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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t <br />
<br />
d r<br />
dt<br />
<br />
d r<br />
dt<br />
<br />
Engineering Mathematics - II - <strong>Vector</strong> Calculus - 2007<br />
<br />
i 2 j<br />
k<br />
6<br />
directional derivative of in the direction of<br />
<br />
c<br />
is<br />
<br />
( i 2 j<br />
k)<br />
(2 i 2 j<br />
2 k).<br />
6<br />
<br />
1<br />
[2 4 2]<br />
6<br />
<br />
4<br />
6<br />
34. Find the equation of the tangent plane <strong>and</strong> the normal line to the surface z =<br />
x 2 + y 2 at (2, -1, 5).<br />
Suggested answer:<br />
x<br />
2<br />
y<br />
2<br />
z<br />
<br />
<br />
2x i 2y j<br />
k<br />
Equation to the normal line at (2, -1, 5) is<br />
x 2<br />
4<br />
<br />
y 1<br />
<br />
2<br />
z 5<br />
1<br />
Equation the tangent plane is 4(x - 2) + (-1)(y + 2)(-1)(z - 5) = 0<br />
i.e., 4x - 2y - z - 5 = 0<br />
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