You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**mrpace****Member**- Registered: 2012-08-16
- Posts: 26

dT/dt = -k(T-T1)

The question then states that this means that T(t)=T1+Ce^(-kt)

I don't see how they got that...can you help please?

Offline

You will get that if you integrate both sides

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

Offline

**mrpace****Member**- Registered: 2012-08-16
- Posts: 26

Agnishom wrote:

You will get that if you integrate both sides

maybe you can show me?

Offline

Someone else might do it. I cannot

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'Humanity is still kept intact. It remains within.' -Alokananda

Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

hi mrpace

Fisrtly, note that T and t represent different variables here.

One method for solving this is called "separation of variables". You have to get all the T terms on one side and all the t terms on the other

Strictly speaking the dT/dt is not separable like this as it represents 0/0, but it works out ok if you quickly put in integration signs.

Now integrate each side. The right hand side is easy as it is just integrate a constant with respect to t. The left hand side is a natural log.

As D is just the constant of integration you can replace it with ln(C), another constant.

Bring the two log terms together using the rules for logs and then re-write the expression with e to get the required expression.

I've left a few steps for you. Post back if you need more help.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

Pages: **1**