Question

a 24 feet ladder rests against a wall. Let θ be the angle between the top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from the wall, how fast does x change with respect to θ when θ = pi/3

Answer 1

Answer to Question #85289 – Math – Calculus

Question

A 24 feet ladder rests against a wall. Let $\theta$ be the angle between the top of the ladder and the wall and let $x$ be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from the wall, how fast does $x$ change with respect to $\theta$ when $\theta = \pi/3$.

Solution

Given that $x$ is the distance from the bottom of the ladder to the wall.

Let the height of the top of the ladder from the base be $y$.

Then the equation is:

$$x^2 + y^2 = 24^2$$ $$x^2 + y^2 = 576$$ $$x^2 = 576 - y^2$$

Differentiating the above equation with respect to $\theta$ we get

$$2x \frac{dx}{d\theta} = 0 - 2y \frac{dy}{d\theta}$$ $$x \frac{dx}{d\theta} = -y \frac{dy}{d\theta}$$ $$\frac{dx}{d\theta} = \left(-\frac{y}{x}\right) \frac{dy}{d\theta}$$

Again we have,

$$\tan \theta = \frac{x}{y}$$ $$\frac{y}{x} = \cot \theta$$

At $\theta = \frac{\pi}{3}$, we have,

$$\frac {y}{x} = \cot \frac {\pi}{3} = \frac {1}{\sqrt {3}}$$ $$y = \frac {x}{\sqrt {3}}$$ $$y ^ {2} = \frac {x ^ {2}}{3}$$

Again,

$$x ^ {2} + y ^ {2} = 5 7 6$$ $$x ^ {2} + \frac {x ^ {2}}{3} = 5 7 6$$ $$\frac {4 x ^ {2}}{3} = 5 7 6$$ $$x ^ {2} = 1 4 4 \times 3$$ $$x = 1 2 \sqrt {3}$$ $$\frac {d y}{d \theta} = \frac {1}{\sqrt {3}} \left(\frac {d x}{d \theta}\right)$$

Then we get,

$$\frac {d x}{d \theta} = \left(- \frac {1}{\sqrt {3}}\right) \left(\frac {1}{\sqrt {3}} \frac {d x}{d \theta}\right)$$ $$\frac {d x}{d \theta} + \frac {1}{3} \frac {d x}{d \theta} = 0$$ $$\frac {4}{3} \frac {d x}{d \theta} = 0$$ $$\frac {d x}{d \theta} = 0$$

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