the volume $v = \\frac{4}{3}\\pi r^{3}$ of a spherical balloon changes with the radius.\na. at what rate…

the volume $v = \\frac{4}{3}\\pi r^{3}$ of a spherical balloon changes with the radius.\na. at what rate (in$^{3}$/in) does the volume change with respect to the radius when $r = 3$ in?\nb. using the rate from part a, by approximately how much does the volume increase when the radius changes from 3 to 3.5 in?\na. at what rate (in$^{3}$/in) does the volume change with respect to the radius when $r = 3$ in?\n$\\square$ in$^{3}$/in\n(type an exact answer in terms of $\\pi$.)

the volume $v = \\frac{4}{3}\\pi r^{3}$ of a spherical balloon changes with the radius.\na. at what rate (in$^{3}$/in) does the volume change with respect to the radius when $r = 3$ in?\nb. using the rate from part a, by approximately how much does the volume increase when the radius changes from 3 to 3.5 in?\na. at what rate (in$^{3}$/in) does the volume change with respect to the radius when $r = 3$ in?\n$\\square$ in$^{3}$/in\n(type an exact answer in terms of $\\pi$.)

Answer

Explanation:

Step1: Differentiate volume formula

The volume formula of a sphere is $V = \frac{4}{3}\pi r^{3}$. Differentiating with respect to $r$ using the power - rule $\frac{d}{dr}(x^{n})=nx^{n - 1}$, we get $\frac{dV}{dr}=4\pi r^{2}$.

Step2: Evaluate derivative at $r = 3$

Substitute $r = 3$ into $\frac{dV}{dr}$. So $\frac{dV}{dr}\big|_{r = 3}=4\pi(3)^{2}=36\pi$.

Step3: Approximate volume increase

The approximate change in volume $\Delta V$ is given by $\Delta V\approx\frac{dV}{dr}\Delta r$. We know $\frac{dV}{dr}=36\pi$ (from part a) and $\Delta r=3.5 - 3=0.5$. Then $\Delta V\approx36\pi\times0.5 = 18\pi$.

Answer:

a. $36\pi$ b. $18\pi$