A child flies a kite at a height of 90 ft , the wind carrying the kite horizontally away from the child at a rate of .
How fast must the child let out the string when the kite is 150 ft away from the child?
The child must let out the string at a rate of when the kite is 150 ft away from the child.
(Simplify your answer.)

Ask by Cervantes Fitzgerald. in the United States

Mar 14,2025

Question

A child flies a kite at a height of 90 ft , the wind carrying the kite horizontally away from the child at a rate of .
How fast must the child let out the string when the kite is 150 ft away from the child?
The child must let out the string at a rate of when the kite is 150 ft away from the child.
(Simplify your answer.)

Ask by Cervantes Fitzgerald. in the United States

Mar 14,2025

UpStudy AI · Accepted Answer

Let $$ x $$ be the horizontal distance between the child and the kite, $$ y = 90 $$ ft be the constant height of the kite, and $$ s $$ be the length of the string. By the Pythagorean theorem, we have

$$
s^2 = x^2 + y^2.
$$

Substituting $$ y = 90 $$ ft,

$$
s^2 = x^2 + 90^2.
$$

Differentiate both sides with respect to time $$ t $$:

$$
2s \frac{ds}{dt} = 2x \frac{dx}{dt}.
$$

Simplify by dividing both sides by $$ 2 $$:

$$
s \frac{ds}{dt} = x \frac{dx}{dt}.
$$

We are given that $$ \frac{dx}{dt} = 30 $$ ft/sec. To find $$ \frac{ds}{dt} $$ when $$ s = 150 $$ ft, first find $$ x $$ when $$ s = 150 $$:

$$
150^2 = x^2 + 90^2.
$$

Calculate:

$$
22500 = x^2 + 8100 \quad \Rightarrow \quad x^2 = 22500 - 8100 = 14400.
$$

Thus,

$$
x = \sqrt{14400} = 120 \text{ ft}.
$$

Now substitute $$ x = 120 $$, $$ s = 150 $$, and $$ \frac{dx}{dt} = 30 $$ ft/sec into the differentiated equation:

$$
150 \frac{ds}{dt} = 120 \times 30.
$$

Solve for $$ \frac{ds}{dt} $$:

$$
\frac{ds}{dt} = \frac{120 \times 30}{150} = \frac{3600}{150} = 24 \text{ ft/sec}.
$$

Thus, the child must let out the string at a rate of $$ \boxed{24 \text{ ft/sec}} $$.
The child must let out the string at a rate of $$ 24 \text{ ft/sec} $$ when the kite is 150 ft away.

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