Question

Girl flies a kite at a height of 300 ft, and the wind carries the kite horizontally away from her at arate of 25 ft/sec. how fast must she let out the string when the kite is 500 ft away from her?

Answer 1

20 ft/sec fast must she let out the string when the kite is 500 ft away from her.

Given

Let

x = distance of girl from the point on the ground directly below the kite at time t.

y = length of string at time t.

How to find the feet?

At time t,

They have a right triangle with horizontal leg of length x.

Vertical leg of length 300, and hypotenuse of length y.

The kite horizontally away from her dx/dt = 25

Find dy/dt when y = 500

By the Pythagorean Theorem,

Since,

x² + 300² = y² and y = 500, x = 400

x² + 300² = y²

2x (dx/dt) = 2y (dy/dt)

2 (400) (25) = 2 (500) (dy/dt)

20,000 dy/dt = 1000 ft/sec

dy/dt = (20,000/1000) ft/sec

dy/dt = 20 ft/sec

So, The fast must she let out the string when the kite is 20 ft/sec.

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https://brainly.com/question/13910918

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Answer 2

20 ft/sec fast must she let out the string when the kite is 500 ft away from her.

Given

Let

x = distance of girl from the point on the ground directly below the kite at time t.

y = length of string at time t.

How to find the feet?

At time t,

They have a right triangle with horizontal leg of length x.

Vertical leg of length 300, and hypotenuse of length y.

The kite horizontally away from her dx/dt = 25

Find dy/dt when y = 500

By the Pythagorean Theorem,

Since,

x² + 300² = y² and y = 500, x = 400

x² + 300² = y²

2x (dx/dt) = 2y (dy/dt)

2 (400) (25) = 2 (500) (dy/dt)

20,000 dy/dt = 1000 ft/sec

dy/dt = (20,000/1000) ft/sec

dy/dt = 20 ft/sec

So, The fast must she let out the string when the kite is 20 ft/sec.

Learn more about feet problem here:

brainly.com/question/13910918

#SPJ4