Rate of change of population after 2.5 hours?

Math Sucks
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Rate of change... derivative?

Homework Statement



the total number of people(in millions) present in a population at a given time is given by the function:

2t(5t+9)^1/2+12

where t represents time (in hours) find the rate in change of the population after 2.5 hours...

Homework Equations





The Attempt at a Solution



im so confused do i find the derivative ?
 
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Math Sucks said:

Homework Statement



the total number of people(in millions) present in a population at a given time is given by the function:

2t(5t+9)^1/2+12

where t represents time (in hours) find the rate in change of the population after 2.5 hours...

Homework Equations





The Attempt at a Solution



im so confused do i find the derivative ?

Sounds like a good idea to me. The derivative is the rate of change.
 


i got 10t+9..
 


10(2.5)+9... 34?
 


Math Sucks said:
i got 10t+9..

Now how did you get that? Your expression looks like 2 t \sqrt{5 t + 9}+12, is that right? Looks like you'll need the product rule and chain rule etc to differentiate it.
 
Last edited:

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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...
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