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) $7$ $\newline$ (B) $\frac{147}{20}$ $\newline$ (C) $20$ $\newline$ (D) $\frac{20}{3}$

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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)

7

\newline

(B)

\frac{147}{20}

\newline

(C)

20

\newline

(D)

\frac{20}{3}

Triangle Description: We're dealing with a right triangle where the ladder is the hypotenuse, the distance from the wall is one leg, and the height of the ladder above the ground is the other leg.
Pythagorean Theorem: Let $x$ be the distance from the bottom of the ladder to the wall, and $y$ be the height of the ladder above the ground. We have $x^2 + y^2 = 29^2$ because of the Pythagorean theorem.
Differentiation: Differentiate both sides with respect to time $t$ to find the rates of change. We get $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ since the length of the ladder doesn't change.
Given Rates: We know $\frac{dy}{dt} = -7$ meters per minute (since the top of the ladder is sliding down), and we need to find $\frac{dx}{dt}$ when $x = 21$ meters.
Equation Simplification: Plug in the values we know: $2\times21\times\frac{dx}{dt} + 2\times y\times(-7) = 0$ .
Calculate y: We need to find $y$ when $x = 21$ . Using the Pythagorean theorem, $y^2 = 29^2 - 21^2$ . Calculate $y^2 = 841 - 441$ .
Solve for $\frac{dx}{dt}$ : $y^2 = 400$ , so $y = 20$ meters (since $y$ is positive as it's a distance).
Final Calculation: Now we can solve for $\frac{dx}{dt}$ : $2 \times 21 \times \left(\frac{dx}{dt}\right) - 2 \times 20 \times 7 = 0$ .
Final Calculation: Now we can solve for $\frac{dx}{dt}$ : $2\times 21\times\frac{dx}{dt} - 2\times 20\times 7 = 0$ .Simplify the equation: $42\times\frac{dx}{dt} - 280 = 0$ .
Final Calculation: Now we can solve for $\frac{dx}{dt}$ : $2\times 21\times\frac{dx}{dt} - 2\times 20\times 7 = 0$ .Simplify the equation: $42\times\frac{dx}{dt} - 280 = 0$ .Add $280$ to both sides: $42\times\frac{dx}{dt} = 280$ .
Final Calculation: Now we can solve for $\frac{dx}{dt}$ : $2\times 21\times\frac{dx}{dt} - 2\times 20\times 7 = 0$ .Simplify the equation: $42\times\frac{dx}{dt} - 280 = 0$ .Add $280$ to both sides: $42\times\frac{dx}{dt} = 280$ .Divide by $42$ to solve for $\frac{dx}{dt}$ : $\frac{dx}{dt} = \frac{280}{42}$ .
Final Calculation: Now we can solve for $\frac{dx}{dt}$ : $2\times 21\times\left(\frac{dx}{dt}\right) - 2\times 20\times 7 = 0$ .Simplify the equation: $42\times\left(\frac{dx}{dt}\right) - 280 = 0$ .Add $280$ to both sides: $42\times\left(\frac{dx}{dt}\right) = 280$ .Divide by $42$ to solve for $\frac{dx}{dt}$ : $\frac{dx}{dt} = \frac{280}{42}$ .Calculate $\frac{dx}{dt}$ : $\frac{dx}{dt} = \frac{20}{3}$ meters per minute.