Question

An oil tanker spills 100,000 cubic meters of oil, which forms a lick that spreads on the water surface in a shape best modeled by a circular disc of radius r meters and thickness h meters. The radius of the circular disc is increasing at a rate of 3m/min. At t=T, the area of the "circular" slick reaches 100 pi m^2.
a) how fast is the area of the slick increasing at t=T?
b) how fast is the thickness of the slick decreasing at t=T?
c) Find the rate of change of the area of the slick with respect to the thickness at t=T?

Answer 1

(a)
Now, A = pi*r^2
dA/dt = 2*pi*r*dr/dt
so at time t, when area is 100 * pi m^2 , radius = 10m and dr/dt = 3 (thats given)
so, dA/dt = 2 * pi * 10 * 3 m^2 / min
therefore, dA/dt = 60 pi m^2/ min

(b)

Disk Volume = Area * Thickness
Volume is constant (assuming density of oil won't change and mass of oil spill was fixed)
so d/dt(Volume) = 0 = hdA/dt + Adh/dt

therefore, dh/dt = -h/A * dA/dt
Now at time = t, area = 100 * pi m^2
therefore h = 100000/(100 * pi) = 1000/pi m

hence dh/dt = - 1000/pi * 1/(100 * pi) * 60 * pi
= - 600/pi m / min

hence rate of decrease of h = 600/pi m /min

(c)

Disk Volume = Area * Thickness
Volume is constant (assuming density of oil won't change and mass of oil spill was fixed)
so d/dt(Volume) = 0 = hdA/dt + Adh/dt

therefore, dh/dt = -h/A * dA/dt

Answer 2

I don't really understand part c, but I get what you did for the other two! Thank you so much! :D

Answer 3

you are right, for part c , we have to find dA/dh .
so V = A * h
differentiate with respect to h
dV/dh = hdA/dh + A
and dV/dh = 0 (because Volume is essentially constant)
hence hdA/dh + A = 0
dA/dh = -A/h

Answer 4

I think for part c) you can use the fact that dA/dh = ( dA/dt ) / ( dh/dt)

Answer 5

@zaphodplaysitsafe are you sure about your answer.
I get
dA/dh = (60 pi ) / ( -600 /pi ) = - 1/10 * pi^2