Isoceles Related Rate Question The legs of an isoceles triangle are increasing at a rate of $16$cm/s while the base of the triangle remains constant at $60$cm. At what rate is the area of the triangle increasing when each leg is $50$cm?
The equation for area is $A = \frac{1}{2}bh$.
Then I took the derivative obtaining
$$
\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{1}{2}\frac{\mathrm{d}b}{\mathrm{d}t}h + \frac{b}{2}\frac{\mathrm{d}h}{\mathrm{d}t}.
$$
I calculated that the height of the triangle would be $40$cm when the legs are $50$cm.
From there,
$$
\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{1}{2}\cdot 0 \cdot 40 + \frac{1}{2} \cdot 60 \frac{\mathrm{d}h}{\mathrm{d}t}.
$$
I don’t know how to get $\mathrm{d}h/\mathrm{d}t$. Thanks.

If you draw out the isosceles triangle and draw a line from the top to the middle of the base, you get the height of the triangle as one of the sides of a right angle triangle (see bottom). Denote this by $h$, and let $l$ be the length of the legs. By Pythagoras, and noting that we have half of the base as one of the other lengths, we have
$$
h^{2} + 30^{2} = l^{2}.
$$
Differentiate this implicitly with respect to $t$ to get
$$
2h\frac{\mathrm{d}h}{\mathrm{d}t} = 2l\frac{\mathrm{d}l}{\mathrm{d}t} \implies \frac{\mathrm{d}h}{\mathrm{d}t} = \frac{l}{h}\frac{\mathrm{d}l}{\mathrm{d}t}.
$$
Now you have an expression for $\mathrm{d}h/\mathrm{d}t$, where $h = \sqrt{l^{2} - 30^{2}}$. You sub in your value for $l$, and sub this into your differential equation, and you should have it.

As a note, if you are uncomfortable with implicit differentiation you could rearrange and use the chain rule on
$$
h = \sqrt{l^{2} - 30^{2}}.
$$
This would give you
$$
\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{1}{2}\big(l^{2} - 30^{2}\big)^{-1/2} \cdot 2l\frac{\mathrm{d}l}{\mathrm{d}t} = \frac{l}{\sqrt{l^{2} - 30^{2}}}\frac{\mathrm{d}l}{\mathrm{d}t},
$$
exactly as above.

Also, when you calculate the derivative of the area, you know that $b$ is kept constant so you could have treated it as a constant and then you wouldn't have to have used the product rule and would have gotten the same answer. This would have just made the steps a little quicker for you.


