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Assignment 6 solutions
Total=5 points
1. Find the tangent plane to the graph of the function
f
(
x,y
) = 2
π

3 cos 4
x
+ 3 sin 2
y
at the point (0
,π
).
Solution:
The tangent plane of
f
at a point (
a,b
) is given by the equation
z
=
f
(
a,b
) + grad
f
(
a,b
)
±
x

a
y

b
²
.
Here (
a,b
) = (0
,π
). The
x
partial derivative of
f
,
∂f
∂x
, is obtained by treating
y
as a constant
and taking the derivative with respect to
x
: here it is given by 0 + 12 sin 4
x
+ 0 = 12 sin 2
x
.
Similarly
∂f
∂y
= 6 cos 2
y
, and so
grad
f
=
±
∂f
∂x
,
∂f
∂y
²
= (12 sin 2
x,
6 cos 2
y
)
.
For the tangent plane (or linear approximation), we evaluate the gradient at the speciﬁed
point (0
,π
), that is, we plug in 0 for
x
and
π
for
y
:
grad
f
(
a,b
) = grad
f
(0
,π
) = (0
,
6)
.
So the tangent plane is given by
z
=
f
(
a,b
) + grad
f
(
a,b
)
±
x

a
y

b
²
=
f
(0
,π
) + (0
,
6)
±
x
y

π
²
=(2
π

3
*
1 + 3
*
0) + (0
,
6)
±
x
y

π
²
=(2
π

3) + 0
*
x
+ (6)(
y

π
)
=

4
π

3 + 6
y.
Alternatively we could have used the equivalent formula
z
=
f
(
a,b
) +
∂f
∂x
(
a,b
)(
x

a
) +
∂f
∂y
(
a,b
)(
y

b
)
for the tangent plane.
1
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View Full Document 2. Find the (2,2)entry of the Jacobian matrix of the function
F
(
x,y
) =
±
x
2
e
y
+ 2
x
sin(
x
y
)
sin(
x
2
)

3
ye

x
²
at the point (2
,
1)
.
Solution:
The Jacobian of
F
=
³
f
g
´
is given by
J
=
∂f
∂x
∂f
∂y
∂g
∂x
∂g
∂y
!
.
The (2,2) entry means the second row, second column ((
i,j
)
th
entry is the entry in the
i
th
row and
j
th
column), that is,
∂g
∂y
.
Here
g
, the second row of
F
, is the function sin(
x
2
)

3
ye

x
. Its
y
partial derivative is
∂g
∂y
=
∂
∂y
(sin(
x
2
)

3
ye

x
) = 0

3
e

x
.
We are asked to evaluate this at the point (2
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This note was uploaded on 09/17/2011 for the course MAT 1332 taught by Professor Munteanu during the Winter '07 term at University of Ottawa.
 Winter '07
 MUNTEANU

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