The velocity field of a flow is given by where x and y are in feet. Determine the fluid speed at points along the x axis; along the y axis. What is the angle between the velocity vector and the x axis at points 15, 52, and 10, 52

There is a part of the question missing and it says;

The velocity field of a flow is given by V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s, where x and y are in feet.

A) At (1,5),angle is -11.31°

B) At (5,2),angle is -68.2°

C) At (1,0), angle is -90°

Explanation:

From the question ;

V = [20y/(x² + y²)^(1/2)]î − [20x/(x² + y²)^(1/2)] ĵ ft/s

Let us assume that u and v are the flow velocities in x and y directions respectively.

Thus we have the expression;

u = [20y/(x² + y²)^(1/2)]

and v = - [20x/(x² + y²)^(1/2)]

Thus, V = √(u² + v²)

V = √[20y/(x² + y²)^(1/2)]² + [-20x/(x² + y²)^(1/2)]²

V = √[400y²/(x²+y²)] +[400x²/(x²+y²)

V = √(400x² + 400y²)/(x²+y²)

Now for the angle;

tan θ = opposite/adjacent

And thus, in this question ;

tan θ = v/u

tan θ = [-20x/(x² + y²)^(1/2)]/ [20y/(x² + y²)^(1/2)]

Simplifying this, we have;

tan θ = - 20x/20y = - x/y

so the angle is ;

θ = tan^(-1)(-x/y)

So let's now find the angle at the various coordinates.

At, 1,5

θ = tan^(-1)(-1/5) = tan^(-1)(-0.2)

θ = -11.31°

At, 5,2;

θ = tan^(-1)(-5/2) = tan^(-1)(-2.5)

θ = -68.2°

At, 1,0;

θ = tan^(-1)(-1/0) = tan^(-1)(-∞)

θ = -90°

At,

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