7. A boat is being pulled toward a dock by a rope attached to its bow through a pirlley on the dock.The pulley is higher than the boat's bow by several feet. Relate the rate that the rope is hauled in with how fast the boat is approaching the dock.

## Step 1<br />This problem involves the concept of related rates in calculus. The situation described in the problem involves a right triangle, where the hypotenuse is the rope, the height difference between the pulley and the boat is one side of the triangle, and the distance between the boat and the dock is the other side.<br /><br />## Step 2<br />We denote the height difference between the pulley and the boat as \(h\), the distance between the boat and the dock as \(x\), and the length of the rope as \(r\). According to the Pythagorean theorem, we have the equation:<br />### \(r^2 = h^2 + x^2\)<br /><br />## Step 3<br />We are interested in the rate at which the boat is approaching the dock, which is the derivative of \(x\) with respect to time, denoted as \(\frac{dx}{dt}\). We can find this by differentiating the equation from Step 2 with respect to time.<br /><br />## Step 4<br />Differentiating the equation from Step 2 with respect to time gives us:<br />### \(2r \frac{dr}{dt} = 2x \frac{dx}{dt}\)<br /><br />## Step 5<br />We can simplify this equation by dividing both sides by 2, giving us:<br />### \(r \frac{dr}{dt} = x \frac{dx}{dt}\)<br /><br />## Step 6<br />We know that the rate at which the rope is hauled in, \(\frac{dr}{dt}\), is negative because the length of the rope is decreasing. We can substitute this into the equation from Step 5 to solve for \(\frac{dx}{dt}\), the rate at which the boat is approaching the dock.<br /><br />## Step 7<br />Substituting \(\frac{dr}{dt}\) into the equation from Step 5 gives us:<br />### \(\frac{dx}{dt} = -\frac{r}{x} \frac{dr}{dt}\)