OK... I was sick and missed this class (Damn my luck!). And the book is no help figuring it out.

 

Related Rates.

 

here's the example problem...

 

Volume of a cone...

 

[math]V=\frac{\pi r^2 h}{3}[/math]

 

it's talking about how if a conical tank of water is being drained... it's all a funtion of t,r,h and V... and it says to implicitly diferentiate...

 

[math]\frac{d}{dt}V=\frac{d}{dt}(\frac{\pi r^2 h}{3})[/math]

 

[math]\frac{dV}{dt}=\frac{\pi}{3} (r^2\frac{dh}{dt}+2rh\frac{dr}{dt})[/math]

 

and then it says that you can see that t is related to r and h!

 

I am REALLY confused.... first off, to that last form... and where is the t?

 

Since the rest of the section is just like this stuff (and I have a test tuesday over this...), I could use a little help!

 

PS... in LaTeX... how can you a)get the fractions smaller ( so that my dy/dx or pi/3 would be the same size as the rest) b)get the parentesis bigger (so that when I paranthesize my big fractions, it reaches from the top to the bottom...)?

All they're basically saying is that h = h(t) and r = r(t), otherwise those derivatives would be zero.

 

As for the LaTeX:

 

a) Use the [imath] tags instead of [math] - it's used to generate inline math. For instance:

 

I like [math]\frac{a}{b}[/math] as a fraction

vs.

I like [imath]\frac{a}{b}[/imath] as a fraction

 

Either that, or you can use \tfrac instead of \frac.

 

b) Use a combination of \left( and \right). So you can have something like:

 

[math]\frac{dV}{dt}=\frac{\pi}{3} \left( r^2\frac{dh}{dt}+2rh\frac{dr}{dt} \right)[/math] (click to see code).

and as I realize that one od my questions got cut off... how did it get dr/dt and dh/dt? In other words, how do you get from the second to last line of latex, to the last?

It's just the differential of V.

 

[math]\frac{d}{dt}\left( \frac{\pi r^2 h}{3} \right) = \frac{\pi}{3} \frac{d}{dt} (r^2 h)[/math]

 

Now, r and h are both functions of time, so you need to use the product rule:

 

[math]\frac{d}{dt} (r^2 h) = h \frac{d}{dt}(r^2) + r^2 \frac{dh}{dt}[/math]

 

To do [imath]\frac{d}{dt}(r^2)[/imath] you'll need to use implicit differentiation. If you haven't covered this yet, you'll need to learn that. But that's equal to [imath]2r \frac{dr}{dt}[/imath].

 

From there the rest follows.

Related rates are fairly easy, just take the derivative with respect to time (usually), so you use the chain rule on anything that changes with time. Therefor an equation relating, say, length of a square to it's area, would be able to tell you how fast the area grows as the rate grows:

 

da/dl * dl/dt

 

or f'(g(x))g'(x)

 

implicit differentiation just means differentiating both sides of the equation with respect to some variable and solving for the derivative. Maybe this helps, maybe not. I'll go on an example real quick, say the length of a square grows at 2 feet per second?

 

so: dl/dt = 2 f/s

 

but it's known that A = l^2

 

so take the derivative of both sides of this equation with respect to t

 

dA/dt = 2l * dl/dt

 

and dl/dt is 2 feet per second... so the area grows at 4l feet per second, which as you can see grows faster as l becomes larger (makes sense eh?). So when one side is 2 feet long, it is growing at 8 f^2 / s

 

The most difficult part of this for most people is understanding the chain rule portion of the work. I hope that I helped!