Question

Calculus Related Rates Question. Help!?

You are in a sea kayak 400 m offshore from point A. You are along a straight sandy beach. A storm is brewing and you wish to go to a shelter 1000m down the beach from A. If you can paddle your kayak at 2 km/h and walk at 3 km/h, how far down the shore should you beach your kayak to reach shelter in the shortest amount of time? Give your answer to the nearest meter.

Answer 1

Please see below.

Explanation:

Note: This is not a related rates problem. It is an optimization problem.

I hope you can sketch a picture of the situation.

Let `x` = the distance between `A` and the landing point.

The distance traveled by kayak is the hypotenuse of a right triangle, so it is `sqrt(x^2+400^2)`

The Time it takes is `"Distance"/"Rate"` which is `sqrt(x^2+400^2)/2`

(Actually it is `sqrt(x^2+400^2)/2000`, but this will work. What is important is the ratio `2:3` which is rate in kayak:rate on beach)

The distance traveled by walking waht is left os the 1000 m total, so it is `1000-x`

The Time it takes is `"Distance"/"Rate"` which is `(1000-x)/3` .

The total time is the sum of the two times.

`T = sqrt(x^2+400^2)/2 + (1000-x)/3`

Domain `[0,1000]`

We want to find `x` to minimize `T` so differentiate, the find and test critical numbers.

`T' = x/(2sqrt(x^2+400^2)) - 1/3`

`T' = 0` at `x = 800/sqrt5`

Test to make sure this is a minimum.

I would use the first derivative test with test points `1` and `900`.

`T` is decreasing left of `x = 800/sqrt5` and increasing on the right.

So there is a minimum at `x = 800/sqrt5`.