# Related Rates.

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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 $r^2 = x^2 + y^2$

Differentiate in time: $2r \frac{dr}{dt} = 2x\frac{dx}{dt}$

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: $v_b = v_r\frac{r}{x}$

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

Here, the problem becomes obvious.

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

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Ok... how do you get from

$v_b = v_r \frac{r}{x}$

to

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

?

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Ok... how do you get from$v_b = v_r \frac{r}{x}$ to $v_b = \frac{1}{cos\theta}v_r$?

Easily,

$\cos{\theta}=\frac{adjacent}{hypotenuse}=\frac{x}{r}$

And so...

$\frac{1}{\cos{\theta}}=\frac{hypotenuse}{adjacent}=\frac{r}{x}$

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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 $\theta\to\frac{\pi}2$. The boat always moves faster than the rope.

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