Please help me with rate questions! its really confusing! A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the front of the boat, which is 9 feet below the level of the pulley. If the rope is pulled through the pulley at a rate of 14 ft/min, at what rate will the boat be approaching the dock when 110 ft of rope is out? The boat will be approaching the dock at ft/min.

|dw:1363095567208:dw|

|dw:1363150094096:dw| So in the picture y represents 9 feet. 110 represents the hypotenuse of a triangle, which is the length of the rope. x represents the distance of the boat away from the pier, and you can find that with the pythagorean theorem. To find the speeds though, you have to realize that it is a right triangle that's changing with respect to time, so you must differentiate the triangle's formula with respect to time. \[x^2+y^2=L^2\] This represents our triangle right? Now lets take the derivative with respect to time. \[2x \frac{ dx }{ dt }+2y\frac{ dy }{ dt }=2L\frac{ dL }{ dt }\] Since none of the variables are t, we are doing implicit differentiation with respect to all the variables. Now what are we looking for? dx/dt since that is the speed of the boat. What about the other two speeds? Well, we know dL/dt is the speed the rope is being pulled in at, 14 ft/min. Now what about the other speed, dy/dt? Think about this a second. What does dy/dt mean? It's the speed that the height of the pulley is changing. It's bolted down to the dock, so its speed is nothing, zip, zero! So lets simplify the equation a bit that we just made: \[ \frac{ dx }{ dt }=\frac{ L }{ x }\frac{ dL }{ dt } \] So it was asking us, what is the speed of the boat when the length of the rope was 110 feet long? You now have an equation for it. You might be thinking, wait, what's x, the distance of the boat from the pier? I already explained it above at the beginning.

Join our real-time social learning platform and learn together with your friends!