Answer
(a) The change in monthly payment per $\$1000$ increase in loan principal is $\$7.1$.
(b)
1. The interest rate increases to $r=6.5\%$
The change in monthly payment is $\$28.845$.
2. The interest rate increases to $r=7\%$
The change in monthly payment is $\$57,69$.
(c) The monthly payment will be reduced by $\$74.2416$ if the length of the loan increases to $24$ years.
Work Step by Step
(a) At $P= \$100,000$, $r=0.06$, and $N=240$, we are given:
$f\left( {100,000,0.06,240} \right) = 716.43$,
$\frac{{\partial f}}{{\partial P}} = 0.0071$, ${\ \ }$ $\frac{{\partial f}}{{\partial r}} = 5769$, ${\ \ }$ $\frac{{\partial f}}{{\partial N}} = - 1.5467$
Suppose that the interest rate and the length of the loan stay the same. An increase in the loan principal by $\$ 1000$ implies that
$\Delta P = \$ 1000$, ${\ \ }$ $\Delta r = 0$, ${\ \ }$ $\Delta N = 0$
Using the linear approximation, Eq. (5) the change in $f$ is estimated to be
$\Delta f \approx \frac{{\partial f}}{{\partial P}}{|_{\left( {100,000,0.06,240} \right)}}\Delta P + \frac{{\partial f}}{{\partial r}}{|_{\left( {100,000,0.06,240} \right)}}\Delta r + \frac{{\partial f}}{{\partial N}}{|_{\left( {100,000,0.06,240} \right)}}\Delta N$
$\Delta f \approx 0.0071\cdot1000 = 7.1$
Thus, the change in monthly payment per $\$1000$ increase in loan principal is $\$7.1$.
(b)
1. Let the loan principal and the length of the loan stay the same. Suppose the interest rate increases to $r=6.5\%$. We have
$\Delta P = 0$, ${\ \ }$ $\Delta r = 0.065 - 0.06 = 0.005$, ${\ \ }$ $\Delta N = 0$
Using the linear approximation, Eq. (5) the change in $f$ is estimated to be
$\Delta f \approx \frac{{\partial f}}{{\partial P}}{|_{\left( {100,000,0.06,240} \right)}}\Delta P + \frac{{\partial f}}{{\partial r}}{|_{\left( {100,000,0.06,240} \right)}}\Delta r + \frac{{\partial f}}{{\partial N}}{|_{\left( {100,000,0.06,240} \right)}}\Delta N$
$\Delta f \approx 5769\cdot0.005 = 28.845$
Thus, the change in monthly payment is $\$28.845$ if the interest rate increases to $r=6.5\%$.
2. Similar to the case above, suppose the interest rate increases to $r=7\%$. We have
$\Delta P = 0$, ${\ \ }$ $\Delta r = 0.07 - 0.06 = 0.01$, ${\ \ }$ $\Delta N = 0$
Using the linear approximation, Eq. (5) the change in $f$ is estimated to be
$\Delta f \approx \frac{{\partial f}}{{\partial P}}{|_{\left( {100,000,0.06,240} \right)}}\Delta P + \frac{{\partial f}}{{\partial r}}{|_{\left( {100,000,0.06,240} \right)}}\Delta r + \frac{{\partial f}}{{\partial N}}{|_{\left( {100,000,0.06,240} \right)}}\Delta N$
$\Delta f \approx 5769\cdot0.01 = 57,69$
Thus, the change in monthly payment is $\$57,69$ if the interest rate increases to $r=7\%$.
(c) Let the loan principal and the interest rate stay the same. If the length of the the loan increases to $24$ years, we have
$\Delta P = 0$, ${\ \ }$ $\Delta r = 0$, ${\ \ }$ $\Delta N = 24\cdot12 - 240 = 48$
Using the linear approximation, Eq. (5) the change in $f$ is estimated to be
$\Delta f \approx \frac{{\partial f}}{{\partial P}}{|_{\left( {100,000,0.06,240} \right)}}\Delta P + \frac{{\partial f}}{{\partial r}}{|_{\left( {100,000,0.06,240} \right)}}\Delta r + \frac{{\partial f}}{{\partial N}}{|_{\left( {100,000,0.06,240} \right)}}\Delta N$
$\Delta f \approx \left( { - 1.5467} \right)\cdot48 = - 74.2416$
Thus, the monthly payment will be reduced by $\$74.2416$ if the length of the loan increases to $24$ years.