Question

The hypotenuse of a right triangle is constantly 8 inches long. One leg is increasing at the

rate of 3in./min. At what rate is the area of the triangle changing when the increasing leg is

4in.

Answer 1

Given

$$x^2+y^2=8^2, x>0, y>0$$

Then

$$x=\sqrt{64-y^2}, 0<y<8$$

The area of the triangle is

$$A=\dfrac{1}{2}xy$$

Substitute

$$A=A(t)=\dfrac{1}{2}y\sqrt{64-y^2}$$

Differentiate with respect to $t$

$$\dfrac{dA}{dt}=\dfrac{1}{2}(\sqrt{64-y^2}+\dfrac{-2y^2}{2\sqrt{64-y^2}})\dfrac{dy}{dt}$$

$$\dfrac{dA}{dt}=(\dfrac{64-y^2-y^2}{2\sqrt{64-y^2}})\dfrac{dy}{dt}$$

$$\dfrac{dA}{dt}=(\dfrac{32-y^2}{\sqrt{64-y^2}})\dfrac{dy}{dt}$$

If $\dfrac{dy}{dt}=3\ in/min, y=4\ in,$ then

$$\dfrac{dA}{dt}=(\dfrac{32-4^2}{\sqrt{64-4^2}})(3)=4\sqrt{3}({in}^2/min)$$

The area of the triangle  is increasing at the rate of $4\sqrt{3}$ in²/min.