Question 1 of 9, Page 2 of 14
Solutions
1. Consider the function
f
(
x, y
) =
x
4
y
3
and the point
P
= (1
,
1
,
1) on its graph.
(a) Write down the equation of the tangent plane at the graph of the function at
the point
P
.
Solution:
The tangent plane has equation, in general,
±
∂f
∂x
,
∂f
∂y
,

1
²³
³
³
³
P
·
(
x

P
) = 0
In this case,
∂f
∂x
= 4
x
3
y
3
which is 4 at (1,1).
∂f
∂y
= 3
x
4
y
2
which is 3 at (1,1).
So the normal vector to the plane is
h
4
,
3
,

1
i
and the tangent plane has
equation:
4(
x

1) + 3(
y

1)

(
z

1) = 0
(b) Using your answer from (a), write down an expression for the change, Δ
z
, in
z
=
f
(
x, y
) depending on Δ
x
and Δ
y
, the change in
x
and
y
, respectively,
near the point
P
= (1
,
1
,
1). Is the function
f
(
x, y
) more sensitive to a change
in
x
or to a change in
y
? Explain.
Solution:
Near the point
P
= (1
,
1
,
1) the tangent plane is a good approximation to the
function. Thus
Δ
z
= 4Δ
x
+ 3Δ
y
hence the function is more sensitive to changes in
x
. That is, small changes
in
x
result in larger diﬀerences in the resulting
z
than small changes in
y
.
(c) Using your answer to (b), ﬁnd the approximate value of
f
(1
.
01
,
1
.
02).
Solution:
Using the tangent plane approximation at (1,1), we ﬁnd
f
(1
.
01
,
1
.
02) =
f
(1
,
1) +
.
01(4) +
.
02(3) = 1 +
.
04 +
.
06 = 1
.
1
since we know using any linear approximation in two variables, this is just the
value of the tangent plane at this point.