Question 160373
A vertical cylindrical storage tank, with diameter 20 feet, is filled with oil
 to a depth of 40 feet.  Sometime later, the oil is drained, decreasing the
 depth a rate of 8 inches per hour.  Write an equation for the volume of oil
 (ft^3) remaining in the tank t hours later as a function of t.
  Draw a geometrically correct sketch, supporting your solution.
:
This is a linear equation so we can find the slope using the time/volume
:
Find the volume at 40 ft, (radius = 10 ft)
V = {{{pi*10^2*40}}}
V = 12566 cu ft (rounded to nearest cu ft)
:
Therefore: t=0; v=12566
:
Find out how many hours to empty the tank
:
At 8 in/hr how long to lower it 40 ft?
t = {{{((40*12))/8}}} = 60 hrs (vol = 0)
Therefore: t=60; v=0
:
Find the slope: m = {{{((0 - 12566))/((60-0))}}} = {{{-12566/60}}} is the slope
:
An equation: V = {{{-12566/60}}}t + 12566
:
You can illustrate this with a graph. V = vertical axis; t = horizontal axis
{{{ graph( 300, 200, -20, 80, -5000, 15000, (-12566/60)x+12566) }}}
:
You can prove this:
t = 30 hr
At 30 hrs the depth would be 30*8" = 240" = 20 feet (40-20)
Find the volume at 20ft
V = {{{pi*10^2*20}}}
V = 6283 cu ft 
:
Using the equation/graph
v = {{{-12566/60}}}(30) + 12566
v = 6283 cu ft
:
Did this sense to you? I am not sure what kind of sketch they have in mind, I'll
leave that up to you.