A kit is 100 ft high. There are 260 ft of string out and it is being let out at a rate of 5 ft/sec. If this results in the kite being carried along horizontally, what is the horizontal speed of the kite?

I know that dh/dt is 5 ft /sec and I know I have to use the Pythagorean Theorem but I'm not sure what to do from here.

1 Answer
Aug 4, 2018

5.42 ft/sec

Explanation:

Let's assume the kite is flying away relative to the kite player, not to the ground. This is a better approach because the speed of the kite player is not given.

Let
x= Horizontal distance of the kite from the player
y= Stationary vertical height of the kite from the kite player = 100 ft
L = The length of the string

Approximate the relationship between x, y and L by an right angle triangle, thus:

L^2=x^2+y^2.

Take derivative with respect to time to get the speeds

2L (dL)/dt =2x (dx)/dt + 2y (dy)/dt

where

dx/dt is the kite speed;

dy/dt =0 because the kite is kept at y= 100 ft;

(dL)/dt = 5(ft)/sec and L= 260 ft

The relationship between kite speed and string releasing rate is thus:

cancel(2)L (dL)/dt =cancel(2)x (dx)/dt + 0

Solve the kite speed dx/dt by substituting x = sqrt(L^2-y^2)

dx/dt= L/x (dL)/dt = L/sqrt(L^2-y^2)(dL)/dt

dx/dt=(260ft)/sqrt((260ft)^2-(100ft)^2)*(5ft)/sec)= 5.42 (ft)/sec