kharmaculpepper
17.12.2019 •
Mathematics
Aballoon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. just when the balloon is 65 feet above the ground, a bicycle passes under it, going 17 ft/sec. how fast is the distance between and bicycle and the balloon changing 3 seconds later?
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Ответ:
DL/dt = 11 feet/sec
Step-by-step explanation:
We have a right triangle between balloon ( h = 65 feet) from the point straight down the ballon, and the bicycle position. Hypothenuse is L the distance between balloon and bicycle
Let x be distance between (vertical from the balloon to ground) then
according to Pythagoras Theorem
L² = x² + h² (1)
and we know
Dx/dt = 17 ft/sec
Dh/dt = 1 ft/sec
We need to find DL/dt
From equation (1)
L² = x² + h²
Differentiating on both sides of the equation
2L DL/dt = 2xDx/dt + 2yDy/dt
L*DL/dt = x*Dx/dt + y*Dy/dt
DL/dt = [ x*Dx/dt + y*Dy/dt ] /L (2)
Now 3 sec after the bicycle passed under the balloon
x = 3 * Dx/dt = 3 * 17 = 51 feet
and h = 65 + 1* 3 = 68 feet
then L = √ √(68)² + (51)² L = √ 4624 + 2601 L = 85 feet
Now plugging in these values in equation (2)
DL/dt = [ 51*17 + 68* 1] / 85
DL/dt = 11 feet/sec
Ответ: