If the volume of the expanding cube is increasing at the rate of $24 cm^3/min$, how fast is the surface area increasing when surface area is $216cm^2$?
My Approach :
$V=l^3$ $$\dfrac {dV}{dt}=3l^2\dfrac {dl}{dt}$$ $$24=3l^2\dfrac {dl}{dt}$$
Also, $S=6l^2$ $$\dfrac {dS}{dt}=12l.\dfrac {dl}{dt}$$
How do I proceed?
On
Use the equation $S = 6l^2$ and the fact that the surface area is $216~\text{cm}^2$ to solve for $l$, then use the equation $$24 = 3l^2~\frac{dl}{dt}$$ to solve for $dl/dt$. Substitute for $l$ and $dl/dt$ to solve for $dS/dt$.
On
The quantity $\frac{V^2}{S^3}$ is the same for all spatial figures of the same shape. For instance, for spheres it is $\frac{(4\pi R^3/3)^2}{(4 \pi R^2)^3}= \frac{1}{36 \pi}$, while for cubes it is $\frac{1}{6^3}$. Therefore $$d\ \log \frac{V^2}{S^3} = 0$$ or $$2 \frac{dV}{V} - 3 \frac{dS}{S}=0$$
From here we get $$d S = \frac{2}{3}\frac{S}{V} d V$$
Since it's about cubes, we have $V= (\frac{S}{6})^{3/2}$. Substitute and get the result.
On
Direct substitution is also possible:
$$\frac{dS}{dt}=6\frac{2}{3}V^{-\frac{1}{3}}\frac{dV}{dt} =4V^{-\frac{1}{3}}\frac{dV}{dt}$$
$$\frac{dS}{dt}=4\sqrt{6}\frac{1}{\sqrt{S}}\frac{dV}{dt}$$
The volume: $V=x^3$. The surface: $S=6x^2$. Given the surface is $216$, we can find the side: $$6x^2=216 \Rightarrow x=6.$$ Given the volume increases at the rate $24$, we can find at what rate the side is increases: $$\frac{dV}{dt}=3x^2\cdot \frac{dx}{dt}=24 \Rightarrow \frac{dx}{dt}=\frac{24}{3\cdot 6^2}=\frac29.$$ Given the volume and surface equation, we can differentiate the volume: $$V=S\cdot \frac{x}{6} \Rightarrow \frac{dV}{dt}=\frac{dS}{dt}\cdot \frac{x}{6}+S\cdot \frac{1}{6}\cdot \frac{dx}{dt}=24 \Rightarrow \\ \frac{dS}{dt}\cdot \frac{6}{6}+216\cdot \frac{1}{6}\cdot \frac{2}{9}=24 \Rightarrow \\ \frac{dS}{dt}=16. $$