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irakobra [83]
3 years ago
9

Andrew is flying a kite at a height of 120ft. While the wind is carrying the kite horizontally away from her at the rate of 24ft

/sec. how fast is the kite string being let out when there is 200ft of line out?

Mathematics
1 answer:
Paha777 [63]3 years ago
3 0

Answer:

The string of the kite is being let out at the rate of 19.2 ft/sec.

Step-by-step explanation:

See the diagram attached.

The vertical height (BK) of the kite is 120 ft.

The kite string is of length (AK) of 200 ft.

Therefore, the horizontal distance (AB) from Andrew to kite is = \sqrt{200^{2} - 120^{2}}  = 160 ft.  

Now, applying the Pythagoras Theorem,

AB² + BK² = AK²

⇒ l² + h² = s² {Where, l is the horizontal length, h is the height and s is the string length}

Differentiating with respect to time t both sides

2l\frac{dl}{dt} + 0 = 2s\frac{ds}{dt} {Since, height of kite is constant}

⇒ 2 \times 160 \times 24 = 2 \times 200 \times \frac{ds}{dt}

⇒\frac{ds}{dt} = 19.2 ft per sec.

Therefore, when the wind is carrying the kite horizontally at the rate of 24 ft/sec, then the string of the kite is being let out at the rate of 19.2 ft/sec. (Answer)

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Step-by-step explanation:

The Perimeter of the full shape is the sum of the lengths of the edges of the parts.  For convenience in referencing them, we'll call the large curve "curve_{big}" and the three smaller curves "curve_1" "curve_2" "curve_3" in order from left to right.

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Substituting back into the main equation above:

P_{overall} = Length(curve_{big})+Length(curve_1)+Length(curve_2)+Length(curve_3)P_{overall}=\frac{1}{2} \pi d_{big} + \frac{1}{2} \pi d_{1} + \frac{1}{2} \pi d_{2} + \frac{1}{2} \pi d_{3}

Note that all terms have common factors of "one-half" and "pi" in them.  These can be factored out:

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Substituting and simplifying (in terms of pi):

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