Question

Oil is dripping onto a surface at the rate of 1
10cm3/s and forms a circular lm which may be
considered to have a uniform depth of 0.1 cm. Find the rate at which the radius of the circular
lm is increasing when this radius is 5 cm.

Answer 1

determine what needs to be solved; and how it relates to the given information

Answer 2

what is a lm?

Answer 3

I undertand the problem, but not quite sure what a "circular lm" is...

General way to do this problem is to get the formula for the volume of the shape (applying the uniform depth) which will give you a two-variable function.

THen take the derivative and for radius = 5.

Answer 4

edit: "Then take the derivative and solve for when the radius = 5"

Answer 5

dr/dt is what to find

dV/dt is given as 10

V = pi r^2 *.1; derive wrt to time (t)

dV/dt = 2pi r * .1 dr/dt

10 = 2pi r * .1 dr/dt

\[\frac{10}{2pi\ r(.1)}=\frac{dr}{dt}\]

Answer 6

lm here is like a big drop... like a pit

Answer 7

when r=5; solve for dr/dt

Answer 8

and i think it reads 110 not 10 ...

Answer 9

does that make sense?

Answer 10

lm was meant to film

Answer 11

Another way we could have looked at it was from the chain rule:
\[\frac{dr}{dt}=\frac{dr}{dV}\frac{dV}{dt}\]
\[V = pi\ r^2(.1)\text{ ; derive with respect to V}\]
\[1 = 2pi\ r(.1)*\frac{dr}{dV}\]
\[\frac{dr}{dV}=\frac{1}{.2pi\ r}\]
\[\frac{dr}{dt}=\frac{1}{.2pi\ r}\frac{110}{1}\]
same results