A ladder 10 m long rests against a vertical wall. Its lower end is pulled away from the wall along the ground at the rate of 1.5 m/s. The rate at which the top of the ladder slides down the wall, when the lower end is 6 m from the wall, is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{9}{8}\) m/s
Let x be the distance of the lower end from the wall and y the height of the top at time t. The ladder is the hypotenuse, so by Pythagoras' theorem
\(x^{2} + y^{2} = 100\)
Differentiating with respect to t:
\( \displaystyle 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \)
\( \displaystyle \frac{dy}{dt} = -\frac{x}{y}\cdot\frac{dx}{dt} \)
When x = 6:
\( y = \sqrt{100 - 36} = 8 \)
With \(\frac{dx}{dt} = 1.5\):
\( \displaystyle \frac{dy}{dt} = -\frac{6}{8} \times \frac{3}{2} \)
\( \displaystyle = -\frac{9}{8} \)
The negative sign shows that y is decreasing, so the top slides down at \(\frac{9}{8}\) m/s.
Why the other options are incorrect
Option (1) 2 m/s — Incorrect. This is \(\frac{8}{6} \times 1.5\): the ratio \(\frac{x}{y}\) is turned upside down.
Option (2) \(\frac{15}{8}\) m/s — Incorrect. This is \(\frac{10}{8} \times 1.5\): it uses the ladder length 10 m in place of the distance x = 6 m.
Option (3) \(\frac{3}{2}\) m/s — Incorrect. The top does not move at the same speed as the lower end; the speeds are in the ratio \(\frac{x}{y} = \frac{6}{8}\).