3.9: RELATED RATES In this type of problems, we are given the rate

advertisement
3.9: RELATED RATES
In this type of problems, we are given the rate of change of one or more certain quantities (with respect to time), and
we are asked for the rate of change of another quantity (also with respect to time).
The basic idea is that if we can figure out how these quantities relate to each other, then we can differentiate the
resulting equality implicitly (with respect to time) and we’ll get another equality relating their rates.
This is a pretty neat thing to be able to do because sometimes we can easily measure or compute a rate -- and via this
method we can get information about the rates we can’t easily measure from the ones we can!
Method Guidelines:
0. READ the problem carefully. Then READ IT AGAIN. Carefully.
1. DRAW a picture/diagram, if one is not provided.
(Caution: Don’t label anything with a number UNLESS that quantity is constant throughout the problem!)
2. LABEL :
a. All quantities whose rates you’re given (with letters, since these vary).
b. The quantity whose rate you want (with a letter, since it varies)
c. Anything else you can (either with a number, if it’s constant -- or with additional letters, if it varies)
3. Write down:
a. All the information that you know and
b. What you want to compute
4. Use the diagram to find an equation relating the quantity whose rate you know to that whose rate you want.
(often needed formulas: Pythagorean Theorem, similar triangles, trig functions, areas/volumes. Review!)
Simplify this relationship as much as possible.
5. DIFFERENTIATE the equality from step 4 with respect to time t.
You’ll need to use Chain Rule, and you’ll get an equation involving two or more derivatives, the one(s) you know
and the one you want.
6. Plug in the values you know (or can compute) into the resulting equation and then solve for the unknown rate.
Examples:
3.9.16: A spotlight on the ground shines on a wall 12 m away. A man 2 m tall walks from the spotlight toward the
building at a speed of 1.6 m/s. How fast is the length of his shadow on the building decreasing when he is 4 m from the
building?
3.9.40: A Ferris wheel with a radius of 10 m is rotating at a rate of one revolution every 2 minutes. How fast is a rider
rising when his seat is 16 m above ground? (i.e.: above the bottom of the wheel)
3.9.37: A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the
camera has to change at the correct rate to keep the rocket in sight. Also, the focusing mechanism has to take into
account the increasing distance between the camera and the rocket. The rocket rises vertically, and its speed is 600
ft/sec when it has risen 3000 ft.
a) How fast is the distance from the camera to the rocket changing at that moment?
b) If the camera is kept aimed at the rocket, how fast is the camera’s angle of elevation changing at that same moment?
Download