How Does the Diameter of a Melting Snowball Change Over Time?

  • Thread starter Thread starter e^(i Pi)+1=0
  • Start date Start date
  • Tags Tags
    Related rates
e^(i Pi)+1=0
Messages
246
Reaction score
1

Homework Statement


A snowball melts such that the volume decreases at a rate of 1cm3/min. At what rate is the diameter decreasing when diameter=10?

I know the answer is \frac{-1cm}{50∏ per ?}. My problem is with the units on the bottom. It was given as seconds, but shouldn't it be per minute? There was no conversion done anywhere in the problem. The answer is small, and it seems like it would fit better as per seconds rather than per minute, but I can't find out how we went from minutes to seconds.

d=derivative
D=diameter

volume = \frac{4∏r^3}{3} = \frac{4∏(\frac{D}{2})^3}{3} = \frac{∏}{6}D^3

\frac{dv}{dt}=\frac{-1cm^3}{min}=-1

\frac{dv}{dt} = \frac{∏}{6}(3D^2)\frac{dD}{dt}

-1 = \frac{∏}{6}(3D^2)\frac{dD}{dt}

-1 = (\frac{D^2∏}{2})(\frac{dD}{dt})

\frac{dD}{dt} = \frac{-1}{(D^2∏)/2}

\frac{dD}{dt} = \frac{-2}{D^2∏}

\left. \frac{dD}{dt}\right|_{D = 10} = \frac{-2}{(10)^2∏} = \frac{-2}{100∏} = \frac{-1}{50∏}
 
Last edited:
Physics news on Phys.org
e^(i Pi)+1=0 said:
My problem is with the units on the bottom. It was given as seconds, but shouldn't it be per minute?
I'd say you are correct. The units should be cm/min, not cm/s.
 
e^(i Pi)+1=0 said:

Homework Statement


A snowball melts such that the volume decreases at a rate of 1cm3/min. At what rate is the diameter decreasing when diameter=10?

I know the answer is \frac{-1cm}{50∏ per ?}. My problem is with the units on the bottom. It was given as seconds, but shouldn't it be per minute? There was no conversion done anywhere in the problem. The answer is small, and it seems like it would fit better as per seconds rather than per minute, but I can't find out how we went from minutes to seconds.

v = \frac{4∏r^3}{3} = \frac{4∏(\frac{d}{2})^3}{3} = \frac{∏}{6}d^3

\frac{dv}{dt}=\frac{-1cm^3}{min}=-1

\frac{dv}{dt} = \frac{∏}{6}(3d^2)\frac{dd}{dt}

-1 = \frac{∏}{6}(3d^2)\frac{dd}{dt}

-1 = (\frac{d^2∏}{2})(\frac{dd}{dt})

\frac{dd}{dt} = \frac{-1}{(d^2∏)/2}

\frac{dd}{dt} = \frac{-2}{d^2∏}

\frac{d(10)}{dt} = \frac{-2}{10^2∏} = \frac{-2}{100∏} = \frac{-1}{50∏}

Your work would be easier to follow if you used upper case D for diameter instead of lower case d. dd/dt is somewhat confusing, while dD/dt is probably less so.

In your last line you have d(10)/dt = some nonzero number. I know what you mean, but this isn't the way to say it, since the derivative of every constant is zero.

I think this is what you meant to say:
$$\left. \frac{dD}{dt}\right|_{D = 10} = \text{whatever}$$
 
Clarified, thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top