a 8-meter ladder is leaning against the wall of a building, and the base of the ladder is sliding away from the building at a rate of 5 meters per second. how fast is the top of the ladder sliding down the wall when the base of the ladder is 6 meters from the wall?



Answer :

Height of the ladder on the wall in decreasing at the rate of ≈ 6m/s

What is differentiation ?

Finding a function's derivative, or rate of change, is the process of differentiation in mathematics. The practical approach of differentiation can be performed utilizing only algebraic operations, three fundamental derivatives, four principles of operation, and an understanding of how to manipulate functions, in contrast to the theory's abstract character.

Calculation

let ym be the height of wall at  which ladder touches

Also let the foot of ladder be xm away from the wall

then by the pythagoras theorem we have ,

[tex]x^{2} +y^{2} =[/tex] 64 [ length of ladder = 8 m ]

⇒ y = [tex]\sqrt{64 - x^{2} }[/tex]

then , the rate of change of height (y) with respect to time (t) is given by

[tex]\frac{dy}{dt}=\frac{-x}{\sqrt{64 - x^{2} }} . \frac{dx}{dt}[/tex]

it is given that [tex]\frac{dx}{dt}[/tex] = 5 m/s

∴ [tex]\frac{dy}{dt}=[/tex] [tex]\frac{-5x}{\sqrt{64 - x^{2} }}[/tex]

now when x = 6m , we have :

[tex]\frac{dy}{dt} = \frac{-5 * 6}{\sqrt{64 - 6^{2} }}[/tex]

  = -30 / 5 .3

  = -5.660377358490566

therefore height of the ladder on the wall in decreasing at the rate of ≈ 6m/s

learn more about differentiation here :

brainly.com/question/24898810

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