Question
A hot-air balloon is 120 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 45 mi hr (66 ft s). If the balloon rises vertically at a rate of 12 ft s, what is the rate of change of the distance between the motorcycle and the balloon 10 seconds later?
Answers
h = 120 + 12 t which is 240 at t = 10
dh/dt = 12
x = 66 * 10 = 660 at t = 10
dx/dt = 66
hypotenuse^2 = z^2 = h^2 + x^2
at t = 10 = 240^2 + 660^2 = 492200
so z= 702 at t = 10
2 z dz/ dt = 2 h dh/dt + 2 x dx/dt
2*702*dz/dt = 2 * 240 * 12 + 2 * 660 * 66
dz/dt = ( 2880 + 43560) / 702 = 66.2
dh/dt = 12
x = 66 * 10 = 660 at t = 10
dx/dt = 66
hypotenuse^2 = z^2 = h^2 + x^2
at t = 10 = 240^2 + 660^2 = 492200
so z= 702 at t = 10
2 z dz/ dt = 2 h dh/dt + 2 x dx/dt
2*702*dz/dt = 2 * 240 * 12 + 2 * 660 * 66
dz/dt = ( 2880 + 43560) / 702 = 66.2