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Question #227723
Find the directional derivative of (x,y,z) = xsin yz at (1,3,0) in the direction of i+2j-k
Expert's answer
f
(
x
,
y
,
z
)
=
x
sin
y
z
f(x,y,z)=x\sin yz
f
(
x
,
y
,
z
)
=
x
sin
yz
∇
f
=
(
sin
y
z
)
i
+
(
x
z
cos
y
z
)
j
+
(
x
y
cos
y
z
)
k
\nabla f=(\sin yz)i+(xz\cos yz)j+(xy\cos yz)k
∇
f
=
(
sin
yz
)
i
+
(
x
z
cos
yz
)
j
+
(
x
y
cos
yz
)
k
∇
f
(
1
,
3
,
0
)
=
0
i
+
0
j
+
3
k
\nabla f(1,3,0)=0i+0j+3k
∇
f
(
1
,
3
,
0
)
=
0
i
+
0
j
+
3
k
∣
i
+
2
j
−
k
∣
=
1
2
+
2
2
+
(
−
1
)
2
=
6
|i+2j-k|=\sqrt{1^2+2^2+(-1)^2}=\sqrt{6}
∣
i
+
2
j
−
k
∣
=
1
2
+
2
2
+
(
−
1
)
2
=
6
u
=
1
6
i
+
2
6
j
−
1
6
k
u=\dfrac{1}{\sqrt{6}}i+\dfrac{2}{\sqrt{6}}j-\dfrac{1}{\sqrt{6}}k
u
=
6
1
i
+
6
2
j
−
6
1
k
D
u
f
(
1
,
3
,
0
)
=
(
0
i
+
0
j
+
3
k
)
(
1
6
i
+
2
6
j
−
1
6
k
)
D_uf(1,3,0)=(0i+0j+3k)(\dfrac{1}{\sqrt{6}}i+\dfrac{2}{\sqrt{6}}j-\dfrac{1}{\sqrt{6}}k)
D
u
f
(
1
,
3
,
0
)
=
(
0
i
+
0
j
+
3
k
)
(
6
1
i
+
6
2
j
−
6
1
k
)
=
−
6
2
=-\dfrac{\sqrt{6}}{2}
=
−
2
6
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#340153
on Dec 2023
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