Studying for final and need help with Related Rates

Printable View

December 18th 2007, 09:34 AM
FalconPUNCH!

Studying for final and need help with Related Rates

I hate word problems and I pretty much suck when it comes to them. Here's one that I got stuck at.

1) A man starts walking north at 4ft/s from a point P. Five minutes later a woman starts walking south at 5ft/s from a point 500ft due east of P. At what rate are the people moving apart 15 min after the woman starts walking?

I don't even know where to begin. My final will be filled with these types of question any tips on how to approach these types of problems?
December 18th 2007, 09:42 AM
colby2152

Tip... draw out what is given in the problem. In related rates & optimization problems, you tend to work with one or two formulas and take a derivative of one of them.
December 18th 2007, 09:45 AM
FalconPUNCH!

Quote:

Originally Posted by colby2152

Tip... draw out what is given in the problem. In related rates & optimization problems, you tend to work with one or two formulas and take a derivative of one of them.

I usually do that, and whenever I can list what x =, y=, and the rates of change, but sometimes I don't know what goes with what.
December 18th 2007, 12:21 PM
Soroban

Hello, FalconPUNCH!

Make a sketch!

Quote:

1) A man starts walking north at 4ft/s from a point P.
Five minutes later a woman starts walking south at 5ft/s from a point 500ft due east of P.
At what rate are the people moving apart 15 min after the woman starts walking?

Code:

M * | * 4t | * | * A * * | * | * 1200 | * | * | * P * - - - - - - - - - * - - - * Q : * | 5t : * | 5t : * | R * - - - - - - - - - - - - - * W 500

The man starts at P and has a 5-minute headstart.
In 5 minutes (300 sec), he walks 1200 feet to point A.
In the next $t$ seconds, he walks $4t$ ft to point M.

The woman starts at Q and walks south at 5 ft/s.
In the next $t$ seconds, she walks $5t$ ft to point W.

Let $x \,=\,MW$
In right triangle MRW, we have: . $MW^2 \:=\:MR^2 + RW^2$
. . That is: . $x^2\:=\:(4t+1200 + 5t)^2 + 500^2 \:=\:(9t+1200)^2 + 500^2$

Differentiate with respect to time: . $2x\left(\frac{dx}{dt}\right) \:=\:2(9t + 1200)$

. . Hence: . $\frac{dx}{dt} \;=\;\frac{9t + 1200}{x}$

In 15 minutes, $t = 900$
Find $x$ at that time, and you're done!
December 18th 2007, 01:15 PM
FalconPUNCH!

Quote:

Originally Posted by Soroban

Hello, FalconPUNCH!

Make a sketch!

Code:

M * | * 4t | * | * A * * | * | * 1200 | * | * | * P * - - - - - - - - - * - - - * Q : * | 5t : * | 5t : * | R * - - - - - - - - - - - - - * W 500

The man starts at P and has a 5-minute headstart.
In 5 minutes (300 sec), he walks 1200 feet to point A.
In the next $t$ seconds, he walks $4t$ ft to point M.

The woman starts at Q and walks south at 5 ft/s.
In the next $t$ seconds, she walks $5t$ ft to point W.

Let $x \,=\,MW$
In right triangle MRW, we have: . $MW^2 \:=\:MR^2 + RW^2$
. . That is: . $x^2\:=\:(4t+1200 + 5t)^2 + 500^2 \:=\:(9t+1200)^2 + 500^2$

Differentiate with respect to time: . $2x\left(\frac{dx}{dt}\right) \:=\:2(9t + 1200)$

. . Hence: . $\frac{dx}{dt} \;=\;\frac{9t + 1200}{x}$

In 15 minutes, $t = 900$
Find $x$ at that time, and you're done!

LOL my sketch didn't look like that. But I understand what to do. Usually most have something to do with triangles, so I'm expecting my professor to do one with triangles. Thanks for helping me out with the example.