A $39$ -meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at $10$ meters per minute. $\newline$ At a certain instant, the bottom of the ladder is $\mathbf{3 6}$ meters from the wall. $\newline$ What is the rate of change of the distance between the top of the ladder and the ground at that instant (in meters per minute)? $\newline$ Choose $1$ answer: $\newline$ (A) $-\frac{25}{6}$ $\newline$ (B) $-12$ $\newline$ (C) $-\mathbf{1 0}$ $\newline$ (D) $-24$

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Grade 8

Solve equations with variables on both sides: word problems

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Q. A

39

-meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at

10

meters per minute.

\newline

At a certain instant, the bottom of the ladder is

\mathbf{3 6}

meters from the wall.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

-\frac{25}{6}

\newline

(B)

-12

\newline

(C)

-\mathbf{1 0}

\newline

(D)

-24

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.
Variable Assignment: Let's call the distance from the wall to the bottom of the ladder $x$ , and the height of the ladder above the ground $y$ .
Pythagorean Theorem: According to the Pythagorean theorem, we have $x^2 + y^2 = 39^2$ , since the ladder is $39$ meters long.
Differentiation: Differentiate both sides of the equation with respect to time $t$ to find the rates of change. We get $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ , because the length of the ladder is constant, so $\frac{d(39^2)}{dt} = 0$ .
Given Rates: We know that $\frac{dx}{dt}$ , the rate at which $x$ is changing, is $10$ meters per minute. We need to find $\frac{dy}{dt}$ .
Finding $y$ : Plug in the values we know: $x = 36$ meters, $\frac{dx}{dt} = 10$ meters/minute. We need to find $y$ before we can solve for $\frac{dy}{dt}$ .
Calculating $y$ : Using the Pythagorean theorem again, we find $y$ by solving $36^2 + y^2 = 39^2$ . This gives us $y^2 = 39^2 - 36^2$ .
Solving for $\frac{dy}{dt}$ : Calculate $y^2$ : $y^2 = 1521 - 1296$ , which gives us $y^2 = 225$ .
Simplify Equation: Take the square root of $y^2$ to find $y$ : $y = \sqrt{225}$ , which gives us $y = 15$ meters.
Solving for $\frac{dy}{dt}$ : Now we can solve for $\frac{dy}{dt}$ using the differentiated Pythagorean theorem: $2\cdot 36\cdot (10) + 2\cdot 15\cdot \left(\frac{dy}{dt}\right) = 0$ .
Final Calculation: Simplify the equation: $720 + 30\left(\frac{dy}{dt}\right) = 0$ .
Final Calculation: Simplify the equation: $720 + 30\frac{dy}{dt} = 0$ . Solve for $\frac{dy}{dt}$ : $30\frac{dy}{dt} = -720$ .
Final Calculation: Simplify the equation: $720 + 30\frac{dy}{dt} = 0$ . Solve for $\frac{dy}{dt}$ : $30\frac{dy}{dt} = -720$ . Divide both sides by $30$ to find $\frac{dy}{dt}$ : $\frac{dy}{dt} = -720 / 30$ .
Final Calculation: Simplify the equation: $720 + 30\frac{dy}{dt} = 0$ . Solve for $\frac{dy}{dt}$ : $30\frac{dy}{dt} = -720$ . Divide both sides by $30$ to find $\frac{dy}{dt}$ : $\frac{dy}{dt} = -720 / 30$ . Calculate $\frac{dy}{dt}$ : $\frac{dy}{dt} = -24$ meters/minute.