1) Let G(x,y) = x 3 + 2xy + 2x and:
x = 2t, y = t 2 Over: (0 < t < 1)
Find the work done in going from A= 0 to B = 1 along the curve.
Solution:
The integral for evaluation is:
W = ò0 1 (x 3 + 2x) dx + (2xy) dy
dx = 2 dt dy = 2t dt
W = ò0 1 2(8 t 3 + 4t) dt + 2 ( 4 t 3 )t dt
W = 4 t 4 + 4t 2 ] 0 1 + 2 t 4 ] 0 1
W = 8 + 2 = 10
2) Two lines have direction cosines:
(1/2, Ö 6/ 4, - Ö 6/ 4 ) and (Ö 6/ 4, 1/4, 3/4)
Show they are orthogonal.
For orthogonality:
cos Θ = 0 = l1 l2+ m1 m2 + n1 n 2
l1 l2+ m1 m2 + n1 n 2 =
(1/2)(Ö 6/ 4) + (Ö 6/ 4)(1/4) + (- Ö 6/ 4 )(3/4) = 0
3) Given the force (vector function):
F = i (2xy) + j (x 2 + 3y 2 ln z) + k (y 3/z)
Find the potential function V from which F was derived.
Soln.
- ¶ V / ¶x = 2xy
- ¶ V / ¶y = x 2 + 3y 2 ln z
- ¶ V / ¶z = y 3/z
dV = (¶ V / ¶x) dx + (¶ V / ¶y) dy + (¶ V / ¶z) dz
So: V = ò (- ¶ V / ¶x) dx + ò (- ¶ V / ¶y) dy + ò (- ¶ V / ¶z) dz
V = ò (- 2xy) dx + ò (- x 2 + 3y 2 ln z) dy + ò (- y 3/z) dz
V = - x 2 y - y 3 ln z - y 3 ln z = - x 2 y - 2y 3 ln z
4)Given a force: F = i (y + z) + j (z + x) + k (x + y)
Find the work done as the point of application moves from (0, 0, 0) to (1,1, 1):
(a) Along the straight line: x = y = z
Soln.
W = òC (y + z) dx + (z + x) dy + (x + y) dz
W = ò0 1 (2x) dx + (2y) dy + (2z) dz
W = 1 + 1 + 1 = 3
(b) Along the curve: x = t, y = t 2 and z = t4
Soln.
dx = dt, dy = 2t dt, dz = 4 t 3 dt
W = ò0 1 (t 2 + t 4 ) dt + 2 (t 4 + t ) t dt + 4 ( t + t 2 ) t 3 dt
W = 0.533 + 1 + 1.467 = 3
(c) Find the direction cosines of the line segment AB (e.g. from (0, 0, 0) to (1,1, 1) ) and the associated respective angles: a, b, g.
Soln.
l = cos a = x2 - x1 / ℓ = 1/ Ö 3
m = cos b = y2 - y1 / ℓ = 1/ Ö 3
u = cos g = z2 - z1 / ℓ = 1/ Ö 3
Thus:
cos a = cos b = cos g = 1/ Ö 3
a = b = g = arc cos (1/ Ö 3) = 0.955 rad = 54. 7 deg
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