# Using Implicit Differentiation to Solve Related Rates Problems

## Key Questions

• There are two values depending on the point.

$\frac{\mathrm{dx}}{\mathrm{dt}} = \left\{\begin{matrix}- 8 \text{ at " (34) \\ 8 " at } \left(- 3 4\right)\end{matrix}\right.$

Let us look at some details.

First, let us find the values of $x$.

By plugging in $y = 4$ into ${x}^{2} + {y}^{2} = 25$,

${x}^{2} + 16 = 25 R i g h t a r r o w {x}^{2} = 9 R i g h t a r r o w x = \pm 3$

Now, let us find some derivatives.

By differentiating with respect to $t$,

$\frac{d}{\mathrm{dt}} \left({x}^{2} + {y}^{2}\right) = \frac{d}{\mathrm{dt}} \left(25\right) R i g h t a r r o w 2 x \frac{\mathrm{dx}}{\mathrm{dt}} + 2 y \frac{\mathrm{dy}}{\mathrm{dt}} = 0$

by dividing by $2 x$,

$R i g h t a r r o w \frac{\mathrm{dx}}{\mathrm{dx}} + \frac{y}{x} \frac{\mathrm{dy}}{\mathrm{dt}} = 0$

by subtracting $\frac{y}{x} \frac{\mathrm{dy}}{\mathrm{dt}}$,

$R i g h t a r r o w \frac{\mathrm{dx}}{\mathrm{dt}} = - \frac{y}{x} \frac{\mathrm{dy}}{\mathrm{dt}}$

Since $y = 4$, $x = \pm 3$, and $\frac{\mathrm{dy}}{\mathrm{dt}} = 6$,

$\frac{\mathrm{dx}}{\mathrm{dt}} = - \frac{4}{\pm 3} \left(6\right) = \pm 8$

12,800cm3s

#### Explanation:

This is a classic Related Rates problems. The idea behind Related Rates is that you have a geometric model that doesn't change, even as the numbers do change.

For example, this shape will remain a sphere even as it changes size. The relationship between a where's volume and it's radius is

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

As long as this geometric relationship doesn't change as the sphere grows, then we can derive this relationship implicitly, and find a new relationship between the rates of change.

Implicit differentiation is where we derive every variable in the formula, and in this case, we derive the formula with respect to time.

So we take the derivative of our sphere:

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

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

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

We were actually given $\frac{\mathrm{dr}}{\mathrm{dt}}$. It's $4 \frac{c m}{s}$.

We are interested in the moment when the diameter is 80 cm, which is when the radius will be 40 cm.

The rate of increase of the volume is $\frac{\mathrm{dV}}{\mathrm{dt}}$, which is what we are looking for, so:

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

$\frac{\mathrm{dV}}{\mathrm{dt}} = 4 \pi {\left(40 c m\right)}^{2} \left(4 \frac{c m}{s}\right)$

$\frac{\mathrm{dV}}{\mathrm{dt}} = 4 \pi \left(1600 c {m}^{2}\right) \left(4 \frac{c m}{s}\right)$

$\frac{\mathrm{dV}}{\mathrm{dt}} = 4 \pi \left(1600 c {m}^{2}\right) \left(4 \frac{c m}{s}\right)$

$\frac{\mathrm{dV}}{\mathrm{dt}} = 12 , 800 \frac{c {m}^{3}}{s}$

And the units even work out correctly, since we should get a volume divided by time.

Hope this helps. 