Question

a spherical balloon is deflated at a rate of 256pi/3 cm^3/sec. at what rate is the radius of the balloon changing when the radius is 8 cm?

Answer 1

What you need to look for is $\frac { dr }{ dt }$ when r=8.

Now:

$Volume\quad of\quad a\quad sphere:\\ \\ V=\frac { 4 }{ 3 } \pi { r }^{ 3 }$

$\therefore \quad \frac { dV }{ dr } =\frac { 4 }{ 3 } \pi \cdot 3{ r }^{ 2 }=4\pi { r }^{ 2 }\\ \\ \therefore \quad \frac { dr }{ dV } =\frac { 1 }{ \frac { dV }{ dr } } =\frac { 1 }{ 4\pi { r }^{ 2 } }$

$And:\\ \\ \frac { dV }{ dt } =-\frac { 256\pi }{ 3 } \\ \\ Therefore:\\ \\ \frac { dr }{ dt } =\frac { dr }{ dV } \cdot \frac { dV }{ dt }$

$\\ \\ =\frac { 1 }{ 4\pi { r }^{ 2 } } \cdot -\frac { 256\pi }{ 3 } \\ \\ =-\frac { 256 }{ 12 } \cdot \frac { \pi }{ \pi } \cdot \frac { 1 }{ { r }^{ 2 } }$

$\\ \\ =-\frac { 64 }{ 3{ r }^{ 2 } } \\ \\ When\quad r=8,\\ \\ \frac { dr }{ dt } =-\frac { 64 }{ 3\cdot { 8 }^{ 2 } } =-\frac { 1 }{ 3 } \\ \\ Answer:\quad -\frac { 1 }{ 3 } \quad cm/sec$