I need help answering this question, Example 3 is shown below the original problem for reference. Thank you!

EXAMPLE 3 An airplane flying north at 640 miles per hour passes over a certain town at noon. A second airplane going east at 600 miles per hour is direct- ly over the same town 15 minutes later. If the airplanes are flying at the same alti- tude, how fast will they be separating at 1:15 P.M.? SOLUTION Step 1: Let / denote the number of hours after 12:15 P.M., y the distance in miles flown by the northbound airplane after 12:15 P.M., x the distance flown by the east- bound airplane after 12:15 P.M., and s the distance between the airplanes. In the 15 y . 160 minutes from noon to 12:15 P.M. the northbound airplane will have flown + = 160 miles, so the distance from the town to the northbound airplane at time / will be y + 160. (See Figure 4.) Step 2: We are given that, for all : > 0, dy/dt = 640 and dx/dt = 600. We want to know ds/dr at / = 1, that is, at 1:15 P.M. Step 3: By the Pythagorean Theorem, Figure 4 s' = x' + (y + 160)3 Step 4: Differentiating implicitly with respect to f and using the Chain Rule, we have 25- as = 2x dx + 2(y + 160) or ds dx + (y + 160) d dy dt Step 5: For all t > 0, dx/dt = 600 and dy/dr = 640, while at the particular instant ( = 1, x = 600, y = 640, and s = V(600) + (640 + 160) = 1000. When we substitute these data in the equation of Step 4, we obtain 25 = (600) (600) + (640 + 160)(640) from which = 872 At 1:15 PM., the airplanes are separating at 872 miles per hour. Now let's see if our answer makes sense. Look at Figure 4 again. Clearly, s is increasing faster than either x or y is increasing, so ds/dr exceeds 640. On the other hand, s is surely increasing more slowly than the sum of x and y; that is, ds/di