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A ladder
20ft. long leans against a vertical wall. If the top slides downward at the rate of
2ft. per sec. Find how fast the ladder end is moving when it is
16ft. from the wall.

A ladder $20 \mathrm{ft}$ . long leans against a vertical wall. If the top slides downward at the rate of $2 \mathrm{ft}$ . per sec. Find how fast the ladder end is moving when it is $16 \mathrm{ft}$ . from the wall.

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Solve proportions: word problems

Full solution

Q. A ladder

20 \mathrm{ft}

. long leans against a vertical wall. If the top slides downward at the rate of

2 \mathrm{ft}

. per sec. Find how fast the ladder end is moving when it is

16 \mathrm{ft}

. from the wall.

Identify Relationship: Identify the relationship between the ladder, wall, and ground. $\newline$ We use the Pythagorean theorem since the ladder, wall, and ground form a right triangle. $\newline$ Let $x$ be the distance from the wall to the bottom of the ladder, and $y$ be the distance from the ground to the top of the ladder. $\newline$ $x^2 + y^2 = 20^2$
Use Pythagorean Theorem: Differentiate the equation with respect to time $t$ to find the rate at which $x$ changes. $\newline$ Differentiating implicitly: $\newline$ $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ $\newline$ Given $\frac{dy}{dt} = -2$ ft/sec (since $y$ is decreasing), $\newline$ $2x\frac{dx}{dt} + 2(16)(-2) = 0$
Differentiate Implicitly: Solve for $\frac{dx}{dt}$ , which is the rate at which the bottom of the ladder moves away from the wall. $\newline$ $2x\left(\frac{dx}{dt}\right) - 64 = 0$ $\newline$ $2x\left(\frac{dx}{dt}\right) = 64$ $\newline$ $\frac{dx}{dt} = \frac{64}{2x}$ $\newline$ We need to find $x$ when $y = 16$ ft. $\newline$ Using the Pythagorean theorem: $x^2 + 16^2 = 20^2$ $\newline$ $x^2 + 256 = 400$ $\newline$ $x^2 = 144$ $\newline$ $x = 12$ ft
Solve for $\frac{dx}{dt}$ : Substitute $x = 12$ ft into the equation for $\frac{dx}{dt}$ . $\frac{dx}{dt} = \frac{64}{(2\cdot12)}$ $\frac{dx}{dt} = \frac{64}{24}$ $\frac{dx}{dt} = 2.67$ ft/sec

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