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Aladsair is offline
Apr20-07, 01:32 AM
P: 2
actually as it turned out:

a(t) = 6 theta

better put as

a(t) = 6 theta(t)

a(t) is just the second derative of theta(t), so

theta''(t) = 6 theta(t)

tossing away the 6 mentally for a second, what function equals its own second derative? e does! so let's play with that and see what we get

if theta(t) = e^t, then theta''(t) = e^t. Okay, but we need to bring that 6 back in. So, theta (t) = e^sqrt6 t, theta'(t) = v(t) = sqrt6 e^sqrt6 t, and v''(t) = a(t) = 6 e^sqrt6 t.

To check it:a(t) = 6 e^sqrt6 t. e^sqrt6 t = theta(t) substitute and we get back to the original a(t) = 6 theta.

yay :) using this I got the right answer for my problem!

But I'm on another problem (last one, promise) with similar premises.

omega(t) = 5 (theta(t))^2 is what I'm given.

So theta'(t) = 5 (theta(t))^2

Toss away the 5, we can deal with it later...but what function equals its derative when you square it?