How Can a Watercraft's Path Be Modeled Without Trigonometry in a Flowing Stream?

[Hendrick]

Homework Statement

A watercraft is crossing the stream to reach the pier. (See attached figure.)

Basically, I have to derive an ordinary differential equation of the path the watercraft travels, which I can then solve using MatLab, etc.
i.e. derive \frac{dy}{dx} in terms of V_{W}, V_{B}, x, y & W only (no trigonometric functions nor \beta).

Homework Equations

V_{W} = speed of stream
V_{B} = craft speed rel. static water
W = stream width

At a point (x,y), the ruling equations for the watercraft are:
\frac{dy}{dt} = V_{W} - V_{B}sin\beta

\frac{dx}{dt} = V_{B}cos\beta

The Attempt at a Solution

\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \frac{V_{W} - V_{B}sin\beta}{V_{B}cos\beta}

I think:
W = V_{B}cos\beta.t
x = cos\beta
y = sin\beta

I know that I have to find an equivalent for cos\beta & sin\beta but I am perplexed at this moment...

- Thanks

 

[kamerling]

you haven't told what the question was. Is the goal to arrive at the pier in the shortest time, or is the boat always pointing towards the pier?
If the boat is always pointing towards the pier, you can get sin(beta) and cos(beta) from x, and y. You have to chose an origin for the coordinates.

 

[Hendrick]

Sorry, yes the watercraft does always point towards the pier (throughout the journey).
Please see the attached figure for a more representative diagram.
What do you mean by choosing an origin?

Thanks

 

[kamerling]

You have to chose where the point with x=0, y=0 is. From what you've written so far this isn't clear.
What you've done so far is ok, but you still need sin(beta) and cos(beta)
I think drawing a triangle with the following sides will help:
1. a line from the boat to the pier
2. a line going through the pier in the direction of the y-axis
3. and a line going through the boat in the direction of the x-axis

 

[Hendrick]

Do you mean like this? (See attached figure.)
Wouldn't the initial position be (0,0)?

Thanks

 

[kamerling]

Do you mean like this? (See attached figure.)

just draw the boat somewhere in the middle of the stream at position (x,y)

Wouldn't the initial position be (0,0)?

It may be a bit easier to use the pier as (0,0). of course you only have to make the substitution u = y + w to get from one to the other.

 

[Hendrick]

How about this? (See attached figure)

Thanks

 

[Hendrick]

I think this diagram is more accurate than my previous one. (See attached figure)

 

[kamerling]

I think this diagram is more accurate than my previous one. (See attached figure)

finding cos(theta) and sin(theta) as a function of x and y should be easy from this rightanled triangle, with all the sides known.

 

[Hendrick]

Hi,

was the hypotenuse correct?
Because if I did it via Pythagoras, it yields an equation which I do not think is equal.
\sqrt{(W-x)^{2}+y^{2}}

 

[kamerling]

Hi,

was the hypotenuse correct?
Because if I did it via Pythagoras, it yields an equation which I do not think is equal.
\sqrt{(W-x)^{2}+y^{2}}

Yes. What do you mean with an equation that is not equal?