A 2.6 m long pipe leans against a wall. If the bottom of the pipe is pulled away from the wall at a rate of 0.2 m/s, how fast is the area of the triangle formed by the pipe, wall, and ground changing

Step 1: Define variables and establish relationships. Let L be the length of the pipe, x be the distance of the lower end from the wall, and y be the height of the top end on the wall. Given: • Pipe length L = 2.6 m (constant). • Rate at which the lower end is pulled away (dx)/(dt) = 0.2 m/s. • We need to find the rate of change of the area (dA)/(dt) when x = 1 m. The pipe, wall, and ground form a right-angled triangle, so by the Pythagorean theorem: x^2 + y^2 = L^2 The area A of this triangle is: A = (1)/(2)xy Step 2: Find the height y when x = 1 m. Substitute x = 1 m and L = 2.6 m into the Pythagorean theorem: 1^2 + y^2 = (2.6)^2 1 + y^2 = 6.76 y^2 = 6.76 - 1 y^2 = 5.76 y = sqrt(5.76) = 2.4 m Step 3: Differentiate the Pythagorean theorem with respect to time t to find (dy)/(dt). 2x(dx)/(dt) + 2y(dy)/(dt) = 0 x(dx)/(dt) + y(dy)/(dt) = 0 Now, substitute x = 1 m, y = 2.4 m, and (dx)/(dt) = 0.2 m/s: (1)(0.2) + (2.4)(dy)/(dt) = 0 0.2 + 2.4(dy)/(dt) = 0 2.4(dy)/(dt) = -0.2 (dy)/(dt) = -(0.2)/(2.4) = -(2)/(24) = -(1)/(12) m/s Step 4: Differentiate the area formula with respect to time t. Using the product rule: (dA)/(dt) = (1)/(2) ( (dx)/(dt)y + x(dy)/(dt) ) Step 5: Substitute all known values into the differentiated area formula. Substitute x = 1 m, y = 2.4 m, (dx)/(dt) = 0.2 m/s, and (dy)/(dt) = -(1)/(12) m/s: (dA)/(dt) = (1)/(2) ( (0.2 m/s)(2.4 m) + (1 m)(-(1)/(12) m/s) ) (dA)/(dt) = (1)/(2) ( 0.48 m^2/s - (1)/(12) m^2/s ) Step 6: Calculate the rate of change of the area. To simplify, convert 0.48 to a fraction: 0.48 = (48)/(100) = (12)/(25). (dA)/(dt) = (1)/(2) ( (12)/(25) - (1)/(12) ) Find a common denominator for 25 and 12, which is 300: (dA)/(dt) = (1)/(2) ( (12 × 12)/(25 × 12) - (1 × 25)/(12 × 25) ) (dA)/(dt) = (1)/(2) ( (144)/(300) - (25)/(300) ) (dA)/(dt) = (1)/(2) ( (144 - 25)/(300) ) (dA)/(dt) = (1)/(2) ( (119)/(300) ) (dA)/(dt) = (119)/(600) m^2/s The rate at which the area is changing is (119)/(600) m^2/s. That's 2 down. 3 left today — send the next one.