A $25$ -meter ladder is sliding down a vertical wall so the distance between the bottom of the ladder and the wall is increasing at $3$ . $5$ meters per minute. $\newline$ At a certain instant, the top of the ladder is $7$ meters from the ground. $\newline$ 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) $-3$ . $5$ $\newline$ (B) $-12$ $\newline$ (C) $-7$ $\newline$ (D) $-24$

Full solution

Q. A

25

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

3

5

meters per minute.

\newline

At a certain instant, the top of the ladder is

7

meters from the ground.

\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)

-3

5

\newline

(B)

-12

\newline

(C)

-7

\newline

(D)

-24

Given Information: We have a $25$ -meter ladder sliding down a wall. We're given the rate at which the bottom moves away from the wall ( $3.5$ m/min) and the height of the top of the ladder above the ground ( $7$ m) at a certain instant.
Pythagorean Theorem: We can use the Pythagorean theorem to find the distance of the bottom of the ladder from the wall at that instant. Let's call this distance 'x'. $\newline$ So, we have $x^2 + 7^2 = 25^2$ .
Calculating Distance: Calculating 'x', we get $x^2 = 25^2 - 7^2 = 625 - 49 = 576$ . $\newline$ So, $x = \sqrt{576} = 24$ meters.
Rate of Change: Now, we need to find the rate of change of the height of the ladder with respect to time, which is $\frac{dy}{dt}$ , where 'y' is the height of the ladder above the ground.
Implicit Differentiation: Using implicit differentiation on the Pythagorean theorem $x^2 + y^2 = 25^2$ , we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$ .
Substitute Values: We know $\frac{dx}{dt} = 3.5$ m/min (the rate at which the bottom moves away from the wall) and at the instant we're considering, $y = 7$ m.
Solving Equation: Plugging in the values, we get $2(24)(3.5) + 2(7) \frac{dy}{dt} = 0$ .
Final Rate Calculation: Solving for $\frac{dy}{dt}$ , we get $\frac{dy}{dt} = -\frac{2(24)(3.5)}{2(7)}$ .
Final Rate Calculation: Solving for $\frac{dy}{dt}$ , we get $\frac{dy}{dt} = -\frac{2(24)(3.5)}{2(7)}$ .Simplifying, $\frac{dy}{dt} = -\frac{168}{14} = -12$ meters per minute.