Question

A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:

(iii) (b) If the foot of the ladder, whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance \( y \) is \( 2 \, \text{m/s} \), then at what rate is the height on the wall \( x \) increasing when the foot of the ladder is 3 m away from the wall?

Hint:

Remember, for the maximum area of the triangle formed by the ladder, wall, and ground, the height \( x \) should be \( \frac{h}{\sqrt{2}} \), which is derived from the condition where the first derivative of the area equals zero.

Answer 1

Given: - The length of the ladder \( h = 5 \, \text{m} \),
- The rate of change of distance \( y \) is \( \frac{dy}{dt} = -2 \, \text{m/s} \),
- The foot of the ladder is 3 m away from the wall, i.e., \( y = 3 \, \text{m} \).
From the Pythagorean theorem: \[ x^2 + y^2 = h^2 \] Differentiating with respect to time \( t \): \[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \] Substitute the given values: \[ 2x \frac{dx}{dt} + 2(3)(-2) = 0 \] \[ 2x \frac{dx}{dt} - 12 = 0 \] \[ x \frac{dx}{dt} = 6 \] Now, solve for \( \frac{dx}{dt} \): \[ \frac{dx}{dt} = \frac{6}{x} \] To find \( x \), use the Pythagorean theorem: \[ x^2 + 3^2 = 5^2 \] \[ x^2 + 9 = 25 \] \[ x^2 = 16 \] \[ x = 4 \] Thus: \[ \frac{dx}{dt} = \frac{6}{4} = 1.5 \, \text{m/s} \] So, the height on the wall is increasing at a rate of \( 1.5 \, \text{m/s} \) when the foot of the ladder is 3 m away from the wall.