Question

Flying a Kite Inge flies a kite at a height of 300 ft, the wind carrying the kite horizontally away at a rate of 25 \(\mathrm{ft} / \mathrm{sec} .\) How fast must she let out the string when the kite is 500 \(\mathrm{ft}\) away from her?

Short answer

Inge must let out the string at a rate of 20 ft/sec when the kite is 500 ft away from her.

Step-by-step solution

Represent the problem as a triangle

The person, the kite, and the point directly above the person on the height of the kite can be thought of as forming a right triangle. The height from the ground to the kite is one leg of the triangle (300 ft), the horizontal distance of the person from the kite is the other leg of the triangle (let it be \(x\)), and the length of the string is the hypotenuse of the triangle (let it be \(y\)). According to Pythagorean Theorem, \(y^2 = x^2 + 300^2\).

Differentiate the equation

Differentiate both sides of the equation with respect to time \(t\), given \(dx/dt = 25 ft/sec\) (rate of change of horizontal distance). The derivative of \(y^2\) with respect to \(t\) is \(2y*(dy/dt)\). The derivative of \(x^2\) with respect to \(t\) is \(2x*(dx/dt)\}. The derivative of \(300^2\) with respect to \(t\) is 0 (since it's constant). Hence, the differentiated equation is \(2y*(dy/dt) = 2x*(dx/dt)\).

Solve for dy/dt

Solve the equation obtained in step 2 for \(dy/dt\), which represents how fast she must let out the string. Substituting \(x = 400 ft\) and \(y = 500 ft\) (from the Pythagorean theorem) and \(dx/dt = 25 ft/sec\), we get \(dy/dt = x*(dx/dt)/y = 400*25/500 = 20 ft/sec\).