Answer
(a) The pedestrian's average waiting time is $29.805$ seconds.
(b) The estimate of the increase in waiting time if $w$ is increased to $27$ ft is $5.79$ seconds.
(c) The estimate of the waiting time if the width is increased to $27$ ft and $R$ decreases to $0.18$ is $31.338$ seconds
(d) The rate of increase in waiting time per $1$-ft increase in width when $w=30$ ft and $R=0.3$ is $13.351$.
Work Step by Step
(a) We are given $f\left( {w,R} \right) = \left( {w/3.5} \right){{\rm{e}}^{wR/3.5}}$. So,
$f\left( {25,0.2} \right) = \left( {25/3.5} \right){{\rm{e}}^{25\cdot0.2/3.5}} \simeq 29.805$
So, the pedestrian's average waiting time is $29.805$ seconds.
(b) We are given $f\left( {w,R} \right) = \left( {w/3.5} \right){{\rm{e}}^{wR/3.5}}$. So, the partial derivatives are
$\frac{{\partial f}}{{\partial w}} = \frac{1}{{3.5}}{{\rm{e}}^{wR/3.5}} + \frac{{wR}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{wR/3.5}}$
$\frac{{\partial f}}{{\partial R}} = \frac{{{w^2}}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{wR/3.5}}$
Using the linear approximation, Eq. (5) the change in $f$ is estimated to be
$\Delta f \approx \frac{{\partial f}}{{\partial w}}\Delta w + \frac{{\partial f}}{{\partial R}}\Delta R$
$\Delta f \approx \left( {\frac{1}{{3.5}}{{\rm{e}}^{wR/3.5}} + \frac{{wR}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{wR/3.5}}} \right)\Delta w + \frac{{{w^2}}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{wR/3.5}}\Delta R$
Let $R$ be constant. If $w$ is increased to $27$ ft, then
$\Delta w = 27 - 25 = 2$, ${\ \ }$ $\Delta R = 0$
At $w=25$ ft and $R=0.2$,
$\Delta f \approx \frac{{\partial f}}{{\partial w}}{|_{\left( {25,0.2} \right)}}\Delta w + \frac{{\partial f}}{{\partial R}}{|_{\left( {25,0.2} \right)}}\Delta R$
$\Delta f \simeq \left( {\frac{1}{{3.5}}{{\rm{e}}^{25\cdot0.2/3.5}} + \frac{{25\cdot0.2}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{25\cdot0.2/3.5}}} \right)\cdot2$
$\Delta f \simeq 5.79$
The estimate of the increase in waiting time if $w$ is increased to $27$ ft is $5.79$ seconds.
(c) From part (b) we get, at $w=25$ ft and $R=0.2$:
$\Delta f \approx \left( {\frac{1}{{3.5}}{{\rm{e}}^{wR/3.5}} + \frac{{wR}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{wR/3.5}}} \right)\Delta w + \frac{{{w^2}}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{wR/3.5}}\Delta R$
$\Delta f \approx \left( {\frac{1}{{3.5}}{{\rm{e}}^{25\cdot0.2/3.5}} + \frac{{25\cdot0.2}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{25\cdot0.2/3.5}}} \right)\Delta w + \frac{{{{25}^2}}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{25\cdot0.2/3.5}}\Delta R$
If the width is increased to $27$ ft and $R$ decreases to $0.18$, we have
$\Delta w = 27 - 25 = 2$, ${\ \ \ }$ $\Delta R = 0.18 - 0.2 = - 0.02$
So, the change in waiting time is
$\Delta f \approx \left( {\frac{1}{{3.5}}{{\rm{e}}^{25\cdot0.2/3.5}} + \frac{{25\cdot0.2}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{25\cdot0.2/3.5}}} \right)\cdot2 + \frac{{{{25}^2}}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{25\cdot0.2/3.5}}\cdot\left( { - 0.02} \right)$
$\Delta f \simeq 1.533$
Thus, the change in waiting time is increased by $1.533$ seconds.
Hence, the estimate of the waiting time if the width is increased to $27$ ft and $R$ decreases to $0.18$ is
$f\left( {25,0.2} \right) + 1.533 \simeq 29.805 + 1.533 = 31.338$ seconds
(d) From part (b) we get the rate of increase in waiting time per $1$-ft increase in width, given by
$\frac{{\partial f}}{{\partial w}} = \frac{1}{{3.5}}{{\rm{e}}^{wR/3.5}} + \frac{{wR}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{wR/3.5}}$
When $w=30$ ft and $R=0.3$ vehicle per second:
$\frac{{\partial f}}{{\partial w}}{|_{\left( {30,0.3} \right)}} = \frac{1}{{3.5}}{{\rm{e}}^{30\cdot0.3/3.5}} + \frac{{30\cdot0.3}}{{{{\left( {3.5} \right)}^2}}}{{\rm{e}}^{30\cdot0.3/3.5}}$
$\frac{{\partial f}}{{\partial w}}{|_{\left( {30,0.3} \right)}} \simeq 13.351$
Thus, the rate of increase in waiting time per $1$-ft increase in width when $w=30$ ft and $R=0.3$ is $13.351$.
