Related Rates stone drop

In summary, the conversation discusses the problem of a stone being dropped into water and forming a circle with a radius of 2m, which slowly expands as the perimeter increases at a rate of 3 m/s. The question asks for the rate at which the radius is increasing at that moment, and the answer is determined to be 0.48 m/s by taking the derivative of the circumference formula and dividing by 2π. An error in arithmetic calculation is discovered and the correct answer is determined to be 0.48 m/s.
  • #1
Schaus
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5

Homework Statement


A stone is dropped into some water and a circle of radius r is formed and slowly expands. The perimeter of the circle is increasing at 3 m/s. At the moment the radius is exactly 2m, what rate is the radius of the circle increasing?
Answer ## \frac {dr} {dt} = 0.48 m/s##

Homework Equations


## C = 2πr##
## A = πr^2##

The Attempt at a Solution


## \frac {dC} {dt} = 3 m/s##
## \frac {dr} {dt} = ?##
## C = 2πr##
Taking the derivative of circumference formula
## \frac {dC} {dt} = 2π \frac {dr} {dt}##
Subbing in my ## \frac {dC} {dt} = 3 m/s##
## 3 = 2π \frac {dr} {dt}##
Dividiving both sides by 2π gives me 4.71 m/s. I have also tried doing this with the Area of a circle formula but I still got 2.36 m/s which is no where near the 0.48 it is supposed to be. Any help would be greatly appreciated.
 
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  • #2
Schaus said:

Homework Statement


A stone is dropped into some water and a circle of radius r is formed and slowly expands. The perimeter of the circle is increasing at 3 m/s. At the moment the radius is exactly 2m, what rate is the radius of the circle increasing?
Answer ## \frac {dr} {dt} = 0.48 m/s##

Homework Equations


## C = 2πr##
## A = πr^2##

The Attempt at a Solution


## \frac {dC} {dt} = 3 m/s##
## \frac {dr} {dt} = ?##
## C = 2πr##
Taking the derivative of circumference formula
## \frac {dC} {dt} = 2π \frac {dr} {dt}##
Subbing in my ## \frac {dC} {dt} = 3 m/s##
## 3 = 2π \frac {dr} {dt}##
Dividiving both sides by 2π gives me 4.71 m/s. I have also tried doing this with the Area of a circle formula but I still got 2.36 m/s which is no where near the 0.48 it is supposed to be. Any help would be greatly appreciated.
Check your arithmetic calculating ##\frac 3 {2\pi}##.
 
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  • #3
That's weird. When I do 3/2pi it gives me 4.71 but if I do 3/6.28 I get my answer. Thanks for the help, I wish I had thought of trying that earlier!
 
  • #4
Schaus said:
That's weird. When I do 3/2pi it gives me 4.71 but if I do 3/6.28 I get my answer. Thanks for the help, I wish I had thought of trying that earlier!

That's because the calculator interprets it as ## \frac{3}{2}\pi##, ## \frac{3\pi}{2}##, or ##\frac{3}{2}(\frac{\pi}{1}) ## as opposed to ## \frac{3}{2\pi}##.

Next time, enter it into the calculator as 3/(2## \pi##)
 
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1. What is a "Related Rates stone drop" problem?

A "Related Rates stone drop" problem is a type of problem in calculus that involves finding the rate of change of one variable with respect to another variable, using the relationship between the two variables. In this specific problem, the stone dropping represents one variable, and the distance of the stone from the ground represents the other variable.

2. How do you solve a "Related Rates stone drop" problem?

To solve a "Related Rates stone drop" problem, you must first identify the variables involved and their relationships. Then, you can use implicit differentiation and the chain rule to find an equation that relates the rates of change of these variables. Finally, you can plug in the given values and solve for the desired rate of change.

3. What are the key concepts needed to solve a "Related Rates stone drop" problem?

To solve a "Related Rates stone drop" problem, you need to have a good understanding of implicit differentiation, the chain rule, and how to set up and solve related rates equations. It is also helpful to have a solid understanding of basic calculus concepts such as derivatives and rates of change.

4. Can you provide an example of a "Related Rates stone drop" problem?

Sure! Here's an example: A stone is dropped from a height of 20 meters. At the same time, another stone is thrown downward from a height of 40 meters with an initial velocity of 10 m/s. How fast is the distance between the two stones increasing after 2 seconds?

In this problem, the variables are the distance of the first stone from the ground (measured in meters) and the distance between the two stones (also measured in meters). Using the given information and the related rates equation, we can find that the distance between the two stones is increasing at a rate of approximately 5.91 m/s after 2 seconds.

5. Are "Related Rates stone drop" problems only applicable to stones or falling objects?

No, "Related Rates stone drop" problems can be applied to a variety of scenarios, not just objects falling due to gravity. They can also be used to solve problems involving rates of change in other situations, such as chemical reactions, population growth, or fluid dynamics.

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