A $10$ -meter ladder is sliding down a vertical wall so the distance between the top of the ladder and the ground is decreasing at $3$ meters per minute. $\newline$ At a certain instant, the bottom of the ladder is $6$ meters from the wall. $\newline$ What is the rate of change of the area formed by the ladder at that instant (in square meters per minute)? $\newline$ Choose $1$ answer: $\newline$ (A) $12$ $\newline$ (B) $-\frac{7}{2}$ $\newline$ (C) $7$ $\newline$ (D) $-6$

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Math Problems

Grade 7

Calculate unit rates with fractions

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Q. A

10

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

3

meters per minute.

\newline

At a certain instant, the bottom of the ladder is

6

meters from the wall.

\newline

What is the rate of change of the area formed by the ladder at that instant (in square meters per minute)?

\newline

Choose

1

answer:

\newline

(A)

12

\newline

(B)

-\frac{7}{2}

\newline

(C)

7

\newline

(D)

-6

Triangle Area Formula: The ladder forms a right triangle with the wall and the ground. The area of a right triangle is $(\frac{1}{2}) \times \text{base} \times \text{height}$ .
Variables and Area Calculation: Let $x$ be the distance from the bottom of the ladder to the wall, and $y$ be the distance from the top of the ladder to the ground. The area $A = (\frac{1}{2}) \times x \times y$ .
Pythagorean Theorem Application: Given that the ladder is $10$ meters long, by the Pythagorean theorem, we have $x^2 + y^2 = 10^2$ .
Differentiation for Rates of Change: Differentiate both sides with respect to time $t$ to find the rates of change: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ .
Given Rate of Change: We know that $\frac{dy}{dt} = -3$ meters per minute (since the top is going down, it's negative), and we need to find $\frac{dx}{dt}$ .
Substitution for Variables: Substitute $x = 6$ meters and $y = \sqrt{(10^2 - 6^2)} = \sqrt{(100 - 36)} = \sqrt{64} = 8$ meters into the differentiated equation.
Equation Simplification: Now we have $2\times 6\times \left(\frac{dx}{dt}\right) + 2\times 8\times (-3) = 0$ .
Solving for $\frac{dx}{dt}$ : Solve for $\frac{dx}{dt}$ : $12\cdot\left(\frac{dx}{dt}\right) - 48 = 0$ .
Final Rate Calculation: Add $48$ to both sides: $12\left(\frac{dx}{dt}\right) = 48$ .
Area Rate of Change Formula: Divide by $12$ : $\frac{dx}{dt} = \frac{48}{12} = 4$ meters per minute.
Substitution for Area Rate: Now we find the rate of change of the area $\frac{dA}{dt} = \frac{1}{2} \cdot (x\cdot\frac{dy}{dt} + y\cdot\frac{dx}{dt})$ .
Area Rate Calculation: Substitute $x = 6$ , $y = 8$ , $\frac{dy}{dt} = -3$ , and $\frac{dx}{dt} = 4$ into the equation for $\frac{dA}{dt}$ .
Area Rate Calculation: Substitute $x = 6$ , $y = 8$ , $\frac{dy}{dt} = -3$ , and $\frac{dx}{dt} = 4$ into the equation for $\frac{dA}{dt}$ .Calculate $\frac{dA}{dt} = \frac{1}{2} \times (6 \times (-3) + 8 \times 4)$ .
Area Rate Calculation: Substitute $x = 6$ , $y = 8$ , $\frac{dy}{dt} = -3$ , and $\frac{dx}{dt} = 4$ into the equation for $\frac{dA}{dt}$ .Calculate $\frac{dA}{dt} = \frac{1}{2} \times (6 \times (-3) + 8 \times 4)$ . $\frac{dA}{dt} = \frac{1}{2} \times (-18 + 32)$ .
Area Rate Calculation: Substitute $x = 6$ , $y = 8$ , $\frac{dy}{dt} = -3$ , and $\frac{dx}{dt} = 4$ into the equation for $\frac{dA}{dt}$ .Calculate $\frac{dA}{dt} = \frac{1}{2} \times (6\times(-3) + 8\times4)$ . $\frac{dA}{dt} = \frac{1}{2} \times (-18 + 32)$ . $\frac{dA}{dt} = \frac{1}{2} \times 14 = 7$ square meters per minute.