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1. 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 [tex]\frac{dy}{dx}[/tex] in terms of [tex]V_{W}[/tex], [tex]V_{B}[/tex], [tex]x[/tex], [tex]y[/tex] & [tex]W[/tex] only (no trigonometric functions nor [tex]\beta[/tex]).
2. Homework Equations
[tex]V_{W} =[/tex] speed of stream
[tex]V_{B} =[/tex] craft speed rel. static water
[tex]W =[/tex] stream width
At a point (x,y), the ruling equations for the watercraft are:
[tex]\frac{dy}{dt} = V_{W}  V_{B}sin\beta[/tex]
[tex]\frac{dx}{dt} = V_{B}cos\beta[/tex]
3. The Attempt at a Solution
[tex]\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \frac{V_{W}  V_{B}sin\beta}{V_{B}cos\beta}[/tex]
I think:
[tex]W = V_{B}cos\beta.t[/tex]
[tex]x = cos\beta[/tex]
[tex]y = sin\beta[/tex]
I know that I have to find an equivalent for [tex]cos\beta[/tex] & [tex]sin\beta[/tex] but I am perplexed at this moment...
 Thanks
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 [tex]\frac{dy}{dx}[/tex] in terms of [tex]V_{W}[/tex], [tex]V_{B}[/tex], [tex]x[/tex], [tex]y[/tex] & [tex]W[/tex] only (no trigonometric functions nor [tex]\beta[/tex]).
2. Homework Equations
[tex]V_{W} =[/tex] speed of stream
[tex]V_{B} =[/tex] craft speed rel. static water
[tex]W =[/tex] stream width
At a point (x,y), the ruling equations for the watercraft are:
[tex]\frac{dy}{dt} = V_{W}  V_{B}sin\beta[/tex]
[tex]\frac{dx}{dt} = V_{B}cos\beta[/tex]
3. The Attempt at a Solution
[tex]\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \frac{V_{W}  V_{B}sin\beta}{V_{B}cos\beta}[/tex]
I think:
[tex]W = V_{B}cos\beta.t[/tex]
[tex]x = cos\beta[/tex]
[tex]y = sin\beta[/tex]
I know that I have to find an equivalent for [tex]cos\beta[/tex] & [tex]sin\beta[/tex] but I am perplexed at this moment...
 Thanks
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