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Gnom [1K]
3 years ago
8

Rewrite the statement in mathematical notation. (Let y be the distance from the top of the ladder to the floor, x be the distanc

e from the base of the ladder to the wall, and t be time.) A ladder is sliding down a wall so that the distance between the top of the ladder and the floor is decreasing at a rate of 6 feet per second. How fast is the base of the ladder receding from the wall?
Mathematics
1 answer:
In-s [12.5K]3 years ago
8 0

Answer:

\frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}

Step-by-step explanation:

Let L be the length of the ladder,

Given,

x = the distance from the base of the ladder to the wall, and t be time.

y = distance from the base of the ladder to the wall,

So, by the Pythagoras theorem,

L^2 = y^2 + x^2

\implies L = \sqrt{y^2 + x^2},

Differentiating with respect to time (t),

\frac{dL}{dt}=\frac{d}{dt}(\sqrt{x^2 + y^2})

=\frac{1}{2\sqrt{x^2 + y^2}}\frac{d}{dt}(x^2 + y^2)

=\frac{1}{2\sqrt{x^2 + y^2}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})

=\frac{1}{\sqrt{x^2 +y^2}}(x\frac{dx}{dt}+y\frac{dy}{dt})

Here,

\frac{dy}{dt}=-6\text{ ft per sec}

Also, \frac{dL}{dt} = 0           ( Ladder length = constant ),

\implies \frac{1}{\sqrt{x^2 +y^2}}(x(-6)+y\frac{dy}{dt})=0

-6x + y\frac{dy}{dt}=0

y\frac{dy}{dt}=6x

\implies \frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}

Which is the required notation.

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