actual truck speed (geometric optics)

Aafour

this is the solution found in the solution manual,



I didn't understand the part of taking the derivative part, that is how does this turn into this

please explain carefully

Redbelly98

This is an example of a "related rates" problem that could be given in 1st-semester calculus.

The chain rule from calculus tells us that

[tex]\frac{ds'}{dt} = \frac{ds'}{ds} \ \frac{ds}{dt}[/tex]

You can work out for yourself what ds'/ds is, and take it from there.

Aafour

i really don't know how to do that. i'll be very appreciated if you help me with this

Redbelly98

Well, you do need to have taken calculus in order to understand this solution. It uses the quotient rule for taking derivatives, which is explained here if you need a review of it:

http://en.wikipedia.org/wiki/Quotient_rule

So, we have an expression for s' in terms of s:

Take the derivative of the right-hand-side of the equation, with respect to s. (f is a constant here.) Remember, you need to use the quotient rule to take the derivative here.

Aafour

if i take the derivitive of this I'll have -fs/(s-f)^2.

what soul i do after that

Redbelly98

That doesn't look right. Can you show your work in how you got that result?

p.s. I'm not sure how familiar you are with our forums, so I'll just mention the following. Our philosophy is for students to do the majority of the work in solving problems, with hints and guidance from the "helpers" like me. We believe this approach makes the student think more and learn the material better.

Just in case you're wondering why I'm not working out the steps for you, that is why

Redbelly98

If it helps, I can review what to do with the formula from the wikipedia article at http://upload.wikimedia.org/math/a/c...7302faea9a.png

To take the derivative of a function g(s) / h(s), the formula is

[tex] \frac{g'(s)h(s) - g(s)h'(s)}{[h(s)]^2}[/tex]

We have

g(s) = fs
h(s) = (s-f)

so you'll need to figure out what g'(s) and h'(s) are, then plug into the formula.