A $29$ -meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at $7$ meters per minute. $\newline$ At a certain instant, the bottom of the ladder is $21$ meters from the wall. $\newline$ What is the rate of change of the distance between the bottom of the ladder and the wall at that instant (in meters per minute)? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{20}{3}$ $\newline$ (B) $20$ $\newline$ (C) $\frac{147}{20}$ $\newline$ (D) $7$

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Grade 8

Solve equations with variables on both sides: word problems

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Q. A

29

-meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at

7

meters per minute.

\newline

At a certain instant, the bottom of the ladder is

21

meters from the wall.

\newline

What is the rate of change of the distance between the bottom of the ladder and the wall at that instant (in meters per minute)?

\newline

Choose

1

answer:

\newline

(A)

\frac{20}{3}

\newline

(B)

20

\newline

(C)

\frac{147}{20}

\newline

(D)

7

Set Up Problem: We got a $29$ -meter ladder against a wall. The top is sliding down at $7$ meters per minute. When the bottom is $21$ meters out, we gotta find how fast it's moving away from the wall.
Identify Variables: Let's call the distance from the wall to the bottom of the ladder $x$ , and the distance from the ground to the top of the ladder $y$ . We're given $y$ is decreasing at $7$ meters per minute, so $\frac{dy}{dt} = -7$ meters per minute.
Apply Pythagoras Theorem: We got a right triangle with the ladder, the wall, and the ground. So, by Pythagoras, we got $x^2 + y^2 = 29^2$ .
Differentiate with Respect to Time: Differentiate both sides with respect to time $t$ to find $\frac{dx}{dt}$ . So, $2x\left(\frac{dx}{dt}\right) + 2y\left(\frac{dy}{dt}\right) = 0$ .
Calculate Height of Ladder: Plug in the values we know: $x = 21$ , $\frac{dy}{dt} = -7$ . We get $2\cdot21\cdot\frac{dx}{dt} + 2\cdot y\cdot(-7) = 0$ .
Substitute Values: We need $y$ , the height of the ladder on the wall when the bottom is $21$ meters out. So, $y^2 = 29^2 - 21^2$ . That's $y^2 = 841 - 441$ .
Calculate Height: Calculate $y^2$ : $y^2 = 400$ . So, $y = 20$ meters (since $y$ is a distance, it's positive).
Solve for dx/dt: Now we got $y$ , so plug it into the differentiated equation: $2 \times 21 \times \frac{dx}{dt} + 2 \times 20 \times (-7) = 0$ .
Simplify Equation: Simplify the equation: $42\left(\frac{dx}{dt}\right) - 280 = 0$ .
Find $\frac{dx}{dt}$ : Solve for $\frac{dx}{dt}$ : $42\left(\frac{dx}{dt}\right) = 280$ . So, $\left(\frac{dx}{dt}\right) = \frac{280}{42}$ .
Find $\frac{dx}{dt}$ : Solve for $\frac{dx}{dt}$ : $42\left(\frac{dx}{dt}\right) = 280$ . So, $\frac{dx}{dt}$ = $\frac{280}{42}$ .Calculate $\frac{dx}{dt}$ : $\frac{dx}{dt}$ = $\frac{280}{42} = \frac{20}{3}$ meters per minute.