NATALIA ROWCZENIO
1.
i)
Let F = F1 I + F2 j + f3 k
be a vector field that div F /= 0
If there exists a vector field G
F = del G
Div F = div ( del G ) = 0
However div F /= 0
Therefore there is a contradiction
2)
Del F = x^2 y + 3ycos(xz)
delta F1 / delta x = x^2 y
delta F2 / delta y = 3ycos(xz)
delta F3 / delta z =0
therefore
F1 = integral ( x ^ 2 y ) delta x
= y integral ( x ^ 2 ) delta x
= ( y x ^ 3 )/3
F2 = integral( 3 y cos (x z) ) dy
= 3 cos (x z) integral (y) dx
= 3 cos (x z) (y ^ 2) / 2
F3 = integral (0) = C
F = < ( ( x ^ 3) y) / 3 , ( 3 / 2) (y ^ 2) cos( x z ), C >
4.
F = del f
Therefore
(delta f) /(delta x) = 1/x
(delta f)/(delta y) = 1 / y
(delta f) / (delta z) = 2z
Therefore f(x, y, z) = f(x, y, z) = f n(x) + fn (y) + z^2
5.
d^2 f / d t^2 + t 2 (d f/ d t) + (10 + s2 + s3 + s4) f = (t – 2) (t – 5)
L{f}=F
L{ d^2 f / d t^2 / d t ^2} = s^2 F(s) – s f0 – f ‘(0)
L { df / dt} = sf(s) – f
Therefore
Integral ( x delta ( t – t1) = x(t)
S^2F – s f(0) – f’(0) – 2sF + 2f(0) + 25 F = 3