Ladder against the wall again? A 12-ft ladder is leaning against a vertical wall when Jack begins pulling the foot of the ladder away from the wall at a rate of 0.2 ft/s. What is the configuration of the ladder al (he instant that the vertical speed of the top of the ladder equals the horizontal speed of the foot of the ladder?

Solution: Step1 To find What is the configuration of the ladder instant that the vertical speed of the top of the ladder equals the horizontal speed of the foot of the ladder Step2 Given that A 12-ft ladder is leaning against a vertical wall when Jack begins pulling the foot of the ladder away from the wall at a rate of 0.2 ft/s. Step3 Step4 Consider the ladder as the hypotenuse of a right angled triangle the wall on which the ladder is leaning as the height of y the triangle and the ground between the ladder and the wall as the base of the triangle say x,’ c = 12 ft. dx/dt = 0.2 ft./s dy/dt = -dx/dt By applying pythagoras theorem we get, c² = x² + y² 12² = x² + y² y² = 12² - x² y = (12² - x²) Differentiate both side we get, 0 = x·dx/dt + y·dy/dt 0 = x·dx/dt - (144 - x²)·dx/dt x = (144 - x²) x² = 144 - x² 2x² = 144 x² = 72 x = 72 = 62 8.485 ft. y = (144 - x²) = (144 - 72) = 72 = 62 8.485 ft Hence height y= 8.485 Also, at the point x= y angle is: tan= y/x = y/y =1 Implies, tan= tan 45 = 45