Question

1. a boy flies a kite which is 120 ft directly above his hand. if the wind carries the kite horizontally at the rate of 30 ft/min, at what rate is the string being pulled out when the length of the string is 150 ft?

Explanation:

Step1: Establish the relationship

Let \(y = 120\) (height of kite), \(x\) be the horizontal distance, and \(z\) be the length of the string. By Pythagorean theorem \(x^{2}+y^{2}=z^{2}\). Since \(y\) is constant (\(\frac{dy}{dt} = 0\)), differentiating with respect to \(t\) gives \(2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}\), which simplifies to \(x\frac{dx}{dt}=z\frac{dz}{dt}\).

Step2: Find \(x\)

Given \(z = 150\) and \(y = 120\), use \(x=\sqrt{z^{2}-y^{2}}\). So \(x=\sqrt{150^{2}-120^{2}}=\sqrt{(150 + 120)(150 - 120)}=\sqrt{270\times30}=\sqrt{8100}=90\).

Step3: Substitute values

We know \(\frac{dx}{dt}=30\) (horizontal speed), \(x = 90\), \(z = 150\). From \(x\frac{dx}{dt}=z\frac{dz}{dt}\), substitute: \(90\times30=150\times\frac{dz}{dt}\). Then \(\frac{dz}{dt}=\frac{90\times30}{150}\).

Answer:

\(18\) ft/min