Chain rule word probblem

[ked]

Assume that a spherical snowball melts in such a way that its radius decreases at a constant rate (that is, the radius is a linear function of time). Suppose it begins as a sphere with a radius 10 cm and takes 2 hours to disappear. What is the rate of change of its volume after 1 hour?

First I did it this way:
\(\displaystyle V(r) = \frac{3}{4}\pi(r)^3\)
\(\displaystyle V'(t) = [\frac{3}{4}\pi(10 - 5t)^3]'\)
but then I realized that gives cm^3*hr instead of cm^3/hr.
Does anyone know how I can express the r in terms of t as a fraction with t in the denominator? Do I need to? Thanks!

 

[Deleted member 4993]

Assume that a spherical snowball melts in such a way that its radius decreases at a constant rate (that is, the radius is a linear function of time).

That means

\(\displaystyle \frac{dr}{dt}\, = \, constant = k\)

then

\(\displaystyle \frac{dV}{dt}\, = \frac{dr}{dt}\,\cdot \frac{dV}{dr}\, = k\cdot \frac{dV}{dr}\,\)


Suppose it begins as a sphere with a radius 10 cm and takes 2 hours to disappear.

so

k = -5 cm/hr <<< fixed the sign

r(t) = 10 - 5t

Now continue....


What is the rate of change of its volume after 1 hour?

First I did it this way:
\(\displaystyle V(r) = \frac{3}{4}\pi(r)^3\) <<<< Incorrect It should be 4/3
\(\displaystyle V'(t) = [\frac{3}{4}\pi(10 - 5t)^3]'\)
but then I realized that gives cm^3*hr instead of cm^3/hr.
Does anyone know how I can express the r in terms of t as a fraction with t in the denominator? Do I need to? Thanks!

 

[galactus]

You seem to be OK. Except that is 4/3 not 3/4 in the volume formula. I assume you just typed it backward.

\(\displaystyle r=10-5t\)

\(\displaystyle V=\frac{4}{3}{\pi}r^{3}\)

\(\displaystyle V=\frac{4}{3}{\pi}(10-5t)^{3}\)

\(\displaystyle \frac{dV}{dt}=4{\pi}(10-5t)^{2}(-5)\)

Now, plug in t=1.

 

[ked]

You seem to be OK. Except that is 4/3 not 3/4 in the volume formula. I assume you just typed it backward.

\(\displaystyle r=10-5t\)

\(\displaystyle V=\frac{4}{3}{\pi}r^{3}\)

\(\displaystyle V=\frac{4}{3}{\pi}(10-5t)^{3}\)

\(\displaystyle \frac{dV}{dt}=4{\pi}(10-5t)^{2}(-5)\)

Now, plug in t=1.

Ok, thanks for clarifying. And yes, it was just a typo. I was concentrating so hard on the LaTex, but I inverted the fraction!

 

[DR._Glockman]

V(t) = (4?/3)[r(t)]³, r(t) = -5t+10, dr/dt=-5

d[V(t)]/dt = 4?[r(t)]²(dr/dt) When t=1, r(t) = 5, and we already know dr/dt = -5

hence dV/dt =4?(5)²(-5) = -500?cm³/hr the volume is changing when time = 1 hour.