A 2.6m pipe leans against a wall. Its bottom end is pulled away from the wall at 0.2 m/s. How fast is the top of the pipe sliding down the wall when its bottom is 0.6m from the wall?

Step 1: Set up the relationship between the pipe, wall, and ground. Let x be the distance of the bottom end of the pipe from the wall, and y be the height of the top end of the pipe on the wall. The length of the pipe is L = 2.6 m. These form a right-angled triangle, so by the Pythagorean theorem: x^2 + y^2 = L^2 x^2 + y^2 = (2.6)^2 Step 2: Differentiate the equation with respect to time t. Differentiating both sides with respect to t: (d)/(dt)(x^2 + y^2) = (d)/(dt)(2.6^2) 2x (dx)/(dt) + 2y (dy)/(dt) = 0 x (dx)/(dt) + y (dy)/(dt) = 0 Step 3: Find the value of y when x = 0.6 m. Given x = 0.6 m and L = 2.6 m: (0.6)^2 + y^2 = (2.6)^2 0.36 + y^2 = 6.76 y^2 = 6.76 - 0.36 y^2 = 6.4 y = sqrt(6.4) m Step 4: Substitute known values and solve for (dy)/(dt). We are given (dx)/(dt) = 0.2 m/s. We found x = 0.6 m and y = sqrt(6.4) m. Substitute these into the differentiated equation: (0.6 m)(0.2 m/s) + (sqrt(6.4) m) (dy)/(dt) = 0 0.12 m^2/s + sqrt(6.4) m (dy)/(dt) = 0 sqrt(6.4) (dy)/(dt) = -0.12 (dy)/(dt) = -(0.12)/(sqrt(6.4)) (dy)/(dt) ≈ -(0.12)/(2.5298) (dy)/(dt) ≈ -0.04743 m/s The negative sign indicates that the top end of the pipe is descending. The top end of the pipe is descending at approximately 0.0474 m/s. That's 2 down. 3 left today — send the next one.