Calculuate rate of change

1. Dec 2, 2007

PlasmaSphere

1. The problem statement, all variables and given/known data

The temparature T of a liquid at a time t minutes is given by the equation
T = 30 + 20e-0.05t, for t > 0

(1)Write down the initial tempatature o the liquid, and find the intial rate of change of temparature.

(2)Find the time at which the temparature is 40

2. Relevant equations

3. The attempt at a solution

When t=0 the initial temparture is 50. I know that to have to find dT/dt, and i and looked at the answer, which says dT/dt = -0.05 x 20e-0.05t, but i cant see how they got that. what happened to the 30? and how did they get to that point?

2. Dec 2, 2007

kevinr

They took the derivative of your T equation. Thus derivative of 30 = 0.

3. Dec 2, 2007

PlasmaSphere

ok, cheers.

I still dont see how we get from T = 30 + 20e-0.05t to dT/dt = -0.05 x 20e-0.05t. i thought that the power had to decrease by 1 when you differenciated it and then multiply by the old power, so 20e-0.05 should become -0.05 x 20e-1.05. but thats not what the answer says, which is why i'm confused.

Last edited: Dec 2, 2007
4. Dec 2, 2007

hotcommodity

If I have a function $$f(x) = e^u$$, where u can be any kind of x term, say .05x for instance, then I know that $$d/dx = e^u u'$$ where u' is the derivative of u, or in this case, the derivative of .05x, which is just .05. Does that make sense?

5. Dec 2, 2007

kevinr

Well there are two because thats how to find derivative of e.

If you had e^2x, than the derivative would be 2e^2x. (2 comes from derivative of 2x).

(if you have e^x, than it is 1e^x = e^x because derivative of x is 1)

So in your problem, what is the derivative of -0.05t? It is -0.05 and thus that is why it comes in the front.

If you are confused ill try to explain better. Let me know!

(Sorry i dont know special writing method yet)

Last edited: Dec 2, 2007
6. Dec 2, 2007

You are trying to apply the power rule, but you have $$e^x$$ not $$x^n$$ the base is e and the exponent is a function,

...chain rule. $$\frac{d}{dx}[e^u]=e^u*\frac{du}{dx}$$ if I remember correctly.

Casey

Edit: I see that like nine others were typing at the same time as me.when it rains it pours! Take your pick :)

7. Dec 2, 2007

PlasmaSphere

Sorry, i eddited my question after you answered it. However i think i get it now, I just forgot that e is its own deriviative.

so when i said; "i thought that the power had to decrease by 1 when you differenciated it and then multiply by the old power, so 20e-0.05 should become -0.05 x 20e-1.05.", thats not correct because you differenciate e differently

so it should become -0.05 x 20e-0.05t

8. Dec 2, 2007

PlasmaSphere

thanx, thats what i needed. I understand now.