Related rates, simple question

In summary, the snowball's volume decreases at a rate of 1cm3/min, causing its diameter to decrease at a rate of -1/50∏ cm/min. There was some confusion about the units, with the correct units being cm/min instead of cm/s. The calculation was performed using the formula for volume of a sphere and taking the derivative with respect to time.
  • #1
e^(i Pi)+1=0
247
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 [itex]\frac{-1cm}{50∏ per ?}[/itex]. 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 = [itex]\frac{4∏r^3}{3} = \frac{4∏(\frac{D}{2})^3}{3} = \frac{∏}{6}D^3[/itex]

[itex]\frac{dv}{dt}=\frac{-1cm^3}{min}=-1[/itex]

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

[itex]-1 = \frac{∏}{6}(3D^2)\frac{dD}{dt}[/itex]

[itex]-1 = (\frac{D^2∏}{2})(\frac{dD}{dt})[/itex]

[itex]\frac{dD}{dt} = \frac{-1}{(D^2∏)/2}[/itex]

[itex]\frac{dD}{dt} = \frac{-2}{D^2∏}[/itex]

[itex]\left. \frac{dD}{dt}\right|_{D = 10} = \frac{-2}{(10)^2∏} = \frac{-2}{100∏} = \frac{-1}{50∏}[/itex]
 
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  • #2
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.
 
  • #3
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 [itex]\frac{-1cm}{50∏ per ?}[/itex]. 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 = [itex]\frac{4∏r^3}{3} = \frac{4∏(\frac{d}{2})^3}{3} = \frac{∏}{6}d^3[/itex]

[itex]\frac{dv}{dt}=\frac{-1cm^3}{min}=-1[/itex]

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

[itex]-1 = \frac{∏}{6}(3d^2)\frac{dd}{dt}[/itex]

[itex]-1 = (\frac{d^2∏}{2})(\frac{dd}{dt})[/itex]

[itex]\frac{dd}{dt} = \frac{-1}{(d^2∏)/2}[/itex]

[itex]\frac{dd}{dt} = \frac{-2}{d^2∏}[/itex]

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

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:
$$\left. \frac{dD}{dt}\right|_{D = 10} = \text{whatever}$$
 
  • #4
Clarified, thanks
 
  • #5
Much better!
 

1. What are related rates?

Related rates refer to the mathematical concept of finding the rate of change of one quantity with respect to another quantity. This is often used in calculus to solve problems involving changing quantities such as distance, velocity, and acceleration.

2. How do you solve related rates problems?

To solve related rates problems, you must first identify the quantities involved and their rates of change. Then, you can use the chain rule to find the relationship between the rates of change. Finally, you can plug in the given values and solve for the unknown rate.

3. What is the chain rule?

The chain rule is a calculus rule that allows you to find the derivative of a composite function. In related rates problems, it is used to find the relationship between the rates of change of different quantities.

4. Can you give an example of a related rates problem?

Sure, a common example of a related rates problem is a ladder leaning against a wall. If the base of the ladder is sliding away from the wall at a rate of 2 feet per second, and the top of the ladder is sliding down the wall at a rate of 1 foot per second, what is the rate of change of the angle between the ladder and the ground?

5. How is related rates used in real life?

Related rates are used in many real-life situations, such as calculating the rate of change of temperature in a room, the rate of change of a population, or the rate of change of water level in a tank. They are also used in physics and engineering to analyze the movement and behavior of objects.

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