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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