Chapter 15 - Differentiation in Several Variables - 15.4 Differentiability and Tangent Planes - Exercises - Page 790: 37

(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.

(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.