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Approximations

Free example problems + complete solutions for typical linear approximation problems in Calculus, designed to help you learn how to solve them routinely.

Update: We now have a much better material to help you learn what’s going with linear approximations, including making use of interactive Desmos graphing calculators so you can see what it is you’re actually doing. Please visit our Introducing Linear Approximations screen to really get this material down for yourself; there are then more practice problems on our Practice Problems: Linear Approximations screen.

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If you just need practice with linear approximation problems for now, previous students have found what’s below super-helpful. And if you have questions, please ask on our Forum! It’s also free for your use.

PROBLEM SOLVING STRATEGY: Approximations
Show/Hide Strategy

Problems of this type will ask you to approximate the value of a function that’s close to a value you immediately know. For instance, you might be asked to approximate $\sqrt{4.02}$, which is close to the value you know of $\sqrt{4} = 2$. Or you could be asked to approximate $\sin(0.1)$, since you know $\sin{(0)} = 0$.

Curve with red dot at (x_0, f(x_0)), and blue dot on the curve a horizontal distance delta-x away

We’ll call the point you know about $x_0$. For instance, if you want to approximate $\sqrt{4.02}$, then $x_0 = 4$. And if you want to approximate $\sin(0.1)$, then $x_0 = 0$. We represent this point that we know about, $(x_o, f(x_0))$, as the red dot in the top figure.

We’ll call the horizontal distance to the point you’re interested in $\Delta x$, so you’re looking for an approximation to $f(x_0 + \Delta x)$. We show this point as the blue dot in the lower figure. So in the examples we’re considering:
\begin{align*}
\sqrt{4.02} &= \sqrt{4 + 0.02} \text{, we have } x_0 = 4 \text{ and } \Delta x = 0.02; \\[8px] \text{and for} \\[8px] \sin(0.1) &= \sin(0 + 0.1) \text{, we have } x_0 = 0 \text{ and } \Delta x = 0.1
\end{align*}

Top figure shows the tangent line at the point (x_0, f(x_0). The lower figure shows the horizontal distance delta-x, and the associated change delta-y equals m * delta-x.

To make the approximation, we replace the function’s actual growth (or decrease) with linear growth (or decrease), which is the same thing as pretending that the function follows the tangent line as you move a small distance away from $x_0$. (See the upper figure.) Said differently, if we imagine walking from the point we know about to the point of interest in order to compute the change, instead of walking from the point along the function’s actual curve (shown in blue), we’re going to walk along the tangent line (shown in green) to get an approximation.

The method works because we can easily compute the slope $m$ of the line tangent to the curve at the point you know about, $x_0$:

$$m = \left. \frac{df}{dx} \right|_{x = x_0}$$

Then computing the approximate growth (or decrease) $\Delta y$ from the value we know about is straightforward (see the lower figure):

\begin{align*}
\Delta y &= m \, \Delta x \\
&= \left. \frac{df}{dx} \right|_{x = x_0}\Delta x
\end{align*}

Putting all the pieces together,
\begin{align*}
f(x_0 + \Delta x) &\approx f(x_0) + \Delta y \\
&\approx f(x_0) + \left. \frac{df}{dx} \right|_{x = x_0}\Delta x
\end{align*}

This approach to estimation is known as a linear approximation since we are replacing the function around the point of interest with a line whose slope equals the tangent to the curve there.

Many students who find the discussion above rather abstract find the method straightforward after practicing the concrete problems below.

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Question 1: Square roots
Without using a calculator, estimate:
(a) $\sqrt{4.04}$
(b) $\sqrt{3.96}$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) Detail
(a) 2.01
(b) 1.99
We have $f(x) = \sqrt{x}$. Then $f(4.04) = f(4 + 0.04)$, and we know $f(4) = 2$. We need to estimate how much the function changes when we move a distance $\Delta x = 0.04$ from $x_0 = 4$.

The slope $m$ of the tangent line at $x=4$ is \begin{align*} \left. \frac{df}{dx} \right|_{x = 4} &= \left[ \frac{d}{dx}\sqrt{x} \right]_{x = 4} \\ \\ &= \frac{1}{2} \left. \frac{1}{\sqrt{x}} \right|_{x = 4} \\ \\ &= \frac{1}{2} \frac{1}{\sqrt{4}} = \frac{1}{2} \, \frac{1}{2} = \frac{1}{4} \end{align*}

Then the approximate change $\Delta y$ is given by \begin{align*} \Delta y &= \left. \frac{df}{dx} \right|_{x = x_0}\Delta x \\ \\ &= \frac{1}{4}(0.04) = 0.01 \end{align*}

Then \begin{align*} f(4.04) &\approx f(4) + \Delta y \\ &\approx 2 + 0.01 = 2.01 \quad \cmark \end{align*}

For purposes of comparison, the actual value, to six decimal places, is 2.009975.
We again have $f(x) = \sqrt{x}$ and $x_0$ = 4, but now $f(3.96) = f(4 – 0.04)$, so $\Delta x = -0.04$.

We can re-use the result we found in part (a), that the slope $m$ of the tangent line at $x=4$ is $$\left. \frac{df}{dx} \right|_{x = 4} = \frac{1}{4}$$

Then the approximate change $\Delta y$ is given by \begin{align*} \Delta y &= \left. \frac{df}{dx} \right|_{x = x_0}\Delta x \\ \\ &= \frac{1}{4}(-0.04) = -0.01 \end{align*}

Then \begin{align*} f(3.96) &\approx f(4) + \Delta y \\ &\approx 2 – 0.01 = 1.99 \quad \cmark \end{align*}

For purposes of comparison, the actual value, to six decimal places, is 1.989975.
[hide solution]
Question 2: sin
Without using a calculator, estimate:
(a) $\sin(0.1)$
(b) $\sin(-0.1)$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) Detail
(a) 0.1
(b) -0.1
We have $f(x) = \sin{x}$. Then $f(0.1) = f(0 + 0.1)$, and we know $f(0) = 0$. We need to estimate how much the function changes when we move a distance $\Delta x = 0.1$ from $x_0 = 0$.

The slope $m$ of the tangent line at $x=0$ is \begin{align*} \left. \frac{df}{dx} \right|_{x = 0} &= \left[ \frac{d}{dx}\sin{x} \right]_{x = 0} \\ \\ &= \left. \cos x \right|_{x=0} \\ \\ &= 1 \end{align*}

Then the approximate change $\Delta y$ is given by \begin{align*} \Delta y &= \left. \frac{df}{dx} \right|_{x = x_0}\Delta x \\ \\ &= (1)(0.1) = 0.1 \end{align*}

Then \begin{align*} f(0.1) &\approx f(0) + \Delta y \\ &\approx 0 + 0.1 = 0.1 \quad \cmark \end{align*} The actual value, to six decimal places, is 0.099833.
We again have $f(x) = \sin{x}$ and $x_0$ = 0, but now $f(-0.1) = f(0 – 0.1)$, so $\Delta x = -0.1$.

We can re-use the result we found in part (a), that the slope $m$ of the tangent line at $x=0$ is $$\left. \frac{df}{dx} \right|_{x = 0} = 1$$

Then the approximate change $\Delta y$ is given by \begin{align*} \Delta y &= \left. \frac{df}{dx} \right|_{x = x_0}\Delta x \\ \\ &= 1(-0.1) = -0.1 \end{align*}

Then \begin{align*} f(-0.1) &\approx f(0) + \Delta y \\ &\approx 0 – 0.1 = -0.1 \quad \cmark \end{align*} The actual value, to six decimal places, is -0.099833.
[hide solution]
Question 3: cos(pi + 1/100) (based on an actual exam question)
Without using a calculator, estimate to two decimal places: $\cos{(\pi + 1/100)}$.
Show/Hide Solution
We have $f(x) = \cos{x}$. We want $f(\pi + 1/100)$, and we know $f(\pi) = -1$. We need to estimate how much the function changes when we move a distance $\Delta x = 1/100$ from $x_0 = \pi$. The slope $m$ of the tangent line at $x= \pi$ is \begin{align*} \left. \frac{df}{dx} \right|_{x = \pi} &= \left[ \frac{d}{dx}\cos{x} \right]_{x = \pi} \\ \\ &= \left. -\sin x \right|_{x=\pi} -\sin(\pi) = 0 \end{align*} Then the approximate change $\Delta y$ is given by \begin{align*} \Delta y &= \left. \frac{df}{dx} \right|_{x = x_0}\Delta x \\ \\ &= (0)\left(\frac{1}{100} \right) = 0 \end{align*} Hence to this level of approximation, there is no change in the function’s value when you move a tiny bit to the right. Finally, then \begin{align*} f(\pi + 1/100) &\approx f(\pi) + \Delta y \\ \\ &\approx -1 + 0 = -1 \quad \cmark \end{align*} The actual value, to six decimal places, is -0.999950.
[hide solution]
Without using a calculator, estimate to two decimal places: $\sqrt{99}$
Show/Hide Solution
We have $f(x) = \sqrt{x}$. Then $f(99) = f(100 – 1)$, and we know $f(100) = 10$. We need to estimate how much the function changes when we move a distance $\Delta x = -1$ from $x_0 = 100$.

The slope $m$ of the tangent line at $x=100$ is \begin{align*} \left. \frac{df}{dx} \right|_{x = 100} &= \left[ \frac{d}{dx}\sqrt{x} \right]_{x = 100} \\ \\ &= \frac{1}{2} \left. \frac{1}{\sqrt{x}} \right|_{x = 100} \\ \\ &= \frac{1}{2} \frac{1}{\sqrt{100}} = \frac{1}{2} \, \frac{1}{10} = (0.5)(0.1) = 0.05 \end{align*}

Then the approximate change $\Delta y$ is given by \begin{align*} \Delta y &= \left. \frac{df}{dx} \right|_{x = x_0}\Delta x \\ \\ &= (0.05)(-1) = -0.05 \end{align*}

Then \begin{align*} f(99) &\approx f(100) + \Delta y \\ &\approx 10 – 0.05 = 9.95 \quad \cmark \end{align*} The actual value, to six decimal places, is 9.949874.
[hide solution]
Question 4: sqrt(99) (based on an actual exam question)
Answer the following:
(a) Estimate $(1.0003)^{100}$.
(b) Show that $(1 + \Delta x)^n \approx 1 + n \, \Delta x$, if $\Delta x$ is small compared to 1.
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) Detail
(a) 1.03
(b) See detailed solution.
We have $f(x) = x^{100}$. Then $f(1.0003) = f(1 + 0.0003)$, and we know $f(1) = 1$. We need to estimate how much the function changes when we move a distance $\Delta x = 0.0003$ from $x_0 = 1$.

The slope $m$ of the tangent line at $x=1$ is \begin{align*} \left. \frac{df}{dx} \right|_{x = 1} &= \left[ \frac{d}{dx}x^{100} \right]_{x = 1} \\ \\ &=100\left. x^{99} \right|_{x = 1} \\ \\ &= 100(1) = 100 \end{align*}

Then the approximate change $\Delta y$ is given by \begin{align*} \Delta y &= \left. \frac{df}{dx} \right|_{x = x_0}\Delta x \\ \\ &= 100(0.0003) = 0.03 \end{align*}

Then \begin{align*} f(1.0003) &\approx f(1) + \Delta y \\ &\approx 1 +0.03 = 1.03 \quad \cmark \end{align*} A numerical computation gives the result as 1.0304499.
We have $f(x) = x^n$. We want $f(1 + \Delta x)$, and we know $f(1) = 1$. We need to estimate how much the function changes when we move a distance $\Delta x $ from $x_0 = 1$, where the problem specifies that $\Delta x$ is small compared to 1 and so we are only moving a tiny distance.

The slope $m$ of the tangent line at $x=1$ is \begin{align*} \left. \frac{df}{dx} \right|_{x = } &= \left[ \frac{d}{dx}x^n \right]_{x = 1} \\ \\ &=n\left. x^{n-1} \right|_{x = 1} \\ \\ &= n(1) = n \end{align*}

Then the approximate change $\Delta y$ is given by \begin{align*} \Delta y &= \left. \frac{df}{dx} \right|_{x = x_0}\Delta x \\ \\ &= n \, \Delta x \end{align*}

Then \begin{align*} f(1 + \Delta x) &\approx f(1) + \Delta y \\ &\approx 1 + n \, \Delta x \quad \cmark \end{align*} This approximation is used very often in physics.
[hide solution]
Question 5: (1.0003)^100 & (1+x)^n
You're going to add a coat of paint of thickness 0.02 cm to a cube of edge-length 10 cm. Approximately how many cubic centimeters of paint will you use?
(This question could state instead, "Use differentials to estimate how much paint you will use.")
Show/Hide Solution
This problem is of a slightly different variety, but the basic idea is the same: we want to compute the approximate change in volume when we increase the cube’s edge-length from $x_0 = 10$ cm to $x = 10.04$ cm. (Remember you add 0.02 cm to each face, increasing the overall length of each side by 0.04 cm.) That is, $\Delta x = 0.04$. And the volume of a cube as a function of its side-length $x$ is $V(x) = x^3$. Then

\begin{align*} \Delta V &\approx \left. \frac{dV}{dx} \right|_{x = x_0}(\Delta x) \\ \\ &\approx \left[ \frac{d}{dx} x^3\right]_{x = 10} (0.04) \\ \\ &\approx \left. 3 x^2 \right|_{x = 10} (0.04) = 3(10)^2(0.04) = 3(100)(0.04) = 12 \text{ cm}^3 \quad \cmark \end{align*}
[hide solution]
Question 6: Adding a coat of paint
You increase a circle's radius by 1%. By approximately what percentage does its area change?
(This question could state instead, "Use differentials to estimate the percentage change in area.")
Show/Hide Solution
Here we have the circle’s area as a function of its radius $A(r) = \pi r^2$. We first want to compute the approximate change $\Delta A$ when we change the radius from $r_0$ to $r = r_0 + 0.01r_0$. Hence $\Delta r = 0.01r_0$.

\begin{align*} \Delta A &\approx \left. \frac{dA}{dr} \right|_{x = r_0}(\Delta r) \\ \\ &\approx \left[ \frac{d}{dr}\pi r^2 \right]_{r = r_0} (0.01 r_0) \\ \\ &\approx \left. 2\pi r \right|_{r = r_0} (0.01 r_0) \\ \\ &\approx 2\pi \, r_0 \, (0.01 r_0) = 0.02 \, \pi r_0^2 \end{align*} To find the percentage change, we divide the change by the original area $A_0 = \pi r_0^2$:

\begin{align*} \% \text{ change} &= \dfrac{\Delta A}{A_0} \\ \\ &= \frac{0.02 \, \pi r_0^2} {\pi r_0^2} \\ \\ &= 0.02 = 2\% \quad \cmark \end{align*}
[hide solution]
Question 7: Increasing a circle's radius
In this question we're going to try to make geometric sense of the differentials associated with a circle's area, and a sphere's volume.
(a) The formula for the area of a circle is $A_{\text{circle}} = \pi r^2$. (i) Find $\dfrac{dA_{\text{circle}}}{dr}$. (ii) The result from (i) should look familiar. What does $\dfrac{dA_{\text{circle}}}{dr}$ represent geometrically? Hint: Look at the result in the form $dA_{\text{circle}} = \_\_\_dr$.
(b) The formula for the surface area of a sphere is $V_{\text{sphere}} = \dfrac{4}{3}\pi r^3$. (i) Find $\dfrac{dV_{\text{sphere}}}{dr}$. (ii) The result from (i) should look familiar. What does $\dfrac{dV_{\text{sphere}}}{dr}$ represent geometrically? Hint: Look at the result in the form $dV_{\text{sphere}} = \_\_\_dr$.
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) Detail
(a) (i) $2 \pi r$; (ii) See detailed solution.
(b) (i) $4 \pi r^2$; (ii) See detailed solution.
(i) \begin{align*} \frac{dA_{\text{circle}}}{dr} &= \frac{d}{dr}\left( \pi r^2 \right) \\ \\ &= 2 \pi r \quad \cmark \end{align*} As described in the text. (ii) We can view the result as $dA_{\text{circle}} = (2\pi r) \, dr$, which we recognize as the circumference of the circle, $2\pi r$, multiplied by the small change in radius $dr$. Geometrically, when we increase the radius a bit, we’re adding the darkened bit to the circle shown in the figure.

If we unwrap that dark bit from the circle and lay it straight, it has length equal to the circumference of the circle, $2\pi r$, and width $dr$. It thus has area $2 \pi r \, dr$. Hence when we increase the radius by the small amount $dr$, we’re adding to the circle the small area $dA = (2\pi r) \, dr$.

Some students find it helpful to imagine a marking pen whose tip has width $dr$. To increase the original circle’s size, you trace the tip around the circle’s circumference, adding the darkened strip shown. You’re thus adding area equal to the length you draw, $2 \pi r$, multiplied by the width of the pen, $dr$.
(i) \begin{align*} \frac{dV_{\text{sphere}}}{dr} &= \frac{d}{dr}\left(\frac{4}{3}\pi r^3 \right) \\ \\ &= 3 \left(\frac{4}{3}\pi r^2 \right) \\ \\ &= 4 \pi r^2 \quad \cmark \end{align*}

(ii) We can view the result as $dV_{\text{sphere}} = (4 \pi r^2) dr$, which we recognize as the surface area of the sphere, $4 \pi r^2$, multiplied by the small change in radius, $dr.$

Geometrically, when we increase the radius a bit, we’re adding a thin “skin” around the surface of the original sphere. You might imagine, for example, placing a piece of paper around a basketball, having cut the paper such that it exactly covers the ball with none left over. The volume you’re then adding to the sphere is the volume of that piece of paper. To calculate that volume, imagine laying the paper flat on the table: it has area equal to the ball’s surface area, $4 \pi r^2$ (since, remember, it exactly covers the ball with none left over). And it has thickness $dr$ since that’s how much we’re increasing the sphere’s radius. The paper thus has tiny volume $dV = (4\pi r^2)dr$.
[hide solution]
Question 8: Differential of a circle's area, and of a sphere's volume
In this question we're going to try to make geometric sense of the differentials associated with a circle's area, and a sphere's volume.
(a) The formula for the area of a circle is $A_{\text{circle}} = \pi r^2$. (i) Find $\dfrac{dA_{\text{circle}}}{dr}$. (ii) The result from (i) should look familiar. What does $\dfrac{dA_{\text{circle}}}{dr}$ represent geometrically? Hint: Look at the result in the form $dA_{\text{circle}} = \_\_\_dr$.
(b) The formula for the surface area of a sphere is $V_{\text{sphere}} = \dfrac{4}{3}\pi r^3$. (i) Find $\dfrac{dV_{\text{sphere}}}{dr}$. (ii) The result from (i) should look familiar. What does $\dfrac{dV_{\text{sphere}}}{dr}$ represent geometrically? Hint: Look at the result in the form $dV_{\text{sphere}} = \_\_\_dr$.
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) Detail
(a) (i) $2 \pi r$; (ii) See detailed solution.
(b) (i) $4 \pi r^2$; (ii) See detailed solution.
(i) \begin{align*} \frac{dA_{\text{circle}}}{dr} &= \frac{d}{dr}\left( \pi r^2 \right) \\ \\ &= 2 \pi r \quad \cmark \end{align*} As described in the text. (ii) We can view the result as $dA_{\text{circle}} = (2\pi r) \, dr$, which we recognize as the circumference of the circle, $2\pi r$, multiplied by the small change in radius $dr$. Geometrically, when we increase the radius a bit, we’re adding the darkened bit to the circle shown in the figure.

If we unwrap that dark bit from the circle and lay it straight, it has length equal to the circumference of the circle, $2\pi r$, and width $dr$. It thus has area $2 \pi r \, dr$. Hence when we increase the radius by the small amount $dr$, we’re adding to the circle the small area $dA = (2\pi r) \, dr$.

Some students find it helpful to imagine a marking pen whose tip has width $dr$. To increase the original circle’s size, you trace the tip around the circle’s circumference, adding the darkened strip shown. You’re thus adding area equal to the length you draw, $2 \pi r$, multiplied by the width of the pen, $dr$.
(i) \begin{align*} \frac{dV_{\text{sphere}}}{dr} &= \frac{d}{dr}\left(\frac{4}{3}\pi r^3 \right) \\ \\ &= 3 \left(\frac{4}{3}\pi r^2 \right) \\ \\ &= 4 \pi r^2 \quad \cmark \end{align*}

(ii) We can view the result as $dV_{\text{sphere}} = (4 \pi r^2) dr$, which we recognize as the surface area of the sphere, $4 \pi r^2$, multiplied by the small change in radius, $dr.$

Geometrically, when we increase the radius a bit, we’re adding a thin “skin” around the surface of the original sphere. You might imagine, for example, placing a piece of paper around a basketball, having cut the paper such that it exactly covers the ball with none left over. The volume you’re then adding to the sphere is the volume of that piece of paper. To calculate that volume, imagine laying the paper flat on the table: it has area equal to the ball’s surface area, $4 \pi r^2$ (since, remember, it exactly covers the ball with none left over). And it has thickness $dr$ since that’s how much we’re increasing the sphere’s radius. The paper thus has tiny volume $dV = (4\pi r^2)dr$.
[hide solution]

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