Question

Math

Posted 5 months ago

is this right?

`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 5 months ago

Solution by Steps

step 1

Apply the Pythagorean theorem to the ladder problem: $9^2 = x^2 + y^2$, where $x$ is the distance from the wall to the bottom of the ladder and $y$ is the distance from the ground to the top of the ladder

step 2

Differentiate both sides of the equation with respect to time $t$ to find the relationship between the rates of change: $0 = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$

step 3

Given that $\frac{dx}{dt} = 2$ meters per second (the rate at which the bottom of the ladder moves away from the wall), substitute $x$ and $\frac{dx}{dt}$ into the differentiated equation: $0 = 2x(2) + 2y \frac{dy}{dt}$

step 4

When $y = 3$ meters, use the Pythagorean theorem to find $x$: $9^2 = x^2 + 3^2$, which gives $x = \sqrt{9^2 - 3^2} = \sqrt{81 - 9} = \sqrt{72}$

step 5

Substitute $x = \sqrt{72}$ and $y = 3$ into the differentiated equation: $0 = 2\sqrt{72}(2) + 2(3) \frac{dy}{dt}$

step 6

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

step 7

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 in calculus and the Pythagorean theorem

Explanation

The rate at which the top of the ladder slides down is found by differentiating the Pythagorean theorem with respect to time and solving for $\frac{dy}{dt}$ when $y = 3$ meters.

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