Suppose the sides of a cube are expanding at a rate of $2$ inches per minute.
How fast is the volume of the cube changing at the moment that the area of the cube’s base is $10$ square inches?
Not sure what the question means by "sides" are expanding , what ever that means I'm not sure how to do it, is it just surface area, or volume or somethings else?
$\endgroup$ 13 Answers
$\begingroup$Start off with the formula for the volume of a cube
$$V=s^3$$
And we know that the sides are increasing at a rate of $s_t=2$. Furthermore, the question gives us the fact that the area of the base is ten inches squared. Therefore $s=\sqrt{10}$. Differentiate the first equation with respect to $t$ to get$$V_t=3s^2 s_t$$And plug in the knowns to get$$V_t=3\left(\sqrt{10}\right)^22=60$$
$\endgroup$ $\begingroup$Response: In response to your question, the question just means that as the length of the sides are increasing, how much is the volume increasing at this specific point? Sides is just simply the length of one edge of the cube.
Let $x$ be the length of the side on the cube, and $V$ be the volume.
Then $V=x^3$
Note that both $x$ and $V$ are independent on time.
Now differentiate the above equation with respect to time variable $t$. $$\dfrac{dV}{dt}=3x^2\dfrac{dx}{dt}$$
Now given that $x=\sqrt{10}$ and $\dfrac{dx}{dt}=2$.
Then the rate of change of volume is $\dfrac{dV}{dt}=3(\sqrt{10})^2(2)=60$
$\endgroup$ 1 $\begingroup$HINT
We have that
- $V=L^3$
- $\frac{dL}{dt}=2$
and by chain rule
$$\frac{dV}{dt}=\frac{dV}{dL}\frac{dL}{dt}$$
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