Question

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) What is the velocity of the top of the ladder when the base is given below

Answer 1

This question is incomplete, the complete question is;

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) What is the velocity of the top of the ladder when the base is given below

7 feet away from the wall

15 feet away from the wall

20 feet away from the wall

Answer:

- When x is 7 feet away from the wall,
`(dr)/(dt)` = -7/12 ft/s

- When x is 15 feet away from the wall,
`(dr)/(dt)` = -3/2 ft/s

- When x is 20 feet away from the wall,
`(dr)/(dt)` = -8/3 ft/s

Step-by-step explanation:

Given the data in the question;

Let
`r`ft and
`x`ft represent the distance of the top of ladder from the ground and the distance of the bottom of ladder from the wall respectively.

Now using the Pythagoras Theorem ( right angled triangle ).

(hypotenuse)² = ( adjacent )² + ( opposite )²

so

r² = ( 25 )² - x²

r² = 625 - x² ------- let this be equation 1

now, we differentiate with respect to t

2r
`(dr)/(dt)` = 0 - 2x
`(dx)/(dt)`

r
`(dr)/(dt)` = -x
`(dx)/(dt)`

`(dr)/(dt)` = -x
`(dx)/(dt)` × 1/r

`(dr)/(dt)` = -x/r
`(dx)/(dt)`

Now, from equation, r² = 625 - x²

r = √( 625 - x² )

Then

`(dr)/(dt)` = -x / √( 625 - x² )
`(dx)/(dt)`

given that, The base of the ladder is pulled away from the wall at a rate of 2 feet per second i.e
`(dx)/(dt)` = 2 ft/s

so;

`(dr)/(dt)` = ( -x / √( 625 - x² ) ) × 2

`(dr)/(dt)` = -2x / √( 625 - x² )

∴

When x is 7 feet away from the wall;

`(dr)/(dt)` = -2(7) / √( 625 - (7)² )

`(dr)/(dt)` = -14 / √( 625 - 49 )

`(dr)/(dt)` = -14 / √576

`(dr)/(dt)` = -14 / 24

we simplify

`(dr)/(dt)` = -7/12 ft/s

Therefore, When x is 7 feet away from the wall,
`(dr)/(dt)` = -7/12 ft/s

When x is 15 feet away from the wall

`(dr)/(dt)` = -2(15) / √( 625 - (15)² )

`(dr)/(dt)` = -30 / √( 625 - 225 )

`(dr)/(dt)` = -30 / √400

`(dr)/(dt)` = -30 / 20

we simplify

`(dr)/(dt)` = -3/2 ft/s

Therefore, When x is 15 feet away from the wall,
`(dr)/(dt)` = -3/2 ft/s

When x is 20 feet away from the wall

`(dr)/(dt)` = -2(20) / √( 625 - (20)² )

`(dr)/(dt)` = -40 / √( 625 - 400 )

`(dr)/(dt)` = -40 / √225

`(dr)/(dt)` = -40 / 15

we simplify

`(dr)/(dt)` = -8/3 ft/s

Therefore, When x is 20 feet away from the wall,
`(dr)/(dt)` = -8/3 ft/s