One car leaves a given point and travels north at 30 mph. Another car leaves the same point at the same time a?

One car leaves a given point and travels north at 30 mph. Another car leaves the same point at the same time and travels west at 40 mph. At what rate is the distance between the two cars changing at the instant when the cars have traveled 2 hours? b.) Suppose that in part a, the second car left one hour later than the first. At what rate is the distance changing at the instant when the second car has traveled 1hr.One car leaves a given point and travels north at 30 mph. Another car leaves the same point at the same time a?a)

y = distance traveled northerly by first car

x = distance traveled westerly by second car

z = distance between the two cars



z = (x^2 + y^2)^(1/2)



dz/dt = (1/2)(x^2 + y^2)^(-1/2)(2xdx/dt + 2ydy/dt)

dz/dt = (x^2 + y^2)^(-1/2)(xdx/dt + ydy/dt)



dy/dt = 30 mph

dx/dt = 40 mph

after two hours, y = 2*30 = 60 mi; x = 2*40 = 80 mi



dz/dt = (80^2 + 60^2)^(-1/2)(80*40 + 60*30)

dz/dt = 50 mph



b)

The first car has already traveled 1*30 = 30 miles. Therefore

z = (x^2 + (y + 30)^2)^(1/2)



dz/dt = (1/2)(x^2 + (y + 30)^2)^(-1/2)(2xdx/dt + 2(y + 30)dy/dt)

dz/dt = (x^2 + (y + 30)^2)^(-1/2)(xdx/dt + (y + 30)dy/dt)



after second car has traveled 1 hr, y = 1*30 = 30 mi; x = 1*40 = 40 mi



dz/dt = (40^2 + (30 + 30)^2)^(-1/2)(40*40 + (30 + 30)*30)

dz/dt = 47 mph