Syllabus Vector Differentiation - Velocity and Acceleration - Gradient ...
Syllabus Vector Differentiation - Velocity and Acceleration - Gradient ...
Syllabus Vector Differentiation - Velocity and Acceleration - Gradient ...
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Engineering Mathematics - II - <strong>Vector</strong> Calculus - 2007<br />
Suggested answer:<br />
<br />
<br />
<br />
<br />
(4x<br />
3<br />
6xy<br />
2<br />
z) i ( 6x<br />
2<br />
yz) j<br />
(12xz<br />
2<br />
3x<br />
2<br />
y<br />
2<br />
) k<br />
<br />
at<br />
(2, 1,2)<br />
<br />
8 i 48 j<br />
84 k<br />
<br />
c 2 i 3 j<br />
6 k<br />
1<br />
. c [8(2) 48( 3)<br />
(84)(6)]<br />
7<br />
376<br />
<br />
7<br />
33. Find the directional derivative of<br />
x<br />
2<br />
y<br />
2<br />
z<br />
2<br />
at (1, 1, -1)<br />
in the direction of the tangent to the curve x = e t , y = 1 + 2sint,<br />
z = t - cost, 1 t 1.<br />
Suggested answer:<br />
x<br />
2<br />
y<br />
2<br />
z<br />
2<br />
<br />
<br />
<br />
<br />
2xy<br />
2<br />
z<br />
2<br />
i 2x<br />
2<br />
yz<br />
2<br />
j<br />
2x<br />
2<br />
y<br />
2<br />
z k<br />
<br />
at<br />
(1, 1,<br />
<br />
1) 2 i 2 j<br />
2 k<br />
<br />
<br />
<br />
r e<br />
t<br />
i (1 2 sin t) j<br />
(t cos t) k<br />
<br />
d r<br />
dt<br />
at t<br />
0<br />
is<br />
<br />
i 2 j<br />
k<br />
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