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*}
Return to Related Rates Problems
Want access to all of our Calculus problems and solutions? Sign in for free with your Google, Facebook or Apple account, or with your dedicated Matheno account (which you can create in 60 seconds). Then visit our Calculus Home screen.
And if you have a Calculus question, please pop over to our Forum and post. Related rates problems can be especially challenging to set up. If you could use some help, please post and we’ll be happy to assist!