Page 162 - Calculus Workbook For Dummies

146 Part III: Differentiation

b. Find an equation that relates your unwanted variable, b, to some other variable in the prob-
lem so you can make a substitution and be left with an equation involving only V and h.
The triangular face of the swill is the same shape of the triangular side of the trough. If you
remember geometry, you know that such similar shapes have proportional sides. So,

b = h
2 . 2 5
. b 2=
2 5 h
b = . h
8
Similar triangles often come up in related rate problems involving triangles, triangular prisms,
and cones.
Now substitute .8h for b in the formula from Step 3a:
V 5 bh
=
8
= 5 $ . h h $
= 4 h 2
4. Differentiate with respect to t.
dv = 8 h dh
dt dt

In all related rates problems, make sure you differentiate (like you do here in Step 4) before you
substitute the values of the variables into the equation (like you do below when you plug 1’8”
into h in Step 5).
dh
5. Substitute all known quantities into this equation and solve for .
dt
You were given that h = 1’8’’ (you must convert this to feet) and you figured out in Step 2
dv
that = - 1, so
dt
2 dh
8 1 $
- 1 = $
3 dt
dh = - 1
dt 40
3
- 3
= ft/min
40
- 9
= inches/min
10
9
Thus, when the swill level drops to a depth of 1’8’’, it’s falling at a rate of ⁄10 inches per minute.
Mmm, mmm, good!
6. Ask whether this answer makes sense.
Unlike the example problem, it’s not easy to come up with a common-sense explanation of
why this answer is or is not reasonable. But there’s another type of check that works here and
in many other related rates problems.
Take a very small increment of time — something much less than the time unit of the rates
used in the problem. This problem involves rates per minute, so use 1 second for your time
increment. Now ask yourself what happens in this problem in 1 second. The swill is leaving
1
the trough at 1 cubic foot / minute; so in 1 second, ⁄60 cubic foot will leave the trough. What
does that do to the swill height? Because of the similar triangles mentioned in Step 3b, when
the swill falls to a depth of 1’8’’, which is ⁄3 of the height of the trough, the width of the surface
2
2
1
of the swill must be ⁄3 of the width of the trough — and that comes to 1 ⁄3 feet. So the surface
1
area of the swill is 1 # 10 feet.
3
Assuming the trough walls are straight (this type of simplification always works in this type of
checking process), the swill that leaves the trough would form the shape of a very, very short
box (“box” sounds funny because this shape is so thin; maybe “thin piece of plywood” is a
better image).