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
Click For Summary

Homework Help Overview

The discussion revolves around a problem involving the melting of a snowball, specifically focusing on how the diameter of the snowball changes over time as its volume decreases at a specified rate. The subject area includes calculus and related rates.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the volume of the snowball and its diameter, using calculus to derive the rate of change of the diameter. There is a focus on unit consistency, particularly regarding whether the rate should be expressed in cm/min or cm/s.

Discussion Status

Some participants have provided clarifications regarding the units of measurement, suggesting that the rate should indeed be in cm/min. There is an ongoing examination of the notation used for diameter and its derivative, with suggestions for clearer representation.

Contextual Notes

Participants note potential confusion arising from the use of different symbols for diameter and its derivative, as well as the implications of treating constants in derivative expressions.

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
 

Similar threads

Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K