# Differentiated - Rates of change

A triangle $$ABC$$ is made out of an elastic piece of string. Vertices A and B are being pulled apart so that the length of the base $$AB$$ is increasing at of $$3 \ cm \ s^{-1}$$ and the height $$h$$ is decreasing at a rate of $$2 \ cm \ s^{-1}$$. Initially , $$AB = 2 cm$$ and $$h=30 cm$$. Find the rate at which the area if the triangle is change when $$AB=26$$ and $$h=26$$

So we are told that $$\frac{d(AB)}{dt} = 3$$ and $$\frac{dh}{dt} = -2$$ and that the area of the traingle is $$\frac 1 2 (AB)h$$. So my thought process was to use the chain rule such that $$\frac{d(area)}{dt} = \frac{d(area)}{dh} \times \frac{dh}{dt}$$

Now, $$\frac{d(area)}{dh} = \frac{d}{dh} \left( \frac {1}{2} (AB)h \right)= \frac 1 2 \frac{d(AB)}{dh}$$

This lead me to say that $$\frac{d(area)}{dh} =-\frac{d(AB)}{dh}$$ And as $$\frac{d(AB)}{dh}=\frac{d(AB)}{dt} \times \frac{dt}{dh}$$ then the rate of change of the area would be $$3/2 \ cm^2 s^{-1}$$ but the answer is 13. Any help would be great

The problem is that $$AB$$ is not a constant, but you have treated it as a constant here
$$\frac{d(area)}{dh} = \frac{d}{dh} \left( \frac {1}{2} (AB)h \right)= \frac 1 2 \frac{d(AB)}{dh}$$
I would think of it this way: Let $$AB=a$$. $$\frac{dA}{dt}=\frac{d(\frac{1}{2}ah)}{dt}$$ By product ruleLINK, $$\frac{dA}{dt}=\frac{d(\frac{1}{2}ah)}{dt}=\frac{1}{2}(a\frac{dh}{dt}+h\frac{da}{dt})=0.5(26\times(3-2))=13.$$