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Question
Hello I am having serious with related rates.

Spike the dog fetches a ball thrown into the lake. He is 20 meters south of the ball and 25 meters west of the ball. He can run 5 meters per second and he can swim 3 meters per second. Spike runs along the bank and jumps into the water at the exact point that minimizes his time to get to the ball. How far from his original position did spike run before he jumped into the water?

Answer
It is unclear which way the coast is running.  Is the shorline running north-south?
That means there is 20 m of beach and the ball is 25 m out to sea.
Also, I will assume Spike is on the shoreline.

Let A be where spike is on the shore.
Let B the point where he jumps in the water.
Let C be the point on the shore where the ball is straight out.

Let the distance from B to C be x.
This makes the distance from A to B be 20-x.
The time is the distance/speed, and that is (20-x)/5.

His time on shore is then (20-x)/5.
The distance in the water is the hypotenuse of the triangle with sides x and 25.
This makes it sqrt(x^2 + 25^2).  Since 25^2 = 625, that is sqrt(x^2 + 625).
The time is sqrt(x^2 + 625)/3.

This makes the total time be t(x) = (20-x)/5 + sqrt(x^2 + 625)/3.
Find t'(x), set it to 0, solve for t, and check to make sure it is the minimum.
Actually, if he swam all the way it would take longer and if he ran down to where he swam straight out it would take longer, so this will be the minimum time.

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