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[NeonBlack]

This is a problem similar to one is Stewert's Early Trancendentals:

 

A boat is being pulled towards a dock by a rope. The rope is reeled in at a constant speed. What is the speed of the boat moving across the top of the water when it is a certain distance away from the dock? (x and y are perpendicular)

 

The solution is fairly straight-forward:

Use the relationship between the distances [math]r^2 = x^2 + y^2[/math]

Differentiate in time: [math]2r \frac{dr}{dt} = 2x\frac{dx}{dt}[/math]

Here, dr/dt is the speed of the rope (constant) and dx/dt is the speed of the boat. I will now call these vr and vb.

Finally: [math] v_b = v_r\frac{r}{x}[/math]

This is the solution given by the book and just about everyone else. But now let's take the angle between the water and the rope and call it theta.

Now: [math]v_b = \frac{1}{cos\theta}v_r[/math]

Here, the problem becomes obvious.

Since cosine < 1 the boat always moves faster than you pull it? Can anyone make sense of this?

[BobbyJoeCool]

Ok... how do you get from

 

[math]v_b = v_r \frac{r}{x}[/math]

 

to

 

[math]

v_b = \frac{1}{cos\theta}v_r

[/math]

 

?

[the tree]

Ok... how do you get from[math]v_b = v_r \frac{r}{x}[/math] to [math]

v_b = \frac{1}{cos\theta}v_r

[/math]?

Easily,

[math]\cos{\theta}=\frac{adjacent}{hypotenuse}=\frac{x}{r}[/math]

And so...

[math]\frac{1}{\cos{\theta}}=\frac{hypotenuse}{adjacent}=\frac{r}{x}[/math]

[D H]

It is a counter-intuitive result. Think about it this way. It takes smaller and smaller changes in the rope length to move the boat by the some fixed distance as [math]\theta\to\frac{\pi}2[/math]. The boat always moves faster than the rope.