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 />
i) Equation to the curve is<br />
<br />
r e<br />
t<br />
<br />
<br />
i<br />
2 cos 3t j<br />
2 sin3t k<br />
<br />
v <br />
<br />
d r<br />
dt<br />
<br />
<br />
e<br />
t<br />
i 6 sin3t j<br />
6 cos 3t k<br />
<br />
a <br />
<br />
d<br />
2<br />
r<br />
dt<br />
2<br />
<br />
<br />
<br />
e<br />
t<br />
i 18 cos 3t j<br />
18 sin3t k<br />
ii) At t = 0,<br />
<br />
v i 6 k,<br />
<br />
a i 18 j<br />
<strong>and</strong><br />
<br />
<br />
| v | 37, | a | 325<br />
04. Find the unit tangent vectors at any point on the curve x = t 2 +1, y = 4t - 3,<br />
z = 2t 2 - 6t. Determine the unit tangent at the point t = 2.<br />
Suggested answer:<br />
Equation to the space curve is<br />
<br />
<br />
<br />
r (t<br />
2<br />
1) i<br />
(4t 3) j<br />
(2t<br />
2<br />
6t) k<br />
<br />
d r<br />
dt<br />
<br />
<br />
2t i 4 j<br />
(4t 6) k<br />
<br />
d r <br />
<br />
dt <br />
t2<br />
<br />
4 i 4 j<br />
2 k<br />
unit tangent<br />
<br />
t <br />
<br />
d r<br />
dt<br />
<br />
d r<br />
dt<br />
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