Question

@kohai who is too self conscious to ask for help.

Answer 1

Stfu Kai

Answer 2

This is English

Answer 3

Is this not private enough, no one comes in here anyways.

Answer 4

Fair

Answer 5

Ok so good lol.

So let's take the full on hardcore looking derivative:\[\LARGE V=\frac{4}{3}\pi r^3\]
We wanna find out how fast the radius is changing when the radius is a certain value. Well, how fast something changes is the change in that value with respect to time, so let's take the derivative of this thing!
\[\LARGE \frac{d}{dt}(V)=\frac{d}{dt}(\frac{4}{3}\pi r^3)\]

Let's do one step at a time, starting from the left, we have: \[\LARGE \frac{d}{dt}(V^1)\] So we use the power rule and the chain rule to get \[\LARGE 1*V^0\frac{dV}{dt}\] See all we did was lower the exponent and subtract by 1 and then multiply by the derivative of the inside, which was V itself. Since anything to the 0 power is 1, really we didn't do anything, I was just showing you how this is all basically consistent. So let's keep going, let's use the product rule for fun.

\[\LARGE \frac{d}{dt}(\frac{4}{3}\pi r^3)=\frac{d}{dt}(\frac{4}{3}\pi)* r^3+\frac{4}{3}\pi *\frac{d}{dt}(r^3)\]

What do we get? The derivative of a constant is just 0, so that whole first term goes away and we just have the second term. We do exactly what we did with V in taing the derivative, except this time it will look slightly different because the exponent on r is 3 instead of 1. \[\LARGE \frac{4}{3}\pi *\frac{d}{dt}(r^3)=\frac{4}{3}\pi *(3r^2\frac{dr}{dt})\] Now look back and see how this is exactly what we did with the derivative of V, because I don't want you thinking that there's some kind of special reasoning going on here.

Answer 6

Questions/comments/whatever ask me in skype if you're too timid to do it here. ;)

Answer 7

Just that response got you a metal.. damn..

Answer 8

Oh shut up -_- lol

Answer 9

No, it makes sense

Answer 10

jbc @kohai

Answer 11

If you're going into microbiology you really must get comfortable being wrong. You can't know everything. I know way too many smart people who would rather look smart than admit something they don't know and actually learn something. I didn't get to be who I am by knowing everything to begin with, I admitted to someone I was wrong and then people helped me out just like I'm helping you right now! =D

Answer 12

wow looooooooooooong response :P

Answer 13

Anyways. What happens after we take the derivative here?

Answer 14

Now we're essentially algebraically solving for what we want and plugging in the given values. Most problems take this route. It usually involves a geometric formula like the pythagorean theorem or something and applying a similar process.

Answer 15

So we're trying to find dr/dt?

Answer 16

Something like that?