# Does anyone have an idea on how to do this Calculus Problem?

Discussion in 'Community Discussion' started by Dabisu, Sep 29, 2007.

1. ### Dabisu macrumors member

Joined:
Jun 8, 2006
#1
Hey, I'm having trouble with this related rates Calculus problem. I just can't figure out a relationship between what's given and what I need to find.

Here it is:

A runner sprints around a circular track of radius 100m at a constant speed of 7 m/s. The runner's friend is standing at a distance 200m from the center of the track. How fast is the distance between the friends changing when the distance between them is 200m?

Any help would be greatly appreciated!

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Australia

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4. ### psychofreak Retired

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May 16, 2006
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London
#4
I always like to start off with a diagram, so while I'm thinking about this, to help myself and others, I made one:

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Jul 11, 2003
6. ### Dabisu thread starter macrumors member

Joined:
Jun 8, 2006
#6
Thanks, Psychofreak. I came up with a similar diagram (yours is better), but I can't figure out what to do with the rate at which the runner is running. Could it maybe have something to do with the section of the circle?

In the reference part of my book it says the section of a circle is equal to radius times theta in radians. That way you have the length of the section. Or maybe something with the circumference?

This problem seems to be a killer. I've been able to do the other related rates problems, but this one is really tough. If no one can easily point out what I'm missing, I think I'll just wait until I can ask my teacher. I only say this because I don't want to take up anyone's time. Thanks for those that helped!

7. ### siurpeeman macrumors 603

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Dec 2, 2006
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the OC
#7
do you have an answer to this? i want to say it's 7/4 times the square root of 15, but i'm not too sure about it.

EDIT: how i arrived at my answer...my diagram used polar coordinates and not rectangular. the equation of the circle is r = 100, and you could make the friend (who's standing) at (200, pi). the runner's coordinates are (100, theta). make an equation from the distance formula (polar), and differentiate the polar distance formula to come up with the answer, d-distance/dt. use the 7m/s to come up with a value for d-theta/dt.

EDIT #2: the answer is both positive and negative, depending on whether the runner is coming or going.

laaaate EDIT #3: you can solve the problem without assigning coordinates altogether. instead, you make a triangle with your three vertices at the runner, the friend who's watching and the center of the circular track. the runner to the center is always 100, and the friend to the center is always 200. your variables would be the distance between the runner and his friend, d, and the angle opposite to that length, theta. you can use law of cosines to come up with: d^2 = 200^2 + 100^2 - 2(200)(100)cos theta. again, you differentiate with dd/dt as your answer. use the 7m/s to figure out d-theta/dt (7/100), leading to your answer of (7√15)/4. it's practically the same as the original method but with less reliance on a coordinate system.

8. ### mduser63 macrumors 68040

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Salt Lake City, UT
9. ### Chundles macrumors G4

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Jul 4, 2005
#9
Mate, if everyone did their own homework and their own research then all the internet forums not related to porn would disappear overnight.

Personally I don't want to wake up with 3% of the internet suddenly gone...

10. ### Dabisu thread starter macrumors member

Joined:
Jun 8, 2006
#10
Yeah, I'd do this problem if I could. It's not that I'm lazy, I just don't know how to do it. I'm so sorry for requesting help. I'm so sorry that I want to know how to solve it so I can get a good grade on my test.

11. ### Dabisu thread starter macrumors member

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Jun 8, 2006
#11
Hey man, YOU GOT IT! Thanks so much! I really did not remember anything about polar coordinates. Again thanks so much and thanks to everyone that contributed!

12. ### hayduke macrumors 65816

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is a state of mind.
#12
Yes.

13. ### mduser63 macrumors 68040

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Salt Lake City, UT
#13
There's a fine line (at least in most teacher's minds) between asking for help and cheating. I suppose it's OK to ask for help as long as you've really tried your best to figure it out. A big part of learning though is struggling to figure things out, and sometimes failing. Good luck on your test, if it helps at all, calculus really is important if you're going into any science or engineering related field.

that gave me a good laugh.

14. ### theyoda3 macrumors member

Joined:
Sep 27, 2006
#14
Whether you were suggesting he was crossing that line or not, he did the right thing in asking for help with the problem. Yes you should figure out problems on your own, but if you don't know what to do then you should always suck up your pride and ask for help. You say how important calculus is in engineering and science, but I'll tell you first hand as an engineer at Cornell that what's even more important is knowing when to ask for help. You won't learn much struggling and failing your way through college. At the same time, if you don't put in the effort to understand what someone has told you when you do get help that will show on your exams. One way to really check and see if you understand material that you learned on your own or that someone helped you with is to explain it to another person. This is one reason why working in groups in strongly encouraged in engineering.

15. ### mduser63 macrumors 68040

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Nov 9, 2004
Location:
Salt Lake City, UT
#15
As an engineer myself, I fully understand that. It does seem that the majority of people coming here are asking for help because they're too lazy to do it themselves. I seem to have been wrong in the OP's case and it seems he was sincerely asking for help so that he could learn better. No problem there, sorry for the comment.