Calculus: Early Transcendentals, Chapter 3, 3.9, Section 3.9, Problem 18

In both cases we must find the rate of change of the hypotenuse of a right triangle. In part (a), the triangle is formed between the first base, second base and the runner; in part (b) the vertices are third base, home plate and the batter.
Let's name the following parameters:
s = distance between home plate and first base.
We call b = 90 ft, the other three sides of the square.
ds/dt = v = 24 ft/s = speed when the batter is moved from the home
h = distance from second base (part a) or third base (part b) to the batter at any time. It is the hypotenuse of our triangles.
According to the theorem of Pythagoras, the hypotenuse h is:
h2 = s2 + b2
h = (s2 + b2)1/2
a)
To find the rate of decrease of h we derive this expression with respect to time:
-(dh/dt) = 1/2(s2 + b2)-1/22s(ds/dt); the minus sign expresses the decrease of h when t increases.
dh/dt = -[1/2(s2 + b2)-1/22s(v)] = -[ sv/(s2 + b2)1/2]
Now we substitute the given values:
dh/dt = -[(s/2)v/((s/2)2 + b2)1/2]= -[(45)(24)/√(2025 + 8100)]
dh/dt = -10.73 ft/s
b)
The procedure for finding the increase of h from third base is the same, but considering the positive sign for the derivative dh/dt. So in this case we have:
dh/dt = 10.73 ft/s