## Related Rates

### Calculus Related Rates Problem:

Given an equation, find the rate.

Not all Related Rates problems are word problems. Here’s a different type that came from a student. The problem gives you an equation, and then asks you to find a rate:

If $y = x^3 + 2x$ and $\dfrac{dx}{dt} = 6$, find $\dfrac{dy}{dt}$ when $x=5.$

### Calculus Solution

We can still use our Related Rates Problem Solving Strategy.

Instead of starting at Step 1, we can start at Step 3:

**3. Take the derivative with respect to time of both sides of your equation. Remember the chain rule. **

\begin{align*}

y &= x^3 + 2x \\[8px]
\dfrac{d}{dt}(y) &= \dfrac{d}{dt}\left(x^3 \right) + \dfrac{d}{dt}(2x) \\[8px]
\dfrac{dy}{dt} &= 3x^2 \dfrac{dx}{dt} + 2 \dfrac{dx}{dt}

\end{align*}

While the derivative of $x^3$ with respect to *x* is

$$\dfrac{d}{dx}x^3 = 3x^2,$$

the derivative of $x^3$ with respect to *time t* is

$$\dfrac{d}{dt}x^3 = 3x^2\dfrac{dx}{dt}.$$

(Recall that that rate is $\dfrac{dx}{dt} = 6$ in this problem.)

The Chain rule term would be clearer if we explicitly wrote the time-dependence of both *y* and *x*. Remember that both quantities *change* as time passes, and so they are functions of time *t*. We could have captured this time-dependence explicitly by writing our original relation as

$$ y(t) = \big[ x(t)\big]^3 + 2 \big[ x(t)\big]$$

to remind ourselves that both *y* and *x* are functions of time *t*. Then when we take the derivative,

$$\frac{d}{dt}y(t) = \frac{d}{dt} \big[ x(t)\big]^3 + 2\dfrac{d}{dt} \big[ x(t)\big]$$

Now the term $\dfrac{d}{dt}y(t)$ *is * the rate $\dfrac{dy(t)}{dt}.$

And looking at the last term, $\dfrac{d}{dt}x(t)$ *is * the rate $\dfrac{dx(t)}{dt}.$

The derivative $\dfrac{d}{dt}\big[\text{(stuff)} \big]^3 = 3 \text{(stuff)}^2 \cdot \dfrac{d}{dt}\text{(stuff)},$ where the extra term at the end is due to the Chain rule. Here $\text{(stuff)} = x(t)$, so we have $\dfrac{d}{dt}\big[ x(t)\big]^3 = 3\big[ x(t)\big]^2 \dfrac{dx(t)}{dt}.$

Most people find that writing the explicit time-dependence $y(t)$ and *x(t)* annoying, and so just write *y* and *x* instead. Regardless, you *must* remember that both *y* and *x* depend on *t*, and so when you take the derivative with respect to time the Chain Rule applies and you have the $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ terms.

**4. Solve for the quantity you’re after.**

We want to find $\dfrac{dy}{dt}$ when $x = 5.$ Recall that the problem told us that $\dfrac{dx}{dt} = 6.$

\begin{align*}

\dfrac{dy}{dt} &= 3x^2 \dfrac{dx}{dt} + 2 \dfrac{dx}{dt} \\[8px]
&= 3(5)^2 (6) + 2 (6) \\[8px]
&= 3(25)(6)+12 \\[8px]
&= 462 \quad \cmark

\end{align*}

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