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A 5-meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at 6 meters per minute.
At a certain instant, the top of the ladder is 3 meters from the ground.
What is the rate of change of the area formed by the ladder at that instant (in square meters per minute)?
Choose 1 answer:
(A) -7
(B) 6
(C) 18
(D) -14

A $5$ -meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at $6$ meters per minute. $\newline$ At a certain instant, the top of the ladder is $3$ meters from the ground. $\newline$ 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) $-7$ $\newline$ (B) $6$ $\newline$ (C) $18$ $\newline$ (D) $-14$

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Calculate unit rates with fractions

Full solution

Q. A

5

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

6

meters per minute.

\newline

At a certain instant, the top of the ladder is

3

meters from the ground.

\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)

-7

\newline

(B)

6

\newline

(C)

18

\newline

(D)

-14

Triangle Information: The ladder, wall, and ground form a right triangle. The ladder's length is the hypotenuse, and it's $5$ meters long. The distance from the top of the ladder to the ground is $3$ meters.
Finding x: Let's call the distance from the bottom of the ladder to the wall $x$ meters. We can use the Pythagorean theorem to find $x$ . $\newline$ $5^2 = 3^2 + x^2$ $\newline$ $25 = 9 + x^2$ $\newline$ $x^2 = 25 - 9$ $\newline$ $x^2 = 16$ $\newline$ $x = 4 \text{ meters.}$
Calculating Area: The area $A$ of the triangle at that instant is $(1/2) \times \text{base} \times \text{height} = (1/2) \times x \times 3$ . $\newline$ $A = (1/2) \times 4 \times 3$ $\newline$ $A = 6$ square meters.
Rate of Change: The rate of change of $x$ is given as $6$ meters per minute. We need to find the rate of change of the area, $\frac{dA}{dt}$ . $\newline$ $\frac{dA}{dt} = \frac{1}{2} \cdot \frac{d(x \cdot 3)}{dt}$ $\newline$ $\frac{dA}{dt} = \frac{1}{2} \cdot 3 \cdot \frac{dx}{dt}$ $\newline$ $\frac{dA}{dt} = \frac{1}{2} \cdot 3 \cdot 6$ $\newline$ $\frac{dA}{dt} = 9$ square meters per minute.

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