good ol P3

A circular oil-slick is increasing in area at a constant rate of 3m^2s^-1. Find the rate of increase of the radius of the slick at the instant when the area is 1200m^2

Reply 1

Gaz031

15

W@t3R

A circular oil-slick is increasing in area at a constant rate of 3m^2s^-1. Find the rate of increase of the radius of the slick at the instant when the area is 1200m^2

I have to go to work but i'll just outline it:
You have Da/Dt. You need to find Dr/dt
Use da/dr and the chain rule:
Dr/dt = Da/dt . dr/da.
a = pir^2

Reply 2

W@t3R

OP

..... yea, attempted that already, i didn't get it.....

Reply 3

dvs

10

dr/dt = (dA/dt) . (dr/dA), where r is the radius of the oil-slick, A is its area, and t is time. You're given dA/dt (the rate of increase in area) and you can find dA/dr by differentiating A=pi.r^2 (since the oil-slick circular) wrt r. Also note that dr/dA=1/(dA/dr).

Reply 4

W@t3R

OP

dvs

dr/dt = (dA/dt) . (dr/dA), where r is the radius of the oil-slick, A is its area, and t is time. You're given dA/dt (the rate of increase in area) and you can find dA/dr by differentiating A=pi.r^2 (since the oil-slick circular) wrt r. Also note that dr/dA=1/(dA/dr).

Yea, u know, I'm still not getting the same answer in the book. Can some1 actually do it? please?

Reply 5

Fermat

8

da/dt = 3 m²/s

a=pir²
r=rt(a/pi)
dr/da = 1/(2rt(api))

dr/dt = dr/da.da/dt
dr/dt = 1/(2rt(api))*3
dr/dt = 3/(2rt(api))
=============

At a = 1200 m²,

dr/dt = 3/(2rt(1200pi))
dr/dt = 0.0244 m/s
dr/dt = 2.44 cm/s
=============

Reply 6

username9816

14

W@t3R

A circular oil-slick is increasing in area at a constant rate of 3m^2s^-1. Find the rate of increase of the radius of the slick at the instant when the area is 1200m^2

dA/dt = 3
A = Pi.r^2 --> dA/dr = 2rPi --> dr/dA = 1/(2rPi)
dr/dt = dr/dA * dA/dt = 3/(2rPi)

When A = 1200 m^2:
1200 = Pi.r^2 ---> r = Sqrt(1200/Pi) = 19.544 m

Hence at A = 1200 m^2:
---> dr/dt = 3/[2(19.544)(Pi)] = 0.0244 ms^-1

Reply 7

Fermat

8

Nima

dA/dt = 3
A = Pi.r^2 --> dA/dr = 2rPi --> dr/dA = 1/(2rPi)
dr/dt = dr/dA * dA/dt = 3/(2rPi)

When A = 1200 m^2:
1200 = Pi.r^2 ---> r = Sqrt(1200/Pi) = 19.544 m

Hence at A = 1200 m^2:
---> dr/dt = 3/[2(19.544)(Pi)] = 0.0244 ms^-1

Which confirms my answer, thanks nima.

good ol P3

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