A circular metal plate is heated so that is diameter is increasing at a constant rate of of $0.005\frac{m}{s}$. At what rate is the area of the surface of the plate increasing when its diameter is $6$ meters?
Answer in $\frac{m^2}{s}$ correct to $2$ decimal places.
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Trick question, diameter increases at a constant rate so therefore the rate is the same at $1$m and at $6$m
There the rate is $0.005\frac{m}{s}$ or $0.01\frac{m}{s}$ to $2$ decimal places.
I just don't know where the $\frac{m^2}{s}$ comes into it, that just doesn't make sense unless you take a derivative but where's that coming from?
Since $A(t) = \frac{\pi}{4}\,D^2(t)$, it follows that $A'(t) = \frac{\pi}{4}\,2\,D(t)\,D'(t) = \frac{\pi}{2}\,D(t)\,D'(t)$.
So when at the time $t = t^*$ you have $D(t^*) = 6\,m$ and $D'(t^*) = \frac{1}{200}\,\frac{m}{s}$, follows: $$A'(t^*) = \frac{\pi}{2} \cdot \left(6\,m\right)\cdot\left(\frac{1}{200}\,\frac{m}{s}\right) = \frac{3\pi}{200}\,\frac{m^2}{s} \approx 0.047\,\frac{m^2}{s}\,. $$