The question I was given was:
The volume of a spherical balloon is increasing at a constant rate of $0.78$ inches per minute. At the instant when the radius is $3.20$ inches, the radius is increasing at a rate of
A) $0.006$ in/min
B) $0.019$ in/min
C) $0.419$ in/min
D) $6.273$ in/min
E) $100.37$ in/min
I know that $dV/dt = 0.78, r = 3.2$, and I am trying to find $dr/dt$, but I am confused on how to start it. Do I find the derivative of the volume of a sphere? (Volume of a sphere: $(4/3)(\pi)(r^3)$)
$\endgroup$3 Answers
$\begingroup$$$V = \frac{4 \pi}{3} r^3$$
so
$$dV = \frac{4 \pi}{3} 3 r^2 dr$$
You know $dV/dt$. Solve for $dr/dt$.
$\endgroup$ $\begingroup$Hint: Think of $V$ and $r$ as being functions of $t$ (time), so that$$V(t) = \frac43\pi \cdot r(t)^3$$and therefore by the chain rule$$V'(t) = 4\pi\cdot r(t)^2 \cdot r'(t)$$You know $r(t_0)$ and $V'(t_0)$ at the instant $t_0$ in question, so this will determine $r'(t_0)$ at that instant as well.
$\endgroup$ $\begingroup$Hint: You know that $V = \frac{4}{3}\pi r^{3}$. Use implicit differentiation to take the derivative of both sides with respect to $t$. You should have $\frac{dV}{dt}$ on the left, and something involving $\frac{dr}{dt}$ on the right.
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