Suppose a ladder that's 10 feet long is (somehow) resting up vertically against a wall. [br][br]The bottom of the ladder is then kicked out so that the base of the ladder is moving away from the wall at a rate of 3 ft/sec. (Go ahead and [color=#38761d][b]kick the ladder[/b][/color]). [br][br][color=#9900ff][b]At what rate is the ladder's height, [i]h[/i], changing[/b][/color] when the bottom of the ladder is 6 feet away from the wall? 9 feet away from the wall? [br][br]Use implicit differentiation to determine the answers to these 2 questions, and then check the approximate values of your your results within the applet. [br][br]In fact, at any time, you can adjust the values of [i]x[/i] and [math]\frac{dx}{dt}[/math]. [br][br]
[color=#9900ff][b]Why is the value of [/b][/color][math]\frac{dh}{dt}[/math][color=#9900ff][b] always negative[/b][/color] (except at [i]x[/i] = 0)? Explain.
How far away does the base of the ladder need to be away from the wall in order for [math]\frac{dx}{dt}=\left|\frac{dh}{dt}\right|[/math]? You can guess-and-check using the applet above. Yet be sure to use calculus to obtain an exact solution!