Question

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Answer 1

It helps if you ask a question.

Answer 2

Consider the following diagram... take a couple minutes to conceptualize it, and then let me know if you agree with the expression I wrote at the bottom right based on the ratios of the triangles. http://dl.dropbox.com/u/345109/junk/lamppost.jpg

Answer 3

It most certainly could be flipped, you just have to make sure that you flip the left hand side as well if you're going to do that. The following two expressions are equivalent. (Though, derivatives with the one I wrote are a ton easier, even though they will both produce the same answers.)\[\frac{s}{h}=\frac{s+d}{H}\Leftrightarrow \frac{h}{s}=\frac{H}{s+d}\]Now, only \(s\) and \(d\) will vary with respect to time (all other variables are constants). Try finding the derivative of both sides with respect to \(s\).

Answer 4

*with respect to time \(t\).

Answer 5

Take the derivative of both sides with respect to time \(t\).

Answer 6

No...there are no \(x\)s and \(y\)s here...that's meaningless.

Answer 7

Exactly!

Answer 8

Imagine as if \(h\) and \(H\) were arbitrary constants, and as if \(s\) and \(d\) were functions of \(t\) (which they are). What I'm asking you to do is evaluate the following in terms of \(\frac{ds}{dt}\) and \(\frac{dd}{dt}\).\[\frac{d}{dt}\left(\frac{s}{h}\right) = \frac{d}{dt}\left(\frac{s+d}{H}\right)\]

Answer 9

Only after you differentiate. If you plug numbers in before you do the differentiation, you WILL get the wrong answer. You MUST differentiate in generic variables first with related rates problems.

Answer 10

You're very close. Your left hand side is correct. There's both an \(s\) and a \(d\) added to each other on the right hand side so you should have both a \(\dfrac{ds}{dt}\) and a \(\dfrac{dd}{dt}\).

Answer 11

Very close...not sure if that's just error in you typing it or not. The correct answer is as follows.\[\frac{\frac{ds}{dt}}{h}=\frac{\frac{ds}{dt}+\frac{dd}{dt}}{H}\]

Answer 12

At this point you're okay to plug in values and solve for what is requested by the problem.