## find rate of change of the volume with respect to time

Limits, differentiation, related rates, integration, trig integrals, etc.
yabo2k
Posts: 19
Joined: Tue Apr 28, 2009 9:09 pm
Contact:

### find rate of change of the volume with respect to time

A spherical snowball is placed in the sun. The sun melts the snowball so that its radius decreases .25 inch per hour. Find the rate of changee of the volume with respect to time at the instant the radius is 4 inches.

(dr/dt)=.25in r=in V=(4/3)(pi)r^3

Then is find the derevative of the Volume fnc
(dV/dt)=(4/3)(pi)(3r^2)(dr/dt)

I got- (dv/dt)= 16in^3(pi) ( I don't know how to put in symbols but pi you know is equal to 3.14......)
Last edited by yabo2k on Wed Apr 29, 2009 2:46 pm, edited 1 time in total.

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: find rate of change of the volume with respect to time

...
I got- (dv/dt)= 16in^3(pi) ( I don't know how to put in symbols but pi you know is equal to 3.14......)
is your rate of change just in in$^3$ ?

yabo2k
Posts: 19
Joined: Tue Apr 28, 2009 9:09 pm
Contact:

### Re: find rate of change of the volume with respect to time

yes it is in (in^3). My answer is (16in^3)(pi). Correct?

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: find rate of change of the volume with respect to time

yes it is in (in^3). My answer is (16in^3)(pi). Correct?
no, it is not in in$^3$... you are missing something.

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
( I don't know how to put in symbols but pi you know is equal to 3.14......)
To format $\pi$ as text, just write "pi", as you've done.
A spherical snowball is placed in the sun. The sun melts the snowball so that its radius decreases .25 inch per hour. Find the rate of changee of the volume with respect to time at the instant the radius is 4 inches.

(dr/dt)=.25in r=in V=(4/3)(pi)r^3
Since the radius is decreasing, rather than increasing, then dr/dt should be negative:

. . . . .$\frac{dr}{dt}\, =\, -0.25\,=\, -\frac{1}{4}$

...with the units being "inches per hour".
Then is find the derevative of the Volume fnc
(dV/dt)=(4/3)(pi)(3r^2)(dr/dt)

I got- (dv/dt)= 16in^3(pi) ( I don't know how to put in symbols but pi you know is equal to 3.14......)
You did the derivation correctly:

. . . . .$\frac{dV}{dt}\, =\, \frac{4}{3}\pi(3r^2)\frac{dr}{dt}$

You know that the radius, at the time in question, is r = 4, so you plugged this in, along with the value for dr/dt:

. . . . .$\frac{dV}{dt}\, =\, \frac{4}{3}\pi (3\times16)\left(-\frac{1}{4}\right)\, =\, -16\pi$

The units on r^2 were obviously "inches squared", and the units on dr/dt were "inches per hour", so the "product" of r^2 and dr/dt has units "inches cubed per hour".

Note the major difference between your answer and the above: the "minus" sign. Since the ball is melting, shouldn't the volume be decreasing, rather than (as you have it) increasing...?