Question

A car starts off 17 miles directly south from the city of Morristown. It travels due west at a speed of 25 miles per hour. After traveling 31 miles, how fast in radians per hour is the angle opposite the westward path changing?

Answer 1

We can solve this problem using the Law of Cosines. Let A be the starting point of the car, B be the point where the car is after traveling 31 miles due west, and C be the city of Morristown. Then, we have:

AB = 31 miles

AC = 17 miles

BC = sqrt(AB^2 + AC^2 - 2*AB*AC*cos(90 degrees)) = sqrt(31^2 + 17^2) miles

Let theta be the angle between AB and AC. Then, we have:

cos(theta) = (AB^2 + AC^2 - BC^2) / (2*AB*AC)

sin(theta) = sqrt(1 - cos^2(theta))

Differentiating both sides with respect to time t, we get:

-sin(theta) d(theta)/dt = (AB d(AB)/dt + AC d(AC)/dt - BC d(BC)/dt) / (2*AB*AC)

d(AB)/dt = 0 (since the car is traveling due west)

d(AC)/dt = 0 (since the city of Morristown is stationary)

d(BC)/dt = -AB d(cos(theta))/dt = AB sin(theta) d(theta)/dt

Substituting these values, we get:

d(theta)/dt = -2*AB*AC / (BC^2 * sin(theta)) * d(BC)/dt

= -2*17*31 / (sqrt(31^2 + 17^2)^2 * sqrt(1 - cos^2(theta))) * (-31 sin(theta))

= 527 / (sqrt(850) * sqrt(1 - cos^2(theta))) * sin(theta)

Substituting cos(theta) = AC/BC = 17/sqrt(850), we get:

d(theta)/dt = 527 / sqrt(850 - 289) * 17 / sqrt(850) * sqrt(1 - 289/850)

= 527 / sqrt(561) * sqrt(561/850)

= 527 / sqrt(850)

= 17.98 radians per hour (approximately)

Therefore, the angle opposite the westward path is changing at a rate of about 17.98 radians per hour.