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 />
<br />
4 i 4 j<br />
2k<br />
<br />
36<br />
<br />
2<br />
3<br />
<br />
i <br />
2<br />
3<br />
1 <br />
j<br />
k<br />
3<br />
05. A particle moves so that its position vector is given by<br />
<br />
r cos t<br />
i<br />
sin t<br />
j,<br />
is<br />
a<br />
cons tan t<br />
show<br />
that<br />
<br />
i) velocity v<br />
is perpendicular to<br />
<br />
r<br />
ii) acceleration<br />
to the distance from the origin.<br />
<br />
a is directed towards the origin <strong>and</strong> has magnitude proportional<br />
<br />
iii) r x v is a cons tan t vector.<br />
Suggested answer:<br />
i)<br />
<br />
r cos t<br />
i sin t<br />
j<br />
<br />
d r<br />
dt<br />
<br />
<br />
sin t<br />
i cos t<br />
j v<br />
Now<br />
<br />
v . r 0<br />
<br />
v r<br />
ii)<br />
<br />
a <br />
<br />
d<br />
2<br />
r<br />
dt<br />
2<br />
<br />
<br />
<br />
2<br />
cos t<br />
i sin t<br />
j<br />
<br />
a <br />
2<br />
r<br />
<br />
| a | 2 <br />
| r |<br />
iii)<br />
<br />
r x v<br />
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