Differential Equations: Finding dy/dx from dx/dt and dy/dt

  • Thread starter djh101
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  • #1
djh101
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Homework Statement


A goose starts in flight a miles due east of its nest. Assume that the goose maintains constant flight speed (relative to the air) so that it is always flying directly toward its nest. The wind is blowing due north at w miles per hour. Figure 8 shows a coordinate frame with the nest at (0,0) and the goose at (x,y). It is easily seen that

[itex]\frac{dx}{dt}[/itex] = -v0cosθ
[itex]\frac{dy}{dt}[/itex] = w - v0sinθ


Show that

[itex]\frac{dy}{dx}[/itex] = [itex]\frac{y - k\sqrt{x^{2} + y^{2}}}{x}[/itex]

where k = w/v0, the ratio of the wind speed to the speed of the goose.

Homework Equations


See Above


The Attempt at a Solution


I don't see how the above can be the solution. x' and y' are constant so dx/dy should just be y/x, shouldn't it?
 

Answers and Replies

  • #2
clamtrox
938
9
I don't see how the above can be the solution. x' and y' are constant so dx/dy should just be y/x, shouldn't it?
What makes you say x' and y' are constants? They both depend on θ.

Use the chain rule to get an expression for dy/dx (this is trivial), then use x=rcosθ, y=rsinθ.
 
  • #3
djh101
160
5
Sorry, I meant their rates of change are constant. Since the goose is always flying directly towards its nest, θ shouldn't change.
 
  • #4
clamtrox
938
9
Since the goose is always flying directly towards its nest, θ shouldn't change.

This would be true is w=0.

Perhaps better way to say it would be that the goose is always trying to fly towards its nest.
 
  • #5
djh101
160
5
I figured out, the wording was just a little confusing. I guess what they meant is that it was always going towards the nest (i.e. forward, not backward) and wen't straight (i.e. no turns on the z-axis), not necessarily in a straight line, though. Thanks for your help, I figured it out after plugging in for sin and cos.
 

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