(CIE 0606/2021/m/22/Q5)
A cube of side $x \mathrm{~cm}$ has surface area $S \mathrm{~cm}^{2}$. The volume, $V \mathrm{~cm}^{3}$, of the cube is increasing at a rate of $480 \mathrm{~cm}^{3} \mathrm{~s}^{-1}$. Find, at the instant when $V=512$,
(a) the rate of increase of $x$,
(b) the rate of increase of $S$.
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*********math solution************* $\begin{array}[t]{ll}\mbox{(a)}&\text { volume of cube }=V=x^{3} \\&\text { surface area of cube }=s=6 x^{2} \\&v=512, \dfrac{d v}{d t}=480, \\&x^{3}=512 \rightarrow x=8, s=6\left(8^{2}\right)=6 \times 64=384 \\&\dfrac{d v}{d x}=3 x^{2}, \quad \dfrac{d s}{d x}=12 x \\&\dfrac{d v}{d t}=\dfrac{d v}{d x} \cdot \frac{d x}{d t} \\&480=3(8)^{2} \cdot \dfrac{d x}{d t} \\&\dfrac{d x}{d t}=\dfrac{480}{3 x+8 \times 8}=\dfrac{5}{2} \\ \mbox{(b)}&\dfrac{d S}{d t}=\dfrac{d S}{d x} \cdot \dfrac{d x}{d t} \\&=12(8) \times \dfrac{5}{2}=240\end{array}$ **********end math solution******************** |
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