Answer:
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.
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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)
