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Question

Refrigeration The temperature TT of food put in a freezer is

T=700t2+4t+10T=\frac{700}{t^2+4 t+10}

where tt is the time in hours. Find the rate of change of TT with respect to tt at each of the following times.
(a) t=1t=1 (b) t=3t=3 (c ) t=5t=5 (d) t=10t=10

Solution

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Answered 2 years ago
Answered 2 years ago
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Given:

T=700t2+4t+10\begin{aligned} T=\frac{700}{t^2+4t+10} \end{aligned}

The rate of change is the derivative of TT.Let u(x)=700u(x)=700 and v(x)=t2+4t+10v(x)=t^2+4t+10, differentiating by quotient rule:

[u(x)v(x)]=[u(x)v(x)u(x)v(x)v(x)2]700t2+4t+10=ddt700t2+4t+10=(ddt700(t2+4t+10)700ddt(t2+4t+10)(t2+4t+10)2)=(0(t2+4t+10)700(ddtt2+4ddtt+ddt10)(t2+4t+10)2)=(700(2t21+41t11+0)(t2+4t+10)2)=(700(2t+41)(t2+4t+10)2)=700(2t+4)(t2+4t+10)2\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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