# Tricky Word Problem

• Jun 1st 2010, 11:23 AM
action812
Tricky Word Problem
A cone shaped water reservoir is 20ft in diameter across the top and 15ft deep. If the reservoir is filled to a depth of 10ft, how much work is required to pump all the water to the top of the reservoir?

I'm having trouble setting this one up. Word problems always seem to get me stumbling backwards...*sigh*
• Jun 1st 2010, 11:58 AM
skeeter
Quote:

Originally Posted by action812
A cone shaped water reservoir is 20ft in diameter across the top and 15ft deep. If the reservoir is filled to a depth of 10ft, how much work is required to pump all the water to the top of the reservoir?

note that the weight of one cubic ft of water is about 62.4 lbs

$
W = \int_0^{10} 62.4\pi\left(\frac{2y}{3}\right)^2 (15-y) \, dy
$
• Jun 1st 2010, 11:59 AM
HallsofIvy
(Wink)
Quote:

Originally Posted by action812
A cone shaped water reservoir is 20ft in diameter across the top and 15ft deep. If the reservoir is filled to a depth of 10ft, how much work is required to pump all the water to the top of the reservoir?

I'm having trouble setting this one up. Word problems always seem to get me stumbling backwards...*sigh*

I will assume that the flat base of the cone is up, though you do not say that. Work is, of course, "force times height" which, here, is "weight times height". The reason this is a calculus problem is that the weight at different heights in the cone is different so we need to analyze it by the weight of water at fixed heights. At height "z" from the bottom, we have to lift the water 15- z feet. A "layer" of water at that height is a thin disk. We need to find the area of that disk.

.
1) Using "similar triangles". Draw a picture. Seen from the side we have a triangle. The entire cone has radius 10 and height 15. At each height, z, The cone below that height has radius r and height z and the triangle below it is similar to the triangle formed by the entire cone- the ratios are equal- $\frac{r}{z}= \frac{10}{15}= \frac{2}{3}$ so r= 2z/3. The area of the disk is [tex](4\pi/9)z^2.

2) Using the equations of the lines forming the sides. Again draw a picture of the cone as seen from the side. Set up a coordinate system with the tip of the cone at (0, 0), the x-axis horiaontal and the z-axis upward. The sides are given by lines passing through (0, 0) and then either (-10, 15) or (10, 15). The first has slope 15/10= 3/2 and so is z= (3/2)x and the other has slope -15/10= -3/2 and so is z= (-3/2)x. From the first x= (2/3)z and the other x= (-2/3)z. The distance between either of those and the center line, the z-axis, is (2/3)z so again, the radius is r= 2z/3 and the area of the disk is $(4\pi/9)z^2$.

Taking the thickness to be "dz" the volume of such a "layer" of water is $(4\pi/9)z^2 dz$. Letting the density of water be " $\delta$" (you can look that numerical value up if it is not given in your textbook- or get it from Skeeter's post!), then the weight is $(4\pi/9)\delta z^2 dz$.

The work done in lifting that 15- z feet is $(4\pi/9)\delta z^2 (15- z)dz$ and so the work done in lifting every layer is the sum of all such layers: $\sum (4\pi/9)\delta z^2(15- z)dz$. That's really a Riemann sum and in the limit, as the "thickness" of each layer goes to 0, becomes the integral
$(4\pi/9)\delta \int_0^{10}z^2(15- z)dz$.
• Jun 1st 2010, 12:54 PM
action812
Ok it makes sense now, are your two integrals the same? Except HOV, moved out the constant while skeeter kept it in the integral?

EDIT:: I did the work, and it works out to be the same answer like I thought. Thank you guys :D