A-Level Differentiation: worked solution

Question

A 3.12 m ladder slides down a vertical wall while its foot moves along the floor at a steady 0.2 m/s (see diagram). Find how fast the top descends when it is 1.2 m above the floor.

This is a connected-rates-of-change problem: name the varying lengths, find an equation linking them, differentiate it with respect to time, then substitute the numbers for the given instant. Let x metres be the distance of the foot of the ladder from the wall and y metres the height of the top above the floor, at time t seconds. The ladder itself stays 3.12 m long. Step 1: link the variables. The wall is vertical and the floor horizontal, so the ladder is the hypotenuse of a right-angled triangle and Pythagoras gives x² + y² = 3.12² Step 2: differentiate both sides with respect to t. Each term needs the chain rule, because x and y are functions of t: 2x (dx/dt) + 2y (dy/dt) = 0 (the right side is constant, so it differentiates to 0) Divide by 2 and make dy/dt the subject: x (dx/dt) + y (dy/dt) = 0 dy/dt = -(x/y)(dx/dt) Step 3: find x at the instant when y = 1.2, using the same Pythagoras relation: x² = 3.12² - 1.2² = 9.7344 - 1.44 = 8.2944 x = sqrt(8.2944) = 2.88 Step 4: substitute x = 2.88, y = 1.2 and dx/dt = 0.2 (the foot moves away at a steady 0.2 m/s): dy/dt = -(2.88/1.2) x 0.2 = -2.4 x 0.2 = -0.48 The negative sign says y is decreasing, i.e. the top is moving down. The top of the ladder descends at 0.48 m/s.