I was trying to explain to someone how I got a solution to OpenStax Calculus Vol 1 section 4.1 (related rates) exercise 43, so first I did a google search to see if there was a posted answer. I found two answers posted, and I disagreed with both of them, and they also disagreed with the OpenStax solutions (which only had a numerical answer and not an explanation).
So I worked it again, and my work agrees numerically with the OpenStax answer and I'm satisfied. If anyone else is searching for the supporting work, this post is an attempt to document that.
First, the problem: A batter hits a ball toward second base at 80 ft/sec and runs toward first base at a rate of 30 ft/sec. At what rate does the distance between the ball and the batter change when the runner has covered one-third of the distance to first base? (Hint: Recall the law of cosines.)
Whatever triangle we construct to visualize this set up has sides along the line from home to 1st base, and from home to 2nd base. The third side of the triangle is from the runner's position to the ball's position. Since a baseball "diamond" is actually a square, we can say the angle between the 1st baseline and the line passing through home and 2nd base is 45 degrees (or Pi/4).
If r is the runner's distance from home and b is the ball's distance from home then Law of Cosines tells us the relationship to d, the distance between runner and ball, is d2=r2+b2-2r*b*Cos(π/4). At the instant the runner is 1/3 the distance to 1st base r=30 (1/3 of 90 feet). This implies time t=1. This tells us b=80. We can calculate d=62.497
If we differentiate that equation relating d, r, and b (with respect to time, as r, b, and d are functions of time) we get d*d' = r*r' + b*b' - Cos(π/4)*(r*b' +b*r'). The problem statement tells us r'=30 and b'=80. Substituting in known values and solving for d' we get d'=62.497.
The solutions manual gives 62.50, which is the same answer to two decimal places precision.
