Related Rates – Pulley Rope

Hello, everystepcalculus.com, a problem in calculus with related rates: a boat being pulled up to a dock, where the dock is higher than the boat problem. Let’s do it, you can see the problem on your screen. Index 8 to get to the menu, we gotta go up to… gonna go up on the cursor so we can go to the bottom of the alphabet to get to the R’s quicker. This is calculus 1 but also it’s in calculus one, two, and three. Related rates and we’re choosing the “Boat/Dock pull”, number 2, and we’re gonna enter our parameters. We have to press alpha before we enter anything into these entry lines here. Alpha, the height is given as alpha 8 meters, the ropes rate of change is equal to alpha 2 meters per second, it’s decreasing because it’s going closer to the Y axes. It’s pulled in, enter distance from the dock is alpha 9 meters. I always show you what you’ve entered, you can change it if you want. I say it’s okay, and I give you the perimeters, now what we’re talking about here, x equals distance from the boat from the dock and y equals the height of the dock above the boat, and L equals the length of the rope, which is the hypotenuse. And of course dL/dt is the rate of change of minus two meters per second. Given. So we find the length of the rope, which is equal to the square root of 145. Pythagorean theorem. And then we formulate the same equation basically but we work out a little bit different because we’re gonna do the derivatives of everything here. So um we do the derivatives you read all this on your paper, you do all the derivatives of the whole function before, just like this, and that’s the answer. Really the answer is square root of 145 times a minus 2 divided by 9, which is dx/dt which is the change of rate of the x axis as the boat is being pulled up the dock. And the approximate answer is -2.6759 meters per second. Have a good one.