PROBLEM SOLVING STRATEGY: Approximations

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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 $\sin{(0)} = 0$.

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*}

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)** $\sqrt{3.96}$**(a)** 2.01**(b)** 1.99

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Solution SummarySolution (a) DetailSolution (b) Detail

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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.

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.

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.

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Question 2: sin

Without using a calculator, estimate:**(a)** $\sin(0.1)$**(b)** $\sin(-0.1)$**(a)** 0.1**(b)** -0.1

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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.

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.

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.

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Question 3: cos(*pi* + 1/100) (based on an actual exam question)

Without using a calculator, estimate to two decimal places: $\cos{(\pi + 1/100)}$.

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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.

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Question 4: sqrt(99) (based on an actual exam question)

Without using a calculator, estimate to two decimal places: $\sqrt{99}$

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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.

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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.

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Question 5: (1.0003)^100 & (1+x)^n

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.**(a)** 1.03**(b)** See detailed solution.

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Solution SummarySolution (a) DetailSolution (b) Detail

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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.

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.

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.

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Question 6: Adding a coat of paint

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.")

(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*}

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\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*}

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Question 7: Increasing a circle's radius

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.")

(This question could state instead, "Use differentials to estimate the percentage change in area.")

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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*}

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\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*}

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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 $dV_{\text{sphere}} = \_\_\_dr$.**(a)** (i) $2 \pi r$; (ii) See detailed solution.**(b)** (i) $4 \pi r^2$; (ii) See detailed solution.

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Solution SummarySolution (a) DetailSolution (b) Detail

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(i)
\begin{align*}
\frac{dA_{\text{circle}}}{dr} &= \frac{d}{dr}\left( \pi r^2 \right) \\ \\
&= 2 \pi r \quad \cmark
\end{align*} (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, $dr$.

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 $dV = (4\pi r^2)dr$.

(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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