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#