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.
