Question

A balloon, which always remains spherical, has a variable diameter 3/2(2x+1). Find the rate of change of its volume with respect to x.

Answer 1

Let V is the volume of sphere and r is the radius of sphere.
we know, volume of sphere, V = 4/3 πr³
Given, diameter = $\frac{3}{2}(2x+1)$
radius , r = diameter/2 = $\frac{3}{4}(2x+1)$
hence, volume of sphere,
V = $\frac{4}{3}\pi\left[\begin{array}{c}\frac{3}{4}(2x+1)\end{array}\right]^3$

so, $\bf{V= \frac{9}{16}\pi(2x+1)^3}$
now differentiate with respect to x,
$\bf{\frac{dV}{dx}=\frac{9}{16}\pi\frac{d(2x+1)^3}{dx}}\\\\=\bf{\frac{9}{16}\pi.3.(2x+1)^{3-1}.2}\\\\=\bf{\frac{27}{8}\pi(2x+1)^2}$

hence , rate of change of sphere is $\bf{\frac{dV}{dx}=\frac{27}{8}\pi(2x+1)^2}$

Answer 2

Answer:

Let V is the volume of sphere and r is the radius of sphere.

we know, volume of sphere, V = 4/3 πr³

Given, diameter =

radius , r = diameter/2 =

hence, volume of sphere,

V =

so,

now differentiate with respect to x,

hence , rate of change of sphere is

Step-by-step explanation: