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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

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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, $$ \frac{dA}{dt}=\frac{d(\frac{1}{2}ah)}{dt}=\frac{1}{2}(a\frac{dh}{dt}+h\frac{da}{dt})=0.5(26\times(3-2))=13. $$

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