A velocity field is given by V = (2x) i + (yt) j m/s, where x and y are in meters and t is in seconds. Find the equation of the streamline passing through (2,-1) and a unit vector normal to the streamline at (2,-1) at t=4s.

a) jyt + jt + 2xi = 4i

b) 2iy + 2i + 8j = 4xj

Explanation:

V = (2x) i + (yt) j

By implicit differentiation:

With V = 0: 2i + dyjt/dy = 0

- dyj/dx = 2i

∴ dy/dx = - 2i/jt

The equation of the streamline passing through (2,-1), using, y - y1 = m (x - x1), where m = dy/dx

y + 1 = - 2i/jt (x - 2)

jt (y + 1) = - 2xi + 4i

jyt + jt + 2xi = 4i

b) Using y - y1 = - 1/m (x - x1), where at unit normal, dy/dx = - 1/m

y + 1 = 4j/2i (x - 2)

2i (y + 1) = 4j (x - 2)

2iy + 2i + 8j = 4xj

Equation of the streamline V = 4i - 4j m/s

Unit Vector n = (4i+4j) / 4√2

Explanation:

Parameters

x=2 and y=-1

V = (2x) i + (yt) j m/s

Substituting (x=2) and (y=-1) into the velocity field V

Therefore V = 4i - tj where t=4s

Equation of the streamline

V = 4i - 4j m/s

Unit vector normal to the streamline n

Note V. n=0 and the velocity only have x - and y - components

Therefore

V. n = (4i - 4j). (nₓi+nyj) = 0 or 4nₓ - 4ny = 0

The unit vector requires that nₓ^2+ny^2=1

Therefore n = (4i+4j) / 4√2