Related Rates

1. Here we are going to assume that the initial volume is negligible and the only thing that matters is the volume 3 seconds after condensation starts to occur. Why? because: $V=\frac{4}{3}(0.1{)}^3=0.0042{\text{ mm}}^3\ll 3{\text{ mm}}^3$​
   In other words, the difference in volume is less than 0.2%. So let’s assume our droplet’s current volume is 3 ${\text{mm}}^3$ . You are going to love what happens next! First, here is what we know:

$V=\frac{4}{3}\pi r^3$

$\frac{dV}{dt}=\frac{1{\text{ mm}}^3}{s}$

$\frac{dr}{dt}=\text{unknown}$

Can you guess what we have to do? Starts with an I. Rhymes with differentiation.

That’s right! Implicit differentiation! How? Let’s take the general derivative with respect to time of the volume function.

$\frac{d}{dt}V=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$​
$\frac{dV}{dt}=\frac{4}{3}\pi \cdot \left(\frac{d}{dr}{r}^3\right)\cdot \frac{dr}{dt}$

Now we can take the derivative with respect to r of the term inside the parentheses.

$\frac{dV}{dt}=\frac{4}{3}\pi \cdot (3r^2)\cdot \frac{dr}{dt}$

$\frac{dV}{dt}=\frac{4}{\cancel{3}}\pi \cdot \cancel{3}{r}^2\frac{dr}{dt}$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

We know two of these unknown variables. We can solve for r after 3 seconds, and find $\frac{dr}{dt}$. First let’s solve for r for the new volume after 3 seconds.

$3{\text{ mm}}^3=\frac{4}{3}\pi r^3$

$r=\sqrt[3]{0}.716=0.895\text{ mm}$

What does this tell us? We can see that in just 3 seconds, the size of the droplet increased 9 times!

However, we know the rate slows down the larger the droplet gets. How much slower? We are about to find out (bolded). Let’s plug in r into our related rates equation.

$\frac{1{\text{ mm}}^3}{s}=4\pi (0.895\text{ mm}{)}^2\cdot \frac{\mathbf{d}\mathbf{r}}{\mathbf{d}\mathbf{t}}$

$\frac{1{\text{ mm}}^3}{4\pi \cdot 0.801}=\frac{\mathbf{d}\mathbf{r}}{\mathbf{d}\mathbf{t}}$​
​
$\frac{\mathbf{d}\mathbf{r}}{\mathbf{d}\mathbf{t}}=0.1\frac{\text{mm}}{s}$

Are we done? No! The question asked us for the rate of diameter increase. Common sense tells us this is just twice the rate of radial increase, but we can prove it mathematically too, using implicit differentiation!

$2r=D$

$\frac{d}{dt}2r=\frac{d}{dt}D$

$2\frac{dr}{dt}=\frac{dD}{dt}$

$\frac{\mathbf{d}\mathbf{D}}{\mathbf{d}\mathbf{t}}=\mathbf{0}.\mathbf{2}\frac{\mathbf{m}\mathbf{m}}{\mathbf{s}}$

b) This part asks us to find what the rate of increase in diameter originally was. We have done most of the work already, so let’s just go back to our related rates equation and plug in a new value for the radius, 0.05 mm (this is half the original diameter given to us in the question).

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

$1\frac{{\text{mm}}^3}{s}=4\pi (0.05{\text{ mm}}^2)\frac{dr}{dt}$

$\frac{1\frac{\text{mm}}{\text{s}}}{4\pi \cdot 0.0025}=\frac{dr}{dt}$

$\frac{dr}{dt}=31.83\frac{\text{mm}}{\text{s}}$

Does this number seem high...? Nope! The droplet really does increase at a size of 31.83 millimeters per second! Why? Because when the droplet is really small, any moderate increase in volume will cause its size to become 3x or even 4x larger! Once the droplet becomes bigger, the same amount of volume ends up having a smaller effect.

For example, if you were living on $1 a day, and someone gave you one dollar, you would be relatively much richer than if you were making $100 per day and someone gave you that same dollar. The same concept applies here. Do we have our answer yet? Nope! Because we are asked for the change in diameter.

$\frac{\mathit{d}\mathit{D}}{\mathit{d}\mathit{t}}\mathbf{=}\mathbf{2}\frac{\mathit{d}\mathit{r}}{\mathit{d}\mathit{t}}\mathbf{=}\mathbf{63.7}\frac{\mathbf{m}\mathbf{m}}{\mathbf{s}}$​

We know that after three seconds, your friend has gone north 15 m based on his constant velocity of 5 m/s. Now we have a right triangle with the adjacent side equal to 10, and the opposite side equal to 15. We can find θ.​

$\theta ={\tan }^{-1}\frac{15}{10}=56.3\text{º}$​

Now how do we find the related rates equation? Let’s look at which two variables are changing. θ is changing and the vertical distance (or opposite side) is changing. Do we know an equation that relates these two? Yes we do!

$\tan \theta =$$\frac{\text{opposite}}{\text{adjacent}}=\frac{\text{N}}{\text{E}}$

Rather than having to say opposite and adjacent all the time, let’s just call these north and east, because that is what they are. Since East doesn’t change, we can treat that as a coefficient to our changing variable, North, and take the implicit derivative with respect to time for both sides of the equation. (sizes below)

$\frac{d}{dt}\tan \theta =\frac{1}{\text{E}}\cdot \frac{d}{dt}(\text{N})$

${\sec }^2(\theta )\cdot \frac{d\theta }{dt}=\frac{1}{\text{E}}\frac{d\text{N}}{dt}$

We know all of these values except for $\frac{d\theta }{dt}$ , which we are trying to solve for. Notice that $\frac{d\text{N}}{dt}$ is just the velocity of your friend running north.

Plugging in these values gives us

$3.25\left(\frac{d\theta }{dt}\right)=\frac{1}{10\cancel{\text{m}}}\cdot 5\cancel{\text{m}}$

$\frac{d\theta }{dt}=\frac{0.154}{\text{s}}$​
​
Our answer is not in degrees, however! It is in radians. Degrees is a unit. Radians are unit-less. We can convert this to degrees by multiplying our answer by $\frac{{180}^{\text{º}}}{\pi }$ , which gives

$\frac{d\theta }{dt}=\frac{8.82\text{º}}{\text{s}}$