2
$\begingroup$

What was the average temperature during that period?

My initial thought was to take the derivative of the problem, plug in 24 for $t$ and solve. I was wrong. This is what I have

$T'=8-t=8-24=-16$ deg.

Should I have taken the integral instead of the derivative?

$\endgroup$
2
$\begingroup$

The average value of $f(t)$ over the interval $[a,b]$ is $$\frac{\int_a^b f(t)\,dt}{b-a}.$$ To see that derivative has not much to do with average value, suppose that $f(t)$ is the constant $K$ over our interval. Then the average value must be $K$. But $f'(t)=0$ for all $t$.

$\endgroup$
  • $\begingroup$ ok so the integral of 49+8t-(1/2)t^2 = 49x+4x^2-0.166667x^3+c. If I take this equation and plug in 24 for the value I get approx 1176. I will multiply that by 1/(b-a) or 1/24 and get an answer of 49 deg which is correct. thank you for your help $\endgroup$ – Tonya May 17 '13 at 23:05
  • $\begingroup$ One integral is $49t+4t^2-\frac{t^3}{6}$. Plug in $24$, take away the (easy) result of plugging in $0$, then divide by $24$. We get $49+4(24)-\frac{24^2}{6}$. $\endgroup$ – André Nicolas May 17 '13 at 23:09
1
$\begingroup$

The derivative gives the rate of change of temperature. Instead, you should compute the integral from 0 to 24 hours and then divide by 24 to get the average temperature.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.