Single Variable Function Derivative W. R. T. Time?

Single Variable Function Derivative W. R. T. Time?
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I was studying calculus and I had doubts about this problem: (this is not homework)

A circular wire expands due to heat so that its radius increases with a speed of $0.01 ms^{-1}$. How rapidly does the area increase when the radius is 2 cm?

The solution goes like this:

Let x be the radius and y the area. Then:

$$y=\pi x^2$$

And then it goes like this: $${dy\over{dt}}=2\pi x {dx\over{dt}}$$ How is this possible to do? y is a single variable function and x is just the independent variable.

This confuseses me a lot.

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1 Answer

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No this is why you should have never been taught that $f(x)$ means $y = ...$. Both $y$ and $x$ are functions of $t$:

$$ y(t) = \pi x^2(t) $$

Now differentiate both sides with respect to $t$--use the chain rule:

$$ \frac{d}{dt}y(t) = \frac{d}{dt}\left(\pi x^2(t)\right) \\ \frac{dy}{dt} = 2\pi x \frac{dx}{dt} $$

Here is another way to look at it:

$$ y = f(x) \\ \frac{dy}{dt} = \frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt} $$

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James H. Sterling
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James H. Sterling

James Sterling reports on renewable energy developments, climate policy, ecological conservation, and green tech innovations around the globe.