Question

Oil spilled from a tanker spreads in a circle whose circumference increases at a rate of 40ft/sec. How fast is the area of the spill increasing when the circumference of the cirlce is 100pi ft

Answer 1

when dealing with rates of change, we wanna ask, what can we derive

Answer 2

ok

Answer 3

Why is it everytime I find a question no one is working on, someone shows up....This person has been waiting for 13 minutes...no response. I show up, and within a minute amistre shows up..... :S

Answer 4

dC/dt = 40

dA/dt = dA/dr * dr/dC * dC/dt ... maybe

Answer 5

im a ninja lol

Answer 6

I'm Mafia

Answer 7

everyone help me

Answer 8

C = 2pi r

C/2pi = r

A = pi (C/2pi)^2

dA/dt = pi 2(C/2pi) (1/2pi) dC/dt

Answer 9

\[\frac{dA}{dt} = \frac{C(40)}{2pi}\]
\[\frac{dA}{dt} = \frac{C(20)}{pi}\]and C = 100 right?

Answer 10

100pi

Answer 11

100pi

Answer 12

so pis cancels and we got 100(20) = 2000 units per time

Answer 13

... huh

Answer 14

huh isnt something i can relate to; your gonna have to be a bit more specific

Answer 15

lol sry can u write out the stops using the equation tab

Answer 16

C=40FT
How fast is the area incraesing when C=100pi ft?

Answer 17

\[C = 2pi\ r\]
\[\frac{C}{2pi} = r\]
\[A = pi \left(\frac{C}{2pi}\right)^2\]
\[A = pi \frac{C^2}{4pi\ pi}\]
\[A = \frac{C^2}{4pi}\]
now to derive

Answer 18

thanks

Answer 19

\[\frac{d}{dt}\left(A = \frac{C^2}{4pi}\right)\]
\[\frac{dA}{dt} = \frac{2C}{4pi}\ \frac{dC}{dt}\]
\[\frac{dA}{dt} = \frac{C}{2pi}\ \frac{dC}{dt}\]
C=100pi and dC/dt = 40
\[\frac{dA}{dt} = \frac{100pi(40)}{2pi}\]
\[\frac{dA}{dt} = 50(40)=2000\]

Answer 20

and this is in feet/sec

Answer 21

yes

Answer 22

thank you sooo much i will review this

Answer 23

A two line Mathematica solution with comments is attached. Confirms amistre64's result.