how to find the rate at which the distance between the two cars is changing. (Round your answer to one decimal place.)

Question

A car leaves an intersection traveling west. Its position 4 sec later is 19 ft from the intersection. At the same time, another car leaves the same intersection heading north so that its position 4 sec later is 29 ft from the intersection. If the speeds of the cars at that instant of time are 8 ft/sec and 14 ft/sec, respectively, find the rate at which the distance between the two cars is changing. (Round your answer to one decimal place.)

()ft/s

Answer 1

the distance between the cars is the hypotenuse of a right triangle with sides = 19 and 29

d^2 = h^2 + b^2 where h = the height of the triangle, and b= base of the triangle, b=19, h= 29

d = sqr(19^2 + 29^2) = sqr1202 = about 34.67

take the derivative with respect to time

2dd' = 2hh' + 2bb' where h' = 14 and b'= 8

divide by 2

dd' = hh' + bb' plug in the values

34.67d' = 29(14) + 19(8) = 406 + 152 = 558

d' = 558/34.67 = 16.095

the distance is changing at a rate of 16.1 feet per second

this type of problem is in textbooks under the top "derivatives" and subsection titled "related rates" Or look it up in the index.