NeonBlack Posted November 10, 2006 Share Posted November 10, 2006 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? Link to comment Share on other sites More sharing options...

BobbyJoeCool Posted November 11, 2006 Share Posted November 11, 2006 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] ? Link to comment Share on other sites More sharing options...

the tree Posted November 11, 2006 Share Posted November 11, 2006 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] Link to comment Share on other sites More sharing options...

D H Posted November 11, 2006 Share Posted November 11, 2006 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. Link to comment Share on other sites More sharing options...

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