Can You Solve This Challenging Related Rates Problem?

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Problem solved :)
 
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Looks good.
 
Really?? Haha, I had no idea what I was doing, honestly. If someone could double check my work, that'd be awesome. I know this is asking alot. :p
 
I worked it out myself. Everything is correct. I would add a couple of notes about part c: You might want to switch your calculator to radian mode if you used a calculator to find arctan (5/12) just to be on the safe side. IMHO, there is never really any reason to choose to work in degree mode unless a problem is written that way or the answer needs to be in degrees. Also, it seems like you used the diagram to obtain the exact value of sec2 \Theta. That was a good choice (better than using any decimal approximation you obtained earlier). Also, not that it matters, but you could express 1/sec2 \Theta as cos2 \Theta if it would make it simpler for you to work with.

Everything was correct, but I wanted you to be able to see these things. Good Job.
 
Thanks Bear B for the input. Yeah, I caught myself doing it in DEG, because I was previously using it for my physics homework.

That is interesting that you point out I could have used the diagram to find the value of sec^2... I didn't do that, but just did right now and obtained the same answer. Thanks for pointing that out though, very interesting.

Thanks everyone. :)
 
In the future I recommend not editing your original post so that the problem disappears. While you are done being helped, there's some chance that someone else might be able to learn something from the discussion of your problem.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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