Calculus question on rates

Question

A car is traveling East at 41 miles per hour and a second car is traveling North at 55 miles per hour. Both are headed for the intersection of the two roads. When the eastbound car is 8 miles from the intersection and the northbound car is 5 miles from the intersection, determine the following:

 

(A) How far apart are the cars?

(B) At what rate are the cars approaching each other?

Answer 1

(A) Let x is distance of 1st car from intersection, y is distance of second car from intersaction,

Then distance S(t) between cars is S(t) = sqrt(x² + y²). And at moment when x = 8 mi, y = 5 mi

this distance S = sqrt(5² + 8²) = sqrt(89) ≈ 9.43 mi

(B) To get rate we differentiate function S(t): S'(t) = (2x·x' + 2y·y')/2sqrt(x² + y²) = (x·x' + y·y')/sqrt(x² + y²).

At moment what we consider rate is equel S' = (8·41 + 5·55)'sqrt(8² + 5²) ≈ 75.26 mi/hr

Answer 2

Part 1

Use the Pythagorean theorem

Part 2

The distance between the cars at any time after time 0 is

sqrt[(8-41t)² + (5-55t)²] = D

The rate you want is dD/dt.