How Fast is the Radius of a Circle Increasing When a Stone is Dropped in Water?

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SUMMARY

The rate at which the radius of a circle increases when a stone is dropped into water is calculated using the relationship between the circumference and radius. Given that the perimeter of the circle increases at 3 m/s and the radius is 2 m, the correct rate of increase of the radius, denoted as dr/dt, is 0.48 m/s. The confusion arose from incorrect arithmetic when calculating 3/(2π), which led to erroneous results of 4.71 m/s and 2.36 m/s.

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  • Understanding of calculus, specifically derivatives
  • Familiarity with the formulas for circumference (C = 2πr) and area (A = πr^2) of a circle
  • Basic arithmetic operations involving π
  • Ability to interpret and manipulate equations
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  • Review calculus concepts related to derivatives and rates of change
  • Practice problems involving the relationship between circumference and radius
  • Learn how to correctly input complex fractions into calculators
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Students studying calculus, particularly those focused on related rates, as well as educators looking for examples of practical applications of derivatives in real-world scenarios.

Schaus
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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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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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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!
 
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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