SECTION
15.4
Differentiability, Linear Approximation, and Tangent Planes
(ET Section 14.4)
673
We now divide by
I
=
x
a
y
b
to obtain
1
I
I
≈
ax
a
−
1
y
b
1
x
I
+
bx
a
y
b
−
1
1
y
I
=
a
−
1
y
b
1
x
x
a
y
b
+
a
y
b
−
1
1
y
x
a
y
b
=
a
1
x
x
+
b
1
y
y
That is,
1
I
I
≈
a
1
x
x
+
b
1
y
y
.
37.
The monthly payment for a home loan is given by a function
f
(
P
,
r
,
N
)
,where
P
is the principal (the initial size
of the loan),
r
the interest rate, and
N
the length of the loan in months. Interest rates are expressed as a decimal: A 6%
interest rate is denoted by
r
=
0
.
06. If
P
=
$100
,
000,
r
=
0
.
06, and
N
=
240 (a 20-year loan), then the monthly
payment is
f
(
100
,
000
,
0
.
06
,
240
)
=
716
.
43. Furthermore, with these values, we have
∂
f
∂
P
=
0
.
0071
,
∂
f
∂
r
=
5
,
769
,
∂
f
∂
N
=−
1
.
5467
Estimate:
(a)
The change in monthly payment per $1,000 increase in loan principal.
(b)
The change in monthly payment if the interest rate increases to
r
=
6
.
5% and
r
=
7%.
(c)
The change in monthly payment if the length of the loan increases to 24 years.
SOLUTION
(a)
The linear approximation to
f
(
P
,
r
,
N
)
is
1
f
≈
∂
f
∂
P
1
P
+
∂
f
∂
r
1
r
+
∂
f
∂
N
1
N
We are given that
∂
f
∂
P
=
0
.
0071,
∂
f
∂
r
=
5769,
∂
f
∂
N
1
.
5467, and
1
P
=
1000. Assuming that
1
r
=
0and
1
N
=
0,
we get
1
f
≈
0
.
0071
·
1000
=
7
.
1
The change in monthly payment per thousand dollar increase in loan principal is $7.1.