Need help with a Calculus word problem?

You are on a camping trip and your camp site is 300ft from the river. Your car is parked 1200ft down river from your camp site and 600ft from the river bank (assume the river runs straight). Your campfire gets out of control and you need water, and you realize that you left your bucket in the car. Assume that you can run 9ft/s through the forest, but slow down to 2ft/s with a bucket full of water.


1. Find an equation for the distance traveled as you go from your car to the river and then to the campsite. Then describe the path that would minimize the distance.


2. Find an equation for the time elapsed as you go from your car to the river to the campsite. Then describe the path that would minimize the time elapsed.|||Draw a diagram. Start with a horizontal line for the river bank. Label 3 points on the line A, B, and C, from left to right. Mark AC as "1200".





Now draw a vertical line down from A to a point E. This is where the campsite is. Mark AE as "300". Next, draw a vertical line down from C to a point F that goes below past the level where E is. This point F will be the car. Mark CF as "600".





To get from the campsite to the car, you'd run EF, the shortest distance. But we don't need this to solve the problems. They're only asking for the time and distance from the car to the river, and from the river to the campsite.





Draw line segment FB. This is the distance you make from the car to the river, to fill up the bucket. Draw line segment BE. This is the path from the river to the campsite.





Notice that ABE and BCF are right triangles. Let AB = x. You should now be able to use the Pythagorean Theorem to get the length of BE in therms of x. Since BC = 1200-x, you should be able to get BF using the Pythagorean theorem on triangle BCF. Add the distances for BE and BF. Use a derivative to find the value of x that gives you the minimal sum.





To find the time elapsed, just use the fact that speed = distance/time, so time is distance over speed. Divide each of the two distances by the speed you're running. Use a derivative again to find the x that gives minimal speed|||You're welcome. Thanks for the points!

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