Question

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.

Answer 1

a) jyt + jt + 2xi = 4i

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

Step-by-step 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

Answer 2

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

Unit Vector n = (4i+4j)/4√2

Step-by-step 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