# Math Help - word problem...

1. ## word problem...

A sphere shrinks at a rate of 1 cm^3 per second. At what rate does the diameter decrease when the diameter is 10 cm? And yes, this is all the info I was given.

2. Originally Posted by driver327
A sphere shrinks at a rate of 1 cm^3 per second. At what rate does the diameter decrease when the diameter is 10 cm? And yes, this is all the info I was given.
Well, for a sphere
$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left ( \frac{d}{2} \right ) ^3$

$V = \frac{1}{6} \pi d^3$

So
$\frac{dV}{dt} = \frac{1}{2} \pi d^2 \frac{dd}{dt}$

The rest you can do yourself.

-Dan

3. Originally Posted by driver327
A sphere shrinks at a rate of 1 cm^3 per second. At what rate does the diameter decrease when the diameter is 10 cm? And yes, this is all the info I was given.
I have little time more before I have to go for work. I just want to practice on this. I'm groping right now, rusty.

To find: rate of decrease of diameter
So dD/dt, where D means diameter

Sphere:
V = (4/3)(pi)(r^3)

in terms of D,
V = (4/3)(pi)[(D/2)^3]
V = (1/6)(pi)(D^3)
Differentiate both sides with respect to time t,
dV/dt = (pi/6)[3 D^2 dD/dt]
dV/dt = (pi/2)(D^2) dD/dt ----------**

So when D = 10cm, and since dV/dt = -1 cu.cm,
-1 = (pi/2)(10^2) dD/dt
-1 = 50pi dD/dt
dD/dt = -1/(50pi)

Therefore, when the diameter is 10cm, the diameter is decreasing at the rate of 1/(50pi) cm/sec. ---------------answer.

No time to review. If there are mistakes, I'll rectify after work---about 11 hrs from now.