Aballoon is rising at a constant speed of 5 ftys. a boy is cycling along a straight road at a speed of 15 ftys. when he passes under the balloon, it is 45 ft above him. how fast is the distance between the boy and the balloon increasing 3 s later?

  13 ft/s

Step-by-step explanation:

t seconds after the boy passes under the balloon the distance between them is ...

  d = √((15t)² +(45+5t)²) = √(250t² +450t +2025)

The rate of change of d with respect to t is ...

  dd/dt = (500t +450)/(2√(250t² +450t +2025)) = (50t +45)/√(10t² +18t +81)

At t=3, this derivative evaluates to ...

  dd/dt = (50·3 +45)/√(90+54+81) = 195/15 = 13

The distance between the boy and the balloon is increasing at the rate of 13 ft per second.

The boy is moving horizontally at 15 ft/s, so his position relative to the spot under the balloon is 15t feet after t seconds.

The balloon starts at 45 feet above the boy and is moving upward at 5 ft/s, so its vertical distance from the spot under the balloon is 45+5t feet after t seconds.

The straight-line distance between the boy and the balloon is found as the hypotenuse of a right triangle with legs 15t and (45+5t). Using the Pythagorean theorem, that distance is ...

  d = √((15t)² + (45+5t)²)

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