Calculus
posted by Jo .
A balloon is rising vertically above a level, straight road at a constant rate of 1 foot/second. Just when the balloon is 65 feet above the ground, a bicycle passes under it going 17 feet/sec. How fast is the distance between the bicycle and balloon increasing 3 seconds later?

let the time passed since the biker passed directly underneath be t seconds
Make a diagram showing the distance between them as d ft
I see it as
d^2 = (17t)^2 + (65+t)^2
when t=3
d^2 = 51^2 + 68^2
d = √7225 = 85
2d dd/dt = 2(17)(17t) + 2(65+t)
divide by 2 and solve for dd/dt
dd/dt = (17(17t) + 65+t)/d
when t=3 and d = 85
dd/dt = (17(51) + 68)/85 = 11 ft/s
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