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 $2\text{cm/min}$, how fast is $y$ changing when $x$ is $4\text{cm}$?

Answer 1

Step-1: Model the given situation mathematically:

It is given that a right triangle has a hypotenuse of fixed length $45\text{cm}$ and variable legs of lengths $x$ and $y$ respectively.

It is also given that $x$ increases at the rate of $2\text{cm/min}$, that is $\frac{dx}{dt}=2\text{cm/min}$.

It is required to find how fast is $y$ changing when $x$ is $4\text{cm}$, that is, ${\overline{)\frac{dy}{dt}}}_{x=4\text{cm}}$.

Step-2: Find an expression for $y$ in terms of $x$.

By the Pythagorean theorem, the length of the legs $x$ and $y$ are related to the length of the hypotenuse $45\text{cm}$ by the relationship, ${\left(45\right)}^2=x^2+y^2$.

Isolate $y$ from this relationship:

$\begin{array}{rcl}{\left(45\right)}^2& =& x^2+y^2\\ & \Rightarrow & {\left(45\right)}^2-x^2=y^2\\ & \Rightarrow & \sqrt{{\left(45\right)}^2-x^2}=y\end{array}$

Thus, $\begin{array}{rcl}\sqrt{{\left(45\text{cm}\right)}^2-x^2}& =& y\end{array}$.

Step-3: Find an expression for the rate of change of $y$ with respect to time $t$.

Differentiate both sides of the equations with respect to time $t$ and find $\frac{dy}{dt}$:

$\begin{array}{l}\begin{array}{rrcl}& \sqrt{{\left(45\right)}^2-x^2}& =& y\\ \Rightarrow & \frac{d}{dt}\left(\sqrt{{\left(45\right)}^2-x^2}\right)& =& \frac{dy}{dt}\\ \Rightarrow & \frac{1}{2}\times \frac{\left(-2x\right)}{\sqrt{{\left(45\right)}^2-x^2}}\times \frac{dx}{dt}& =& \frac{dy}{dt}\\ \Rightarrow & \frac{-x}{\sqrt{{\left(45\right)}^2-x^2}}\times \frac{dx}{dt}& =& \frac{dy}{dt}\end{array}\end{array}$

Thus, $\begin{array}{rcl}\frac{-x}{\sqrt{{\left(45\right)}^2-x^2}}\times \frac{dx}{dt}& =& \frac{dy}{dt}\end{array}$.

Step-4: Find the rate of change of $y$ with respect to time $t$, when $x=4\text{cm}$.

Find ${\overline{)\frac{dy}{dt}}}_{x=4\text{cm}}$ by substituting the given value $\frac{dx}{dt}=2\text{cm/min}$ and $x=4\text{cm}$:

$\begin{array}{rcl}\frac{-x}{\sqrt{{\left(45\right)}^2-x^2}}\times \frac{dx}{dt}& =& \frac{dy}{dt}\\ & \Rightarrow & \frac{-x}{\sqrt{{\left(45\right)}^2-x^2}}\times 2=\frac{dy}{dt}\\ & \Rightarrow & \frac{-\left(4\right)}{\sqrt{{\left(45\right)}^2-{\left(4\right)}^2}}\times 2={\overline{)\frac{dy}{dt}}}_{x=4}\\ & \Rightarrow & \frac{-\left(4\right)}{\sqrt{\left(45-4\right)\left(45\text{+}4\right)}}\times 2={\overline{)\frac{dy}{dt}}}_{x=4}\\ & \Rightarrow & \frac{-\left(4\right)}{\sqrt{\left(41\right)\left(49\right)}}\times 2={\overline{)\frac{dy}{dt}}}_{x=4}\\ & \Rightarrow & \frac{-\left(4\right)}{7\sqrt{41}}\times 2={\overline{)\frac{dy}{dt}}}_{x=4}\\ & \Rightarrow & \frac{-8}{7\sqrt{41}}\mathrm{cm}/\min ={\overline{)\frac{dy}{dt}}}_{x=4}\end{array}$

Hence, $y$ is changing at the rate of ${\overline{)\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}}}_{\mathbf{x}\mathbf{=}\mathbf{4}\mathbf{\text{cm}}}\mathbf{=}\frac{\mathbf{-}\mathbf{8}}{\mathbf{7}\sqrt{\mathbf{41}}}\mathbf{\text{cm/min}}$ when $x$ is $4\text{cm}$.

The negative sign means that $y$ decreases as time $t$ increases.

Also, the expression can be approximated ${\overline{)\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}}}_{\mathbf{x}\mathbf{=}\mathbf{4}\mathbf{\text{cm}}}\mathbf{\approx }\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{176}\mathbf{}\mathbf{\text{cm/min}}$.