Question

Math

Posted 7 months ago

`A ladder 9 meters in length is leaning against a vertical wall on level ground. As the bottom end of the ladder is moved away from the wall at a constant rate of 2 meters per second, the top end slides downward along the wall. How fast, in meters per second, will the top end of the ladder be sliding downward at the moment the top end is 3 meters above the ground?`

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Answer from Sia

Posted 7 months ago

Solution by Steps

step 1

To find the rate at which the top of the ladder slides down, we use the Pythagorean theorem for a right triangle formed by the ladder, the wall, and the ground: $x^2 + y^2 = 9^2$, where $x$ is the distance from the wall to the bottom of the ladder and $y$ is the height of the ladder on the wall

step 2

Differentiate both sides of the equation with respect to time $t$ to relate the rates of change: $\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(9^2)$

step 3

Apply the chain rule to differentiate $x^2$ and $y^2$: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

step 4

Plug in the given values $y = 3$ meters and $\frac{dx}{dt} = 2$ meters per second into the differentiated equation: $2x(2) + 2(3)\frac{dy}{dt} = 0$

step 5

Solve for $x$ using the Pythagorean theorem: $x = \sqrt{9^2 - y^2} = \sqrt{81 - 3^2} = \sqrt{81 - 9} = \sqrt{72}$

step 6

Substitute $x = \sqrt{72}$ into the equation from step 4: $2\sqrt{72}(2) + 2(3)\frac{dy}{dt} = 0$

step 7

Solve for $\frac{dy}{dt}$: $4\sqrt{72} + 6\frac{dy}{dt} = 0$ leads to $\frac{dy}{dt} = -\frac{4\sqrt{72}}{6} = -\frac{2\sqrt{72}}{3}$

step 8

Simplify to find the rate at which the top of the ladder slides down: $\frac{dy}{dt} = -\frac{2\sqrt{72}}{3}$ meters per second

Answer

$\frac{dy}{dt} = -\frac{2\sqrt{72}}{3}$ meters per second

Key Concept

Related rates and the Pythagorean theorem

Explanation

We use related rates to connect the rate at which the bottom of the ladder moves away from the wall with the rate at which the top of the ladder slides down the wall. By differentiating the Pythagorean theorem with respect to time, we can relate these rates and solve for the desired rate.

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