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Consider the flow described by the velocity field V = A(1 + Bt)i + Ctyj, with A = 1 m/s, B = 1 s^-1, and C = 1 s^-2. Coordinates are measured in meters. Plot the pathline traced out by the particle that passes through the point (1, 1) at time t = 0. Compare with the streamlines plotted through the same point at the instants t = 0. 1 and 2 s.

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MrRoyal

Answer:

x = (1 + t)lny/t

Explanation

Velocity Field

V = A(1 + Bt)i + Ctyj

Let u = A(1 + By)I

v = Cty

When t=0,x=1 and y=1.

u = A(1 + Bt)

u = dx/dt = A(1 + Bt)

Integrate both side with respect to t.

x = At + ½Bt² + c1

t=0,x=1 and c1 = 1.

So,

x = At + ½Bt² + 1

v = Ct

v = dy/dt = Ct

dy/y = integral of Ct

Integrate both sides with respect to t

Iny = ½Ct² + c2

t = 0, y = 1, c2 = 0

So,

lny = ½Ct² ----- make t the subject of formula

t = √(2lny/C)

x = At + ½Bt² + 1

So,

x = A√(2lny/C) + ½B(2lny/C) + 1

x = √2lny + lny + 1 --------- This is the path line

dx/u = dy/v --------- For the stream line

dx/(A(1 + Bt)) = dy/(C + y)

A = B = C = 1

So, we have

t/(1+t) integral of dx = integral of 1/y dy

tx/(1 + t) = lny + c3

At (1,1) t = 0

We have

0.1 = ln1 + c3

So, c3 = 0

For x

x = (1 + t)lny/t

.........

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