Page 144 - Calculus Workbook For Dummies

128 Part III: Differentiation

4. Differentiate with respect to time. Now you’ve got what you need to finish
the problem.
This is a lot like implicit differentiation
because you’re differentiating with 2 h dh + 2 b db = 0
respect to t but the equation is in terms dt dt
of h and b. 2 1 - 20 + ^h 2 17 .97h db = 0
h
^ ^
dt
2
2
h + b = 18 2 db 40
dh db dt = 35 .94 . . 1 11 feet/sec
2 h + 2 b = 0
dt dt
6. Ask yourself whether your answer is
5. Substitute known values for the rates reasonable.
and variables in the equation from Yes, it does make sense. Lean a yardstick
Step 4, and then solve for the thing against a wall so the bottom of it is about
you’re asked to determine.
4 or 5 inches from the wall. Then push
db
You’re trying to determine , so you the bottom of the yardstick the 4 or 5
dt
have to plug numbers into everything else. inches to the wall. You’ll see that the top
But, as often happens, you don’t have a would barely move up. Right triangles
number for b, so use a formula to get the with a fixed hypotenuse like this one
number you need. This will usually be the always work like that. If one leg is much
same formula you already used. shorter than the other, the short leg can
change a lot while the long leg barely
2
2
h + b = 18 2 changes. It’s a by-product of the
2
2
1 + b = 18 2 Pythagorean Theorem.
b = ! 323 . ! 17 .97 feet
(Obviously, you can reject the negative
answer.)