Related Rates: What's Wrong Here?

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AbilityInfinity

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A bit of a challenge problem on related rates is below. I am having trouble understanding why my equation is incorrect and yields the wrong answer.

A man starts walking north at 4 ft/sec from point P. Five minutes later a woman starts south 5 ft/sec from a point 500 feet east of P. At what rate are the people moving apart 15 minutes after the woman starts walking? (Ans. 12.676 ft/sec)

My teacher used the same method to solve this equation as I did. I am almost certain I made no mathematical mistakes, 3 friends checked my work. My teachers equation is: (M + W)^2 + 500^2 = D^2 where M is the distance the man traveled, W is the distance the women traveled and D is the distance between the two people.

My equation is the same as above except M and W are substituted with M = 4t and W = 5(t-5), where t is the time traveled. Note: I made sure to convert t(min) to t(sec). Since the statements above are both true, I don't understand why the derivative of the two equations is different after I plug in all the variables except dD/dt. Any help?

EDIT: I know how to solve this problem, but I dont understand why the equation is what it is. There is no need for me to share my mathematics, please don't ask me to like 9/10 other posts.

On a side note, when one takes the derivative of y they get dy/dx, is that because x is on the horizontal axis? In the equation above dD/dt is present, does that mean if I graphed that line, t would be on the x axis?

Thanks!
 
Last edited:
AbilityInfinity said:
EDIT: I know how to solve this problem, but I dont understand why the equation is what it is. There is no need for me to share my mathematics, please don't ask me to like 9/10 other posts.
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You do not get to make this determination. If you want to control things, you should consider paying for the arrangement. Please stop abusing volunteers. Volunteers are far to valuable.

Since you DID share quite a bit of your work, it's rather odd that you declined to share, as required by the usage rules you should have read prior to submitting a question.

Anyway, W = 5(t-5), '5' appears to be in minutes, inside the parentheses, even though '5' is seconds outside the parentheses. Are you SURE you converted ALL the units consistently? If you showed more of your work, I would be able to tell.
 
tkhunny said:
Anyway, W = 5(t-5), '5' appears to be in minutes, inside the parentheses, even though '5' is seconds outside the parentheses. Are you SURE you converted ALL the units consistently? If you showed more of your work, I would be able to tell.
Click to expand...

You didn't answer my question and you didn't show your work, so I'm not sure how to help you.
 
One thing that may be giving you a fit is that the given answer is incorrect. It is NOT 12.676 ft/sec.
 
Hello, AbilityInfinity!

Sorry, I don't understand your dilemma.
I'll show you the way I approach "headstart" problems.

A man starts walking north at 4 ft/sec from point P.
Five minutes later a woman starts south 5 ft/sec from a point 500 feet east of P.
At what rate are the people moving apart 15 minutes after the woman starts walking?
(Answer: 12.676 ft/sec) .??
Click to expand...

Code: 
    B *
      | *
   4t |   *
      |     *
    A *       *
      |         *  D
 1200 |           *
      |             *
    P *---------------*-------* Q
                        *     |
                          *   | 5t
                            * |
                              * C
      : - - - -  500  - - - - :
The man has a 5-minute (300-second) headstart.
He walks 1200 feet from point \(\displaystyle P\) to point \(\displaystyle A.\)
In the next \(\displaystyle t\) seconds, he walks \(\displaystyle 4t\) feet to point \(\displaystyle B.\)

In the same \(\displaystyle t\) seconds, the woman walks \(\displaystyle 5t\) feet
from point \(\displaystyle Q\) to point \(\displaystyle C\), where \(\displaystyle PQ = 500\) feet.

Let \(\displaystyle D\) = the distance \(\displaystyle BC.\)

We have: .\(\displaystyle D^2 \;=\;(9t+1200)^2 + 500^2 \;=\;81t^2 + 21,\!600t + 1,\!690,\!000
\)


Differentiate with respect to time:

. . .\(\displaystyle 2D\dfrac{dD}{dt} \;=\;162t + 21,\!600 \quad\Rightarrow\quad \dfrac{dD}{dt} \;=\;\dfrac{81t + 10,\!800}{D}\) .[1]


When \(\displaystyle t = 15\text{ minutes} = 900\text{ seconds,}\)
. . \(\displaystyle D^2 \:=\:81(900^2) + 21,\!600(900) + 1,\!690,\!000 \:=\:86,\!740,\!000 \)

Hence: .\(\displaystyle D\:\approx\:9313.43\)


Substitute into [1]: . \(\displaystyle \dfrac{dD}{dt} \;=\;\dfrac{81(900) + 10,\!800}{9313.43} \;\approx\;8.987 \text{ ft/sec}\)
 
soroban said:
Hello, AbilityInfinity!...
Click to expand...

Soroban, that was incredibly helpful! As it turns out I did get the correct answer, just the way you did! I knew how to do the problem all along, but the answer my teacher gave me was wrong.

Thanks, I appreciate it.
 
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