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Question

# Refrigeration The temperature $T$ of food put in a freezer is$T=\frac{700}{t^2+4 t+10}$where $t$ is the time in hours. Find the rate of change of $T$ with respect to $t$ at each of the following times. (a) $t=1$ (b) $t=3$ (c ) $t=5$ (d) $t=10$

Solution

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Given:

\begin{aligned} T=\frac{700}{t^2+4t+10} \end{aligned}

The rate of change is the derivative of $T$.Let $u(x)=700$ and $v(x)=t^2+4t+10$, differentiating by quotient rule:

\begin{aligned} \left[\frac{u(x)}{v(x)}\right]'&=\left[\frac{u'(x)\cdot v(x)-u(x)\cdot v'(x)}{v(x)^2}\right]\\ \frac{700}{t^2+4t+10}&=\frac{d}{dt} \frac{700}{t^2+4t+10}\\ &=\left(\frac{\frac{d}{dt}700\cdot (t^2+4t+10)-700 \cdot \frac{d}{dt}(t^2+4t+10)}{(t^2+4t+10)^2}\right)\\ &=\left(\frac{0\cdot (t^2+4t+10)-700 \cdot (\frac{d}{dt}t^2+4\frac{d}{dt}t+\frac{d}{dt}10)}{(t^2+4t+10)^2}\right)\\ &=\left(\frac{-700 \cdot (2t^{2-1}+4\cdot 1t^{1-1}+0)}{(t^2+4t+10)^2}\right)\\ &=\left(\frac{-700 \cdot (2t+4\cdot 1)}{(t^2+4t+10)^2}\right)\\ &=\frac{-700 \cdot (2t+4)}{(t^2+4t+10)^2}\\ \end{aligned}

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