Question

Consider a right triangle with a hypotenuse of fixed length $45\text{cm}$ and variable legs of lengths $x$ and $y$ respectively. If $x$ increases at the rate of $2$$\mathrm{cm}/\min$, how fast is $y$ changing when $x$ is $4\text{cm}$?

Answer 1

Finding the rate of change of $y$:

Using Pythagorean theorem, we have

$$\begin{gathered} {45}^2=x^2+y^2----\left(i\right) \\ y=\sqrt{{45}^2-x^2} \\ =\sqrt{{45}^2-4^2}\left[x=4cm\right] \\ =\sqrt{2025-16} \\ \Rightarrow y=\sqrt{2009}---\left(ii\right) \end{gathered}$$

Differentiating the above equation w.r.t., we get ;

$$\begin{gathered} \frac{d}{dt}(x^2+y^2)=\frac{d}{dt}{\left(45\right)}^2 \\ \Rightarrow 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 \\ \Rightarrow 2y\frac{dy}{dt}=-2x\frac{dx}{dt} \\ \Rightarrow y\frac{dy}{dt}=-x\frac{dx}{dt}---\left(1\right) \end{gathered}$$

$x$ increases at the rate of $2$ $\text{cm/min}$,

Therefore $\frac{dx}{dt}=2\text{cm/min}\text{and}x=4\text{cm}$

Substituting the above values in equation (1), we get :

$$\begin{gathered} y\frac{dy}{dt}=-4\times 2 \\ \Rightarrow \frac{dy}{dt}=\frac{-4\times 2}{y} \\ =\frac{-8}{\sqrt{2009}}\left[\because y=\sqrt{2009}\right] \\ \Rightarrow \frac{dy}{dt}=-0.178\approx -0.18 \end{gathered}$$

Negative sign implies the function is decreasing.

Hence, $y$ is decreasing at the rate of $-0.18\text{cm/min}$when $x=4\text{cm}$.