Related Rates Kite Problem

Question

A student flies a kite at exactly 20 meters altitude. The kite begins to drift away at 2 meters per second. The kite continues to fly at altitude 20m. How fast is the distance from the student to the kite changing after 3 seconds?

Answer 1

Y=20 <--- that's it

dX/dt = 2

X = 2t + c

Assuming x(0) = 0, then c=0, so X=2t

The distance as a function of time is d(t) = sqrt((2t)^2 + 400) = sqrt( 4t^2 + 400)

dD/dt = (1/2)(8t)/ sqrt(4t^2 + 400) =

(4t)/sqrt(4t^2 + 400) =

(4t)/ 2 * sqrt(t^2 + 100) =

(2t)/ sqrt(t^2 + 100)

After 3 seconds,

6/sqrt(109)

6 * sqrt(109)/ 109

=0.574695771132690835583-

Answer 2

Draw a right triangle with a vertical side of length 20, a horizontal side labelled x and hypotenuse labelled y

x = distance the kite drifts horizontally t seconds after being directly above the student

y = distance between the student and the kite at time t

Given: dx/dt = 2

Find: dy/dt when t = 3

From the Pythagorean Theorem, x² + 20² = y²

Differentiate implicitly with respect to t to obtain: 2x(dx/dt) = 2y(dy/dt)

When t = 3, x = 2(3) = 6 and 6² + 20² = y^2. So, y = √436

Therefore, 12(2) = 2√436(dy/dt)

dy/dt = 12/√436 = 12 / [2√109] = 6/√109 m / sec ≈ 0.575 m /sec