Related rates in a rectangle with diagonal growth

Key concept: This is a classic related-rates setup: a right triangle hidden inside a rectangle, with the diagonal, length, and width all changing over time. The key is to connect the variables with Pythagorean theorem and differentiate correctly.

Step by step

Set up the relationship

Let the rectangle’s length be $x$, width be $y$, and diagonal be $z$. Because the diagonal forms a right triangle with the sides, we use

$z^2 = x^2 + y^2.$

This is the only geometric equation you need. The given rates are $\frac{dz}{dt}=1$ in/hr and $\frac{dx}{dt}=\frac14$ in/hr. You are asked for $\frac{dy}{dt}$ when $x=8$ and $y$ is the given width value from the problem statement.

Before differentiating, make sure every symbol is a function of time. That is what makes this a related-rates problem rather than a static geometry problem.

Differentiate with respect to time

Differentiate both sides of $z^2=x^2+y^2$ with respect to $t$:

$2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}.$

Now divide by 2:

$z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}.$

This step is where many students get confused. There is no isolated $2\frac{dy}{dt}$ term; the derivative of $y^2$ is $2y\frac{dy}{dt}$ by the chain rule. The variable $y$ stays attached to its derivative.

Solve for the width rate

Rearrange to isolate $\frac{dy}{dt}$:

$y\frac{dy}{dt}=z\frac{dz}{dt}-x\frac{dx}{dt},$

so

$\frac{dy}{dt}=\frac{z\frac{dz}{dt}-x\frac{dx}{dt}}{y}.$

If the width is $6$ in and the length is $8$ in, then the diagonal is

$z=\sqrt{8^2+6^2}=10.$

Substitute:

$\frac{dy}{dt}=\frac{10(1)-8(1/4)}{6}=\frac{10-2}{6}=\frac{8}{6}=\frac43\text{ in/hr}.$

So the width is increasing at $\boxed{\frac43\text{ in/hr}}$.

Why the marked step matters

The confusion usually comes from skipping the chain rule. When a side length changes with time, its square changes as well, so the derivative must include the variable itself. The same logic works for any related-rates problem involving area, volume, or geometry inside a right triangle.

Pitfall alert

A common mistake is to treat $\frac{d}{dt}(y^2)$ as $2\frac{dy}{dt}$ instead of $2y\frac{dy}{dt}$. That drops the current width value and gives the wrong rate. Another frequent error is using the diagonal formula correctly but forgetting to compute $z$ from the actual side lengths before substituting the rates. Always solve for the missing geometric quantity first, then plug into the differentiated equation.

Try different conditions

If the problem instead asked: “The diagonal increases at 1 in/hr, the length increases at 1/4 in/hr, and the width is 6 in while the length is 8 in. Find the width rate,” the method is identical. Use $z^2=x^2+y^2$, differentiate, and substitute $z=10$. The only change is the numerical value of $y$, which makes the final rate computable. If the width were different, the same formula would still apply, but the answer would change because the denominator is $y$.

Further reading

Pythagorean theorem, chain rule, related rates