Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 18: 29

Answer

(a) The length of the rod at temperature, $T=90^\circ C$ is $L=40.0248$ cm. (b) The length of the rod at temperature, $T=50^\circ C$ is $L=64.9597$ cm. (c) The required function is $L(T)=0.000806T+64.9194$ cm.

Work Step by Step

(a) Here, $\alpha={1.24\times 10^{-5}}^{\circ} C^{-1}$, $L_{0}=40$ cm, $T_{0}=40^\circ C$, and $T=90^\circ C$, so, we have $\Delta T=T-T_{0}=50^{\circ}C$. As we have $\Delta L=\alpha L_{0}\Delta T$, thus, we get $\Delta L=1.24\times 10^{-5}\times 40\times 50=0.0248$ cm. Thus, length increases by $\Delta L=0.0248$ cm, so, the length at temperature, $T=90^\circ C$ is $L=L_{0}+\Delta L=40.0248$ cm. (b) Here, $\alpha={1.24\times 10^{-5}}^{\circ} C^{-1}$, $L_{0}=65$ cm, $T_{0}=100^\circ C$, and $T=50^\circ C$, so, we have $\Delta T=T-T_{0}=-50^{\circ}C$. As we have $\Delta L=\alpha L_{0}\Delta T$, thus, we get $\Delta L=1.24\times 10^{-5}\times 65\times (-50)=-0.0403$ cm. Thus, length decreases by $\Delta L=-0.0403$ cm, so, the length at temperature, $T=50^\circ C$ is $L=L_{0}+\Delta L=64.9597$ cm. (c) The length $L$ varies linearly with respect to $T$, since, slope of the tangent, $\dfrac{\Delta L}{\Delta T}=\alpha L_{0}$, which is a constant. Here, $\alpha ={1.24\times 10^{-5}}^{\circ} C^{-1}$, $L_{0}=65$ cm and $T_{0}=100^\circ C$. Slope of the function will be $\dfrac{\Delta L}{\Delta T}=\alpha L_{0}=1.24\times 10^{-5}\times 65=0.000806\text{ cm}/^\circ C$. Thus, using slope and point form, we get $L-65=0.000806(T-100)$ $\Rightarrow L=0.000806T-0.0806+65$ $\Rightarrow L=0.000806T+64.9194$ Thus, in function notation, we have $L(T)=0.000806T+64.9194$, which is the required function.
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