linear approximation centered at

r=5

cm, so we evaluate the derivative at this point:

V'(5) = 4\pi (5^2) = 100\pi

. The change in radius is

dr = 5.1 - 5 = 0.1

cm. The approximate increase in volume is

dV = V'(5)dr = (100\pi)(0.1) = 10\pi

^3

Question 2

The side length of a cube is measured to be $10$ cm with a possible measurement error of $\pm 0.03$ cm. Use differentials to estimate the maximum possible relative error in calculating the volume of the cube.

$\pm 0.003$
$\pm 0.009$ (correct answer)
$\pm 0.3$
$\pm 9$ cm $^3$

Explanation: The volume of a cube is

V=s^3

. The relative error in the volume is

\frac{\Delta V}{V}

, which can be approximated by the differential ratio

\frac{dV}{V}

. The differential

dV

is given by

dV = V'(s)ds = 3s^2 ds

. Therefore, the relative error is

\frac{dV}{V} = \frac{3s^2 ds}{s^3} = \frac{3 ds}{s}

. We are given the side length

s=10

cm and the error in the side length

ds = \pm 0.03

cm. Plugging these values in, the relative error is

\frac{3 (\pm 0.03)}{10} = \frac{\pm 0.09}{10} = \pm 0.009

. The maximum possible relative error is the magnitude of this value,

0.009

Question 3

Let $L(x)$ be the linearization of the function $f(x) = e^{2x}$ at $x=0$ . When using $L(0.1)$ to approximate the value of $f(0.1)$ , is the approximation an overestimate or an underestimate, and why?

An underestimate, because the graph of $f(x)$ is concave up at $x=0$ . (correct answer)
An overestimate, because $f(x)$ is an increasing function for all $x$ .
An overestimate, because the graph of $f(x)$ is concave down at $x=0$ .
An underestimate, because $f(0.1)$ is greater than $f(0)$ .

Explanation: The relationship between a linear approximation and the actual function value is determined by the function's concavity at the point of tangency. To find the concavity of

f(x) = e^{2x}

x=0

, we examine the second derivative. The first derivative is

f'(x) = 2e^{2x}

. The second derivative is

f''(x) = 4e^{2x}

. At

x=0

f''(0) = 4e^{0} = 4

. Since

f''(0) > 0

, the graph of

f(x)

is concave up at

x=0

. For a concave up function, the tangent line lies below the curve, which means the linear approximation will be an underestimate of the actual function value.

Question 4

The linear approximation for a differentiable function $f(x)$ at $x=a$ is given by the line $L(x) = -\frac{1}{2}x + 3$ . If it is also known that $f(a)=1$ , what is the value of $a$ ?

$1$
$6$
$-4$
$4$ (correct answer)

Explanation: The tangent line

L(x)

x=a

must pass through the point of tangency

(a, f(a))

. This means that

L(a) = f(a)

. We are given the equation for

L(x)

and the value of

f(a)

. Substituting

x=a

into the equation for

L(x)

gives

L(a) = -\frac{1}{2}a + 3

. Setting this equal to

f(a)=1

, we get the equation

1 = -\frac{1}{2}a + 3

. To solve for

a

, we subtract 3 from both sides:

1 - 3 = -\frac{1}{2}a

, which simplifies to

-2 = -\frac{1}{2}a

. Multiplying both sides by

-2

gives

a=4

Question 5

The function $g(x)$ has a local linear approximation at $x=3$ given by $L(x) = 5 - 2(x-3)$ . Let $h(x) = (g(x))^2$ . Which of the following is the local linear approximation for $h(x)$ at $x=3$ ?

$10 - 4(x-3)$
$25 - 4(x-3)$
$25 - 20(x-3)$ (correct answer)
$5 - 20(x-3)$

Explanation: When you encounter a problem asking for the linear approximation of a composite function like

h(x) = (g(x))^2

, you need to use the chain rule along with the given linear approximation. From the given information,

L(x) = 5 - 2(x-3)

tells us that

g(3) = 5

and

g'(3) = -2

. To find the linear approximation of

h(x) = (g(x))^2

x = 3

, you need

h(3)

and

h'(3)

. First,

h(3) = (g(3))^2 = 5^2 = 25

. Next, using the chain rule:

h'(x) = 2g(x) \cdot g'(x)

, so

h'(3) = 2g(3) \cdot g'(3) = 2(5)(-2) = -20

. Therefore, the linear approximation is

h(3) + h'(3)(x-3) = 25 - 20(x-3)

. Looking at the wrong answers: Choice A gives

10 - 4(x-3)

, which incorrectly uses

h(3) = 2g(3) = 10

instead of

(g(3))^2 = 25

, and also miscalculates the derivative. Choice B gives

25 - 4(x-3)

, which correctly finds

h(3) = 25

but incorrectly computes

h'(3) = 2g'(3) = -4

, forgetting to multiply by

g(3)

in the chain rule. Choice D gives

5 - 20(x-3)

, which correctly finds the derivative but uses

h(3) = g(3) = 5

instead of squaring it. Study tip: When finding linear approximations of composite functions, always apply the chain rule carefully to find the derivative, and don't forget that the constant term comes from evaluating the original function, not its derivative.

Question 6

Let $f(x) = x^5 + 2x + 1$ and let $f^{-1}$ be the inverse function of $f$ . Use a linear approximation to estimate the value of $f^{-1}(3.8)$ .

$\frac{34}{35}$ (correct answer)
$-0.4$
$\frac{36}{35}$
$1.4$

Explanation: We need to find a linear approximation for

f^{-1}(x)

near

x=3.8

. First, we find a convenient point for our approximation. By inspection,

f(1) = 1^5 + 2(1) + 1 = 4

. This means

f^{-1}(4)=1

. We will center our approximation at

b=4

. The linearization of

f^{-1}

b=4

L(x) = f^{-1}(4) + (f^{-1})'(4)(x-4)

. We need the derivative of the inverse function,

(f^{-1})'(4) = \frac{1}{f'(f^{-1}(4))} = \frac{1}{f'(1)}

. The derivative of

f(x)

f'(x) = 5x^4 + 2

. So,

f'(1) = 5(1)^4 + 2 = 7

. Therefore,

(f^{-1})'(4) = \frac{1}{7}

. The approximation is

L(3.8) = 1 + \frac{1}{7}(3.8 - 4) = 1 + \frac{1}{7}(-0.2) = 1 - \frac{0.2}{7} = 1 - \frac{2}{70} = 1 - \frac{1}{35} = \frac{34}{35}

Question 7

Let $L(x)$ be the linear approximation of the function $f(x) = x \cos(x)$ at $x=\frac{\pi}{2}$ . Which of the following is the value of $L(\frac{\pi}{2} - 0.1)$ ?

$0$
$0.05\pi$ (correct answer)
$-0.1$
$0.05\pi - 0.1$

Explanation: The linear approximation of

f(x)

a=\frac{\pi}{2}

L(x) = f(\frac{\pi}{2}) + f'(\frac{\pi}{2})(x-\frac{\pi}{2})

. First, find

f(\frac{\pi}{2}) = \frac{\pi}{2}\cos(\frac{\pi}{2}) = \frac{\pi}{2} \cdot 0 = 0

. Next, find the derivative

f'(x)

using the product rule:

f'(x) = 1 \cdot \cos(x) + x \cdot (-\sin(x)) = \cos(x) - x\sin(x)

. Evaluate the derivative at

x=\frac{\pi}{2}

f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) - \frac{\pi}{2}\sin(\frac{\pi}{2}) = 0 - \frac{\pi}{2}(1) = -\frac{\pi}{2}

. Now, assemble the approximation for

x = \frac{\pi}{2} - 0.1

L(\frac{\pi}{2} - 0.1) = 0 + (-\frac{\pi}{2}) ( (\frac{\pi}{2} - 0.1) - \frac{\pi}{2}) = -\frac{\pi}{2}(-0.1) = 0.05\pi

Question 8

The position of a particle along the x-axis is given by $p(t) = \sin(\frac{\pi t}{4})$ . Use the tangent line at $t=1$ to approximate the position of the particle at $t=1.1$ .

$\frac{1}{2} + \frac{\sqrt{3}\pi}{80}$
$\frac{\sqrt{2}}{2} + \frac{\sqrt{2}\pi}{40}$
$\frac{\sqrt{2}}{2} + \frac{\sqrt{2}\pi}{80}$ (correct answer)
$\frac{\sqrt{2}}{2} - \frac{\sqrt{2}\pi}{80}$

Explanation: This question tests linear approximation using tangent lines, a fundamental application of derivatives. When you need to estimate a function value near a known point, the tangent line provides the best linear approximation. To find the tangent line at

t=1

, you need both the function value and the derivative at that point. First, find

p(1) = \sin(\frac{\pi \cdot 1}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}

. Next, find the derivative:

p'(t) = \cos(\frac{\pi t}{4}) \cdot \frac{\pi}{4}

. At

t=1

p'(1) = \cos(\frac{\pi}{4}) \cdot \frac{\pi}{4} = \frac{\sqrt{2}}{2} \cdot \frac{\pi}{4} = \frac{\sqrt{2}\pi}{8}

. The tangent line equation is

L(t) = p(1) + p'(1)(t-1) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}\pi}{8}(t-1)

. For

t=1.1

L(1.1) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}\pi}{8}(0.1) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}\pi}{80}

. This confirms answer C is correct. Answer A uses

\frac{1}{2}

instead of

\frac{\sqrt{2}}{2}

, which would come from incorrectly evaluating

\sin(\frac{\pi}{4})

\sin(30°)

rather than

\sin(45°)

. Answer B has the wrong denominator in the linear term (40 instead of 80), likely from forgetting to multiply by the change in

t

(0.1). Answer D has the correct calculation but wrong sign, suggesting a sign error when computing the tangent line. Remember: linear approximation problems always follow the pattern

f(a) + f'(a)(x-a)

. Double-check your trigonometric values and arithmetic carefully.

Question 9

A curve is defined by the equation $x^3 + y^3 = 9xy$ . Use the tangent line to the curve at the point $(4,2)$ to approximate the y-coordinate of a point on the curve with x-coordinate $4.1$ .

$0.400$
$1.875$
$2.080$
$2.125$ (correct answer)

Explanation: First, we find the slope of the tangent line at

(4,2)

using implicit differentiation on

x^3 + y^3 = 9xy

. Differentiating with respect to

x

gives

3x^2 + 3y^2 \frac{dy}{dx} = 9y + 9x \frac{dy}{dx}

. Solving for

\frac{dy}{dx}

yields

\frac{dy}{dx} = \frac{9y - 3x^2}{3y^2 - 9x} = \frac{3y - x^2}{y^2 - 3x}

. At the point

(4,2)

, the slope is

m = \frac{3(2) - 4^2}{2^2 - 3(4)} = \frac{6-16}{4-12} = \frac{-10}{-8} = \frac{5}{4}

. The equation of the tangent line (the linearization) at

(4,2)

L(x) = y_0 + m(x-x_0) = 2 + \frac{5}{4}(x-4)

. To approximate the y-coordinate at

x=4.1

, we evaluate

L(4.1) = 2 + \frac{5}{4}(4.1-4) = 2 + \frac{5}{4}(0.1) = 2 + \frac{0.5}{4} = 2 + 0.125 = 2.125

Question 10

Let $h(x) = \ln(f(x))$ , where $f(x)$ is a differentiable function with $f(2)=e^3$ and $f'(2)=-2e^3$ . Use a linear approximation centered at $x=2$ to estimate the value of $h(1.9)$ .

$2.8$
$3.2$ (correct answer)
$3 + 0.2e^3$
$3 - 0.1/e^3$

Explanation: The linear approximation of

h(x)

a=2

L(x) = h(2) + h'(2)(x-2)

. First, we find

h(2) = \ln(f(2)) = \ln(e^3) = 3

. Next, we find the derivative

h'(x)

using the chain rule:

h'(x) = \frac{1}{f(x)} \cdot f'(x) = \frac{f'(x)}{f(x)}

. At

x=2

, the slope is

h'(2) = \frac{f'(2)}{f(2)} = \frac{-2e^3}{e^3} = -2

. Now we can approximate

h(1.9)

L(1.9) = h(2) + h'(2)(1.9 - 2) = 3 + (-2)(-0.1) = 3 + 0.2 = 3.2