A ladder 15 meters long is leaning against a vertical wall. The top of the ladder is sliding down the wall at a rate of 3 meters per second. How fast is the bottom of the ladder sliding away from the wall when the top is 9 meters above the ground?
Guide On Rating System
Vote
Let \( x \) denote the distance between the bottom of the ladder and the wall. Since the ladder is 15 meters long, we have that \( \sqrt{15^2 - x^2} \) represents the height of the ladder above the ground. Taking derivatives with respect to time, we obtain
\[ \frac{d}{dt} \sqrt{15^2 - x^2} = \frac{d}{dt}(\textnormal{Top of ladder}) = -3, \]
since the top of the ladder is sliding down the wall at a rate of 3 meters per second. Applying the chain rule, we have
\[ \frac{d}{dt} \sqrt{15^2 - x^2} = \frac{1}{2\sqrt{15^2 - x^2}}(-2x \frac{dx}{dt} ) = -3. \]
When the top is 9 meters above the ground, we have \( x = \sqrt{15^2 - 9^2} \), and hence
\[ \frac{d}{dt} \sqrt{15^2 - x^2} = \frac{1}{2\sqrt{15^2 - (\sqrt{15^2 - 9^2})^2}} \frac{dx}{dt}. \]
Simplifying, it follows that
\[ -\frac{27}{\sqrt{15^2 - (\sqrt{15^2 - 9^2}})^2} = \frac{dx}{dt}. \]
Simplifying further, we obtain
\[ \frac{dx}{dt} = -\frac{27}{\sqrt{15^2 - 12^2}} = -\frac{27}{9} = \boxed{-3}. \]
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99