Question

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.

Answer 1

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`