A 5 foot long pole leans against a wall. The bottom is 3 feet from the wall. How much farther should the bottom be pulled away from the wall so that the top mov

Question

I dont understand how to solve

Answer 1

the pole forms a right triangle, 3 by 4 by 5.

the pole length is the hypotenuse

the most the bottom of the pole can be pulled is 2 feet

IF the question involves movement of the top of the pole

this is a related rates problem. A calculus textbook will have a section on "related rates" which is about derivatives

call the pole length L=5, the ground length = G = 3

and the height of the top of the pole H=4

L^2 = G^2 + H^2

take the derivative

2LL' = 2GG'+ 2HH' plug in the values and solve for H' which is the movement or rate of change of the height of the top of the pole. The problem likely gave G' as some value. L'=0, the pole length doesn't change

H'= GG'/H = 3G'/4

The most the bottom of the pole can move from the wall is 2 more feet. At the same time as it moves 2 feet, the top of the pole will move 4 feet.

If the bottom of the pole moves 1 foot, the top of the pole will move 1 foot down. Just calculate that distance using the Pythagorean Theorem. If the bottom moves from 3 feet from the wall to 4 feet from the wall, the top will move from 4 feet above the wall to 3 feet above the wall. 5^2 = 3^2 + 4^2

and 5^2 = 4^2 + 3^2

A move 1 foot horizontally for the bottom of the pole

corresponds to a 1 foot move down for the top of the pole.

a move 1/2 foot left for the pole bottom means a move down by the top of the pole = 4-sqr[5^2 -(3.5)^2) = 4-sqr(25-12.25) = 4-sqr12.75 = about 4-3.5707 feet = 0.4293 feet

half foot left for the bottom means 0.4293 feet down for the pole top