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Here's my question:

A metal bar is heated to a certain temperature and then the heat source is removed. At time t minutes after the heat source is removed, the temperature, x degrees Celcius, of the metal bar is given by $x = \dfrac{280}{1+0.02t}$ At what rate is the temperature decreasing 100 minutes after the removal of the heat source?

I'm guessing the chain rule needs to be employed; as in $\dfrac{dx}{dt} = \dfrac{dx}{du}\dfrac{du}{dt}$ but couldn't figure out exactly how? Help appreciated, thanks!

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The chain rule is not needed, you can use the quotient rule if you like. To use the chain rule, write $x(t)=280(1+0.02 t)^{-1}$. The derivative is then $280\cdot\bigl(-(1+0.02t)^{-2}\bigr)\cdot 0.02 =-5.6(1+0.02t)^{-2}$. –  David Mitra May 6 '12 at 17:19
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Deniz, once you get the answer that you were looking for, please accept the answer with the "check" sign to the left of the question. This not only gives you a higher acceptance rate, but it also gives credit to the people giving the best answer. –  Nico Bellic May 6 '12 at 18:19

1 Answer 1

up vote 3 down vote accepted

I guess that if you want to apply the chain rule then what you want to do is to write your function as

$$ x(t) = 280(1 + 0.02t)^{-1} $$

and then it's derivative can be found by applying the chain rule as follows. Following your notation you can let $u = 1 + 0.02t$ and then you have $x = 280u^{-1}$. Therefore

$$ \frac{dx}{dt} = \color{green}{\frac{dx}{du}} \color{red}{\frac{du}{dt}} = \color{green}{-280u^{-2}} \color{red}{(0.02)} = \frac{-280(0.02)}{u^2} = \frac{-5.6}{(1 + 0.02t)^2} $$

Then with the derivative in hand I guess you can figure out the rate at which the temperature is decreasing 100 minutes after.

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