# Derivatives in Action, Change in radius per time of circle.

1. Jul 17, 2015

### christian0710

Hi, I'm trying to understand how Differentiation and

integral works in practice, and would really appreciate

some help interpreting this calculation-

If we have a circle with Area A=pi*r^2

1) If i want to find the change in Area with respect to

dA/dr= 2pi*r

2) If I'm told that the area of the circle changes with

time

dA/dt = 0.03 (in)^2/sec

And i wanted to find how the radius changes with time,

would this then be the right conclusion

dr/dt = dr/dA*dA/dt = (1/dA/dr)*dA/dt = 1/(2pi*r)*0.03

So the change in radius with respect to time decreases as

the radius increases. So is it correct to leave the

expression like this, or should i express r in the equation

as time, t, since it says dr/dt?

If i want to find the radius at a specific time when the

area increases as
dA/dt = 0.03 (in)^2/sec would it the be correct to

integrate

dr/dt= 0.03/(2pi*r)

to give me a function r which depends on t?

If i want to express r as t in the eqation
dr/dt= 1/(2pi*r)*0.03 , how exactly would i do it:

I could isolate r in A=pi*r^2 --> sqrt(A/pi) and

substitute sqrt(A/pi) into the equation and then to

substitute t into A, i guess i would integrate dA/dt = 0.03

(in)^2/sec to get A=A(t)=0.03*t and then substitute A=0.03t

like this

dr/dt= 0.03/(2pi*(sqrt(0.03*t/pi)

I don't know if I'm breaking any rules here? OR if my

reasoning is wrong?

2. Jul 17, 2015

### Dr. Courtney

Is this a homework problem or just general interest?

3. Jul 17, 2015

### GiuseppeR7

The first formula is wrong:
A=2 Pi r(t)^2
dA/dt= 2 Pi 2 r(t) (dr/dt)