Finding Rate of Change and Time of Temperature Change in a Liquid System

[PlasmaSphere]

Homework Statement

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

Homework Equations

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 can't see how they got that. what happened to the 30? and how did they get to that point?

 

[kevinr]

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

 

[PlasmaSphere]

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

ok, cheers.

I still don't 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 that's not what the answer says, which is why I'm confused.

 

[hotcommodity]

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

 

[kevinr]

Well there are two because that's 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 don't know special writing method yet)

 

[Saladsamurai]

ok, cheers.

I still don't 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 it should be -0.05 x 20e^-1.05. but that's not what the answer says.

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

...chain rule. [tex]\frac{d}{dx}[e^u]=e^u*\frac{du}{dx}[/tex] 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 :)

 

[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.", that's not correct because you differenciate e differently

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

 

[PlasmaSphere]

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

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

thanx, that's what i needed. I understand now.