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I am starting calculus and was given the following problem:

In a right triangle, leg $x$ is increasing at a rate of $2m/s$ while leg $y$ is decreasing so that the area is always $6m^2$. How fast is the hypothenuse $z$ changing when $x = 3m$?

I have differentiated $x^2+y^2=z^2$ in regards to time $t$, giving:

$(x(dx/dt) + y(dy/dt))/z = dz/dt$

and I know that I need to find $dz/dt$. However, I don't know how to find $dy/dt$ to input in the above equation. Is there any better way to solve this problem? If not, how do I go about solving it?

oscar6721
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1 Answers1

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You're doing perfectly well so far. You immediately know $x' = 2$ and $x=3$. Clearly we also have that

$$ \begin{align} &A = \frac{1}{2}xy \\ \implies & 6 = \frac{1}{2}(3)y \end{align} $$

thus $y$ is easy to find. We know that the area is constant with time therefore, by the product rule,

$$A' = \dfrac{d}{dt}\bigg[\frac{1}{2}xy\bigg] = \frac{1}{2}x'y + \frac{1}{2}xy' = 0 \text{ }$$

and thus it is easy to solve for $y'$ since we already know $x,x',y$.

Now, we just use $x^2+y^2=z^2$ to find $z$ and then we can compute $z'$ using the formula you derived.

vb628
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