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, [tex]\frac{dr}{dt}[/tex] = -7/12 ft/s
- When x is 15 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -3/2 ft/s
- When x is 20 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -8/3 ft/s
Explanation:
Given the data in the question;
Let [tex]r[/tex]ft and [tex]x[/tex]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 [tex]\frac{dr}{dt}[/tex] = 0 - 2x [tex]\frac{dx}{dt}[/tex]
r [tex]\frac{dr}{dt}[/tex] = -x [tex]\frac{dx}{dt}[/tex]
[tex]\frac{dr}{dt}[/tex] = -x [tex]\frac{dx}{dt}[/tex] × 1/r
[tex]\frac{dr}{dt}[/tex] = -x/r [tex]\frac{dx}{dt}[/tex]
Now, from equation, r² = 625 - x²
r = √( 625 - x² )
Then
[tex]\frac{dr}{dt}[/tex] = -x / √( 625 - x² ) [tex]\frac{dx}{dt}[/tex]
given that, The base of the ladder is pulled away from the wall at a rate of 2 feet per second i.e [tex]\frac{dx}{dt}[/tex] = 2 ft/s
so;
[tex]\frac{dr}{dt}[/tex] = ( -x / √( 625 - x² ) ) × 2
[tex]\frac{dr}{dt}[/tex] = -2x / √( 625 - x² )
∴
When x is 7 feet away from the wall;
[tex]\frac{dr}{dt}[/tex] = -2(7) / √( 625 - (7)² )
[tex]\frac{dr}{dt}[/tex] = -14 / √( 625 - 49 )
[tex]\frac{dr}{dt}[/tex] = -14 / √576
[tex]\frac{dr}{dt}[/tex] = -14 / 24
we simplify
[tex]\frac{dr}{dt}[/tex] = -7/12 ft/s
Therefore, When x is 7 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -7/12 ft/s
When x is 15 feet away from the wall
[tex]\frac{dr}{dt}[/tex] = -2(15) / √( 625 - (15)² )
[tex]\frac{dr}{dt}[/tex] = -30 / √( 625 - 225 )
[tex]\frac{dr}{dt}[/tex] = -30 / √400
[tex]\frac{dr}{dt}[/tex] = -30 / 20
we simplify
[tex]\frac{dr}{dt}[/tex] = -3/2 ft/s
Therefore, When x is 15 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -3/2 ft/s
When x is 20 feet away from the wall
[tex]\frac{dr}{dt}[/tex] = -2(20) / √( 625 - (20)² )
[tex]\frac{dr}{dt}[/tex] = -40 / √( 625 - 400 )
[tex]\frac{dr}{dt}[/tex] = -40 / √225
[tex]\frac{dr}{dt}[/tex] = -40 / 15
we simplify
[tex]\frac{dr}{dt}[/tex] = -8/3 ft/s
Therefore, When x is 20 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -8/3 ft/s
