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On January 21 2009 13:24 ninjafetus wrote:Question 2: sqrt(3) miles. (see edit) MS paint awesomeness: EDIT: oh, fuck, I put the pig 3 miles from the road instead of 4... the same idea works, but change sqrt(9 + x^2) to sqrt (16 + x^2) and rework it. x should end up being, uhh... sqrt(16/3).
Ah, ok. It makes sense to me now. Thanks~
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hmmm... for the 2nd question I got an answer I wasn't expecting but the math kinda got big so either I messed up or it is a "messy" problem.
I got the answer to be 2.3094 miles to the right of point P.
This is what I did.
1. Name distance from P toward destination S (for Sallys) wheverever you intersect the road as L. This is what we are trying to solve for.
2. Express the distance traveled on the road or through the field in terms of L so you only have 1 variable. Lets have distance in mud be y and distance on road be z.
3. z = 7-L and x=(L^2 + 16)^.5 (remember that .5 is just the square root)
4. Write equation for time so you can optimize it. That is just z/3 + x/6 since time = distance/velocity.
5. Take derivative and set to 0. Don't forget chain rule on x. It gets messy here.
6. Solve for L. I got L to be 2.3094. or the square root of 16/3
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I think there is an error in the paint version above my post. It looks like they set distance to P as 3. Thats how they got 9 while in my equation I got 16.
EDIT: Ahh crap. I thought I was awesome but I just noticed the edit. So I wasn't first. But at least you have 2 independent answers both coming to sqrt or 16/3 so you can be confident.
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i think i am understanding it more now : )
<3 TL <3
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PM me if you ever want calc/optimization help but don't wanna make a new thread. I enjoy helping with this kind of stuff--I used to tutor for free.
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EPT![[image loading]](http://img.photobucket.com/albums/v40/ninjafetus/math.jpg)
