# The path of crossing a stream

1. Apr 15, 2008

### Hendrick

1. The problem statement, all variables and given/known data
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 $$\beta$$).

2. Relevant 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:
$$\frac{dy}{dt} = V_{W} - V_{B}sin\beta$$

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

3. 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

#### Attached Files:

• ###### River crossing.bmp
File size:
277.1 KB
Views:
62
2. Apr 15, 2008

### 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.

3. Apr 15, 2008

### 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

#### Attached Files:

• ###### River crossing.bmp
File size:
277.1 KB
Views:
48
Last edited: Apr 15, 2008
4. Apr 15, 2008

### 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

5. Apr 15, 2008

### Hendrick

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

Thanks

#### Attached Files:

• ###### River crossing triangle.bmp
File size:
277.1 KB
Views:
49
6. Apr 15, 2008

### kamerling

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

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.

7. Apr 16, 2008

### Hendrick

How about this? (See attached figure)

Thanks

#### Attached Files:

• ###### River crossing triangle2.bmp
File size:
294 KB
Views:
49
8. Apr 16, 2008

### Hendrick

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

#### Attached Files:

• ###### River crossing triangle3.bmp
File size:
294 KB
Views:
44
9. Apr 16, 2008

### kamerling

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.

10. Apr 16, 2008

### Hendrick

Hi,

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

11. Apr 16, 2008

### kamerling

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

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