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Question

At 12:00PM a FedEx truck is 100 miles east of a UPS truck. The FedEx truck is driving west at a 55 mi/h and the UPS truck is driving north at 75 mi/h. How fast is the distance between these trucks changing at 2:00PM?

I have 87.46.

Answer 1

Position of U: v_ut in y direction and 0 in x, and position of F: (100-v_Ft) in x and 0 in y

Expression for d = ( (75t)² + (100 - 55t)²)^1/2

d(d)/dt = (1/2 / d ) (11250t + (100 - 55t)(-55)) t = 2 hrs and d = 150.33 miles.

d(d)/dt = 76.66 mi/hr

A nicer way is to calculate the results for 2 hours: we get the d above and the positions (-10,0) for F and (0, 150) for U.

d² = (x_u - x_F)² + (y_U - y_F)² implicitly differentiating to 2d d(d)/dt = 2x_F dx_F/dt + 2y_U dy_U/dt which leads to the same equation: d(d)/dt = (1/2d) *( 2(-10)(-55) + 2(150)(75)) which leads to the same expression

Hopefully, that clears things up.

Answer 2

I get about 80 but don't take that as gospel.

d^2 = y^2 + x^2

2dd' = 2yy' + 2xx'

dd' = yy' + xx'

(150.3) d' = 150(75) + (-10)(55)

d' = about 79?