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I have a related rates problem I couldn't really start which is in the study guide for my tomorrow related rates quiz. The question is :

The radius r of a sphere is increasing at a constant rate. At the time when the surface area and the radius of a sphere are increasing at the same numerical rate, what is the radius of the sphere?

The answer of this question according to my teacher is :

$r = \frac{1}{8\barwedge}$ or $0.040$ units

How would i go about solving this problem? I am clueless now.

3 Answers 3

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We know that the surface area of the sphere is $4\pi r^2$.

We also know that the increasing rate is the same, so applying the derivatives of the radius $r$ which is $\frac d {dr} r =1$ and the surface area of the sphere which is $\frac d {dr} 4\pi r^2=8\pi r$ and setting them equal we find $1=8\pi r$, so $r=\frac 1 {8\pi}$.

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    Wow.. i didnt know the derivative was 1 when it means constant.. I thought by constant it meant grew at constant rate.. so derrivative can be 2 and grow at a constant rate of 2 inches. You sure its 1? – 2017-02-05
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    I think you misunderstood. The **growth** is constant, not the radius. The derivative tells you how much it grows; if its growth is constant, then the derivative is constant – 2017-02-05
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    Giu : Right then can't the derivative be of any constant like 2 or 3 or 4? – 2017-02-06
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    Yes, for example if you took the diameter of the sphere, which is $2r$, you'd have had $2$ as derivative – 2017-02-06
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For the surface area of a sphere, we have that:

$$\mathcal{S}_\text{sphere}\left(\text{r}\right)=4\pi\cdot\text{r}^2\tag1$$

So, for the derivative of $\mathcal{S}_\text{sphere}$ with respect to $\text{r}$ we get:

$$\frac{\text{d}\mathcal{S}_\text{sphere}\left(\text{r}\right)}{\text{d}\text{r}}=\frac{\text{d}}{\text{d}\text{r}}\left(4\pi\cdot\text{r}^2\right)=4\pi\cdot\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}^2\right)=4\pi\cdot2\cdot\text{r}=8\pi\cdot\text{r}\tag2$$

And for the derivative of $\text{r}$ with respect to $\text{r}$ we find that it is $1$.

So, we get:

$$\frac{\text{d}\mathcal{S}_\text{sphere}\left(\text{r}\right)}{\text{d}\text{r}}=\frac{\text{d}}{\text{d}\text{r}}\left(\text{r}\right)=8\pi\cdot\text{r}=1\space\Longleftrightarrow\space\text{r}=\frac{1}{8\pi}=0.03978873577297384\dots\tag3$$

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I find the following approach a little easier to understand because it follows exactly the procedure that is used in many other related rates problems:

The surface area of the sphere is $S = {4\pi r^2}$. The derivative with respect to time is: $$ \frac {dS}{dt} = 8\pi r \frac {dr}{dt} $$

But the question/problem tells us that the rate of change of surface area is the same as the rate of change of radius, therefore:

$$ \frac {dS}{dt} = \frac {dr}{dt} $$

So, substituting, we have $$ \frac {dr}{dt} = 8\pi r \frac {dr}{dt} $$

Or, if you prefer, we can simplify the notation like so: $$ r^{'}=8\pi rr^{'} $$

Dividing both sides by $r^{'}$ (or $dr\over dt$), we have $$1=8\pi r$$

Solving for $r$, we have $$r= \frac{1}{8\pi}$$

or approximately 0.0397887.