1. ## average temperature

Hi everyone,

Could you please tell me if this is correct?

If a cup of coffe has temperature 95 C in a room where the temperature is 20 C, then according to Newton's Law of Cooling, the temperature of the coffee after t minutes is T(t)=20+75e^-t/50. What is the average temperature of cofee during the first half hour?

u=-t/50
du=-1/50dt
-50du=-1/50dt

int. 20+75e^-t/50 from 0 to 30
20t-3750e^u from 0 to 30
-1458.044

Thank you very much

2. Originally Posted by chocolatelover
int. 20+75e^-t/50 from 0 to 30
20t-3750e^u from 0 to 30
-1458.044

Thank you very much
1) The coefficient of the exponential is $\displaystyle -\frac{75}{50}$ not $\displaystyle -75 \cdot 50$.

2) You forgot to divide by the time range.

-Dan

3. Hi,

1) The coefficient of the exponential is not .
Could you please show me how you got -75/50?

I would then have 1/30(20t-75/50e^-t/50) right?

Thank you very much

4. Hi,

1) The coefficient of the exponential is not .
Could you please show me how you got that?

I got 19.973C. Does that look right?

Thank you very much

5. Originally Posted by chocolatelover
Hi,

Could you please show me how you got -75/50?

I would then have 1/30(20t-75/50e^-t/50) right?

Thank you very much
Sorry. I screwed up the coefficient.

$\displaystyle T_{ave} = \frac{1}{30 - 0} \int_0^{30}(20 + 75e^{-t/50})~dt$

Let $\displaystyle u = -\frac{t}{50} \implies dt = -50 du$ <-- Typically we keep the negative sign in the exponent, but this is fine too.
$\displaystyle T_{ave} = \frac{1}{30} \int_0^{-30/50}(20 + 75e^{u})~(-50du)$

$\displaystyle T_{ave} = -\frac{50}{30} \int_0^{-3/5}(20 + 75e^{u})~du$ <-- Your expression here was slightly off: you didn't multiply the 20 by -50.

$\displaystyle T_{ave} = -\frac{5}{3} \int_0^{-3/5}(20 + 75e^{u})~du$

$\displaystyle T_{ave} = \frac{5}{3} \int_{-3/5}^0(20 + 75e^{u})~du$

$\displaystyle T_{ave} = \frac{5}{3} (20(0)+ 75e^0) - \frac{5}{3} \left (20 \cdot \frac{-3}{5} + 75e^{-3/5} \right )$

$\displaystyle T_{ave} = \frac{5}{3} \cdot 75 - \frac{5}{3} (-12 + 75e^{-3/5})$

$\displaystyle T_{ave} =76.3985$

-Dan

PS By the way, you should be able to tell that your answer is incorrect. The temperature can't go below 75 C, so the average temperature can't be that low either.

6. Thank you very much

Can you explain why you changed the limits? Wouldn't it stay from 0 to 30?

Thank you very much

7. Originally Posted by chocolatelover
Thank you very much

Can you explain why you changed the limits? Wouldn't it stay from 0 to 30?

Thank you very much
No because you have changed your variable by a factor of 50. So your limits should change as well to reflect that.

-Dan

8. No because you have changed your variable by a factor of 50. So your limits should change as well to reflect that
One more question, does this only happen when you are finding the average or do you also have to change the limits when you have a numberdu when solving an integral? How come for the upper limit you divided by -50 instead of multiplying?

Thank you very much

9. Originally Posted by chocolatelover
One more question, does this only happen when you are finding the average or do you also have to change the limits when you have a numberdu when solving an integral? How come for the upper limit you divided by -50 instead of multiplying?

Thank you very much
Any time that you use a substitution in a definite integral you need to alter the limits. In the general case we have:
$\displaystyle \int_a^b f(kx)~dx$

We obviously want to use a substitution of $\displaystyle u = kx \implies du = k~dx$. Now, we see that any value of x corresponds to a value of u such that $\displaystyle x = \frac{u}{k}$. The limits, belonging to a span of the x axis, are "x" values. So we need to change them to "u" values. Hence:
$\displaystyle \int_a^b f(kx)~dx = \int_{a/k}^{b/k} f(u)~\frac{du}{k}$

-Dan