# Related Rates stone drop

1. Apr 6, 2017

### Schaus

1. The problem statement, all variables and given/known data
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 $\frac {dr} {dt} = 0.48 m/s$
2. Relevant equations
$C = 2πr$
$A = πr^2$
3. 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.

2. Apr 6, 2017

### LCKurtz

Check your arithmetic calculating $\frac 3 {2\pi}$.

3. Apr 6, 2017

### Schaus

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. Apr 6, 2017

### Rio Larsen

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$\pi$)

Last edited: Apr 6, 2017