Derivative of P(t) for Equilibrium Point Problem

pyrosilver
Messages
38
Reaction score
0

Homework Statement


I was sick today, and here is what my friend told me. I don't quite understand the question, but apparently we talked about equilibrium points, and for homework, we have to take the derivative of P(t)= 1/((ab^t)+(1/c)), put it in terms of P, and get P'(t)=(P-6)(-k)P.


Homework Equations





The Attempt at a Solution



I'm utterly confused on how to do this.
 
Physics news on Phys.org
what are you differentiating with respect to?
 
pyrosilver said:
derivative of P(t)= 1/((ab^t)+(1/c))


well it is in respect to t so if you can find the derivative of b^t than it is -(a(derivative of b^t))/((ab^t)+(1/c))^2 ... assuming that a, b and c are all constants
 
So
P(t)= ((ab)^t+ (1/c))^{-1}

Can you differentiate that with respect to t?

It will help to notice that since
P(t)= ((ab)^t+ (1/c))^{-1}
1/P= (ab)^t+ 1/c so that (ab)^t= 1/P- 1/c
 
almost ... it is a(b)^t otherwise agree with that.
 
No, it is (ab)^t, I forgot to specify the a and b going together.

Thanks HallsofIvy! Very helpful :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top