Question

A rod of length 41 m with an end A on the floor and another end B on the wall perpendicular to the floor is sliding away horizontally from the wall at the rate of 3 ft/min. When the end B is at the height of 9 ft from the floor, then the rate at which the area of the triangle formed by the rod with wall and floor changes at that instant is (in ft/min)

• A. \( \frac{1519}{6} \)
• B. \( \frac{1618}{3} \)
• C. \( \frac{1600}{3} \)
• D. \( \frac{1509}{6} \)

Hint:

In related rates problems, identify the quantities that are changing and the rates that are given. Find a static equation that relates the quantities (e.g., Pythagorean theorem, area formula). Differentiate this equation with respect to time, and then substitute the instantaneous values to solve for the unknown rate.

Answer 1

The Correct Option is A

Let x be the distance of end A from the wall, and y be the height of end B from the floor. The rod, wall, and floor form a right-angled triangle.
By the Pythagorean theorem, \( x^2 + y^2 = 41^2 = 1681 \).
We are given that the end A is sliding away from the wall, so \( \frac{dx}{dt} = 3 \) ft/min.
The area of the triangle is \( A = \frac{1}{2}xy \). We need to find \( \frac{dA}{dt} \).
Differentiate the area equation with respect to time using the product rule:
\( \frac{dA}{dt} = \frac{1}{2} \left( \frac{dx}{dt} y + x \frac{dy}{dt} \right) \).
We need to find the values of x and \( \frac{dy}{dt} \) at the instant when \(y=9\).
When \(y=9\), we find x from the Pythagorean relation: \( x^2 + 9^2 = 41^2 \implies x^2 + 81 = 1681 \implies x^2 = 1600 \implies x = 40 \).
To find \( \frac{dy}{dt} \), we differentiate the Pythagorean relation with respect to time:
\( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \).
\( x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \).
Substitute the known values at the given instant: \( x=40, y=9, \frac{dx}{dt}=3 \).
\( (40)(3) + (9) \frac{dy}{dt} = 0 \).
\( 120 + 9 \frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{120}{9} = -\frac{40}{3} \).
Now, substitute all values into the derivative of the area equation:
\( \frac{dA}{dt} = \frac{1}{2} \left( (3)(9) + (40)(-\frac{40}{3}) \right) \).
\( \frac{dA}{dt} = \frac{1}{2} \left( 27 - \frac{1600}{3} \right) = \frac{1}{2} \left( \frac{81 - 1600}{3} \right) = \frac{1}{2} \left( \frac{-1519}{3} \right) = -\frac{1519}{6} \).
The rate of change is \( -\frac{1519}{6} \). The question asks for the rate, and the options are positive, so we take the magnitude, assuming it means rate of decrease. Or there might be a sign error in the question, e.g., A moving towards the wall. Assuming the magnitude is asked, it matches.