Question

Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 9 mi^2/hr . How rapidly is radius of the spill increasing when the area is 10 mi^2 ?

Answer 1

3 mi/hr (square root of 9, because square of radius increases in proportion to area)

Answer 2

wait, I'm on crack.

Answer 3

me too.

Answer 4

me too.

Answer 5

\[A=\pi r^2\]
\[A'=2 \pi r r'\]
\[r'=\frac{A'}{2\pi r}\]

Answer 6

\[A'=9\] so \[r'=\frac{9}{2\pi r}\]

Answer 7

if
\[A=10\] then \[r=\sqrt{\frac{10}{\pi}}\]

Answer 8

plug that in and get the answer
\[r'=\frac{9}{2\pi}\times \sqrt{\frac{\pi}{10}}\]

Answer 9

im lost is that the answer you got? because when i pluged it in it said it was wrong

Answer 10

A = pi*r^2

dA/dt = 2*pi*r*dr/dt

dr/dt = A'/(2*pi*r) = 9/(2*pi*10) = 9/20pi = 0.1432

Answer 11

oh wait...no

Answer 12

no i don't think so

Answer 13

satellite is right

Answer 14

you are told area is 10 yes? and
\[A=\pi r^2\] making
\[r=\sqrt{\frac{10}{\pi}}\]

Answer 15

you obviously entered it into the calculator incorrectly

Answer 16

do you want a decimal for this?

Answer 17

yes please

Answer 18

= 0.8029

Answer 19

that is what i get as well

Answer 20

yea that was perfect thanks