A round balloon is being blown up with helium so that t seconds after starting, the radius is 2t cm. Remember that for a sphere with a radius r, V=(4/3)(pi)r^3 and SA= 4(pi)r^2. What is the equation for V'(r)?

To find the derivative of the volume function V(r) with respect to r, we first express V in terms of r. Given that the radius of the balloon is a function of time, r = 2t, we substitute this expression for r into the volume formula V(r).

V(r) = (4/3)πr^3

Substitute r = 2t:

V(t) = (4/3)π(2t)^3

Now, we differentiate V(t) with respect to t using the chain rule:

dV/dt = dV/dt * d(r^3)/dt

dV/dt = (4/3)π * 3(2t)^2 * 2

Simplify:

dV/dt = 8πt^2 * 4

dV/dt = 32πt^2

Now, we need to find dV/dr, which is the derivative of V with respect to r. We use the chain rule again:

dV/dr = dV/dt * dt/dr

Given that r = 2t, dt/dr = 2.

Substitute dV/dt:

dV/dr = 32πt^2 * 2

dV/dr = 64πt^2

Finally, we substitute r = 2t back into the equation:

dV/dr = 64π(2t)^2

dV/dr = 256πt^2

So, the equation for V'(r) is:

V'(r) = 256πt^2

To find the equation for $V'(r)$, we first need to express the volume $V$ in terms of the radius $r$. Given that the radius $r$ of the balloon is a function of time $t$ (i.e., $r = 2t$), we need to express $V$ in terms of $t$.

The volume $V$ of a sphere with radius $r$ is given by $V = \frac{4}{3} \pi r^3$.

Since $r = 2t$, we can substitute $2t$ for $r$ in the equation for $V$:

$V = \frac{4}{3} \pi (2t)^3$

Now, let's simplify this expression:

$V = \frac{4}{3} \pi (8t^3)$ $V = \frac{32}{3} \pi t^3$

Next, we'll find the derivative of $V$ with respect to $r$, denoted as $V'(r)$, using the chain rule:

$V'(r) = \frac{dV}{dr} = \frac{dV}{dt} \cdot \frac{dt}{dr}$

First, we find $\frac{dV}{dt}$, the derivative of $V$ with respect to $t$:

$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{32}{3} \pi t^3 \right)$ $\frac{dV}{dt} = \frac{32}{3} \pi \cdot 3t^2$ $\frac{dV}{dt} = 32 \pi t^2$

Now, we need to find $\frac{dt}{dr}$, the derivative of $t$ with respect to $r$. Since $r = 2t$, we can express $t$ in terms of $r$ and then differentiate:

$r = 2t$ $t = \frac{r}{2}$

Now, differentiating $t$ with respect to $r$:

$\frac{dt}{dr} = \frac{d}{dr} \left( \frac{r}{2} \right)$ $\frac{dt}{dr} = \frac{1}{2}$

Finally, we can find $V'(r)$ by multiplying $\frac{dV}{dt}$ and $\frac{dt}{dr}$:

$V'(r) = \frac{dV}{dt} \cdot \frac{dt}{dr}$ $V'(r) = 32 \pi t^2 \times \frac{1}{2}$ $V'(r) = 16 \pi t^2$

Since $t = \frac{r}{2}$, we can substitute this expression for $t$ back into the equation:

$V'(r) = 16 \pi \left( \frac{r}{2} \right)^2$ $V'(r) = 16 \pi \left( \frac{r^2}{4} \right)$ $V'(r) = 4 \pi r^2$

So, the equation for $V'(r)$ is $V'(r) = 4 \pi r^2$.