Report

By: Courtney Kent Carpentry/PDT A carpenter is cutting down a 200 foot tree on a construction site; the sun is directly overhead. At the moment when the tree makes an angle of 30° with the horizontal, its shadow is lengthening at the rate of 50 feet/sec. How fast is the angle changing at that moment? Step1: Draw a diagram portraying the problem. Step 2: Fill in the diagram with the given information. The rate at which the shadow is lengthening is equal to the derivative of side A. The tree is a constant , it remains 200ft. The tree is equal to side C. The angle created when side A and C meet is 30°. What the problem is trying to ask for is dθ/dt. B 30° 50ft/sec A cos(θ)=a/c (-sin θ)(dθ/dt)= (c)(da/dt)- (a)(dc/dt) (c²) (-sin θ)(dθ/dt)= (c)(da/dt) (c²) (-sin θ)(dθ/dt)= 200(50) (200²) (-sin θ)(dθ/dt)= 1 4 (-sin (30))(dθ/dt)= 1 4 (0.988)(dθ/dt)= 1 4 (0.988)(dθ/dt)= (0.25) (0.988) (0.988) (dθ/dt)= 0.253° sec Step3:Use the derivative of cosine to find the missing information. Chain rule (first derivative time the derivative of what is on the inside) is used in this process as well as the quotient rule (low dihigh- high dilow/ the square of what is below) The second part of the chain rule(A•dc/dt) cancels because A is a constant and therefore does not have a derivative. Step 4:Solve the equation for dθ/dt. At the moment when the tree makes an angle of 30° with the horizontal, and its shadow is lengthening at the rate of 50 feet/sec the angle changing 0.253° per second at that moment. A construction worker pulls a five meter plank up the side of a building under construction by means of a rope tied to one end of the plank. Assume the opposite end of the plank follows a path perpendicular to the wall of the building and the worker pulls the rope at a rate of 0.15 meters per second. How fast is the end of the plank sliding along the ground when it is 2.5 meters from the wall of the building? Step1: Draw a diagram portraying the problem. Step 2: Fill in the diagram with the given information. The distance from the building, 2.5 meters, is side A. The rate at which the rope is being pulled is 0.15 meters per second. This is the derivative of side B The plank is equal to side C. The plank is a constant , it remains 5 meters. What the problem is trying to ask for is the derivative of side A. 0.15m/sec B 2.5 meters A (a)²+ (b)²= (c)² (2.5)²+ (b)²= (5)² 6.25 + (b)²= 25 -6.25 -6.25 (sqrt)(b)²= (sqrt of 18.75) b = 4.330127 (a)²+ (b)²= (c)² (2a)(da/dt)+(2b)(db/dt)=(2c)(dc/dt) (2(2.5)(da/dt)+(2(4.330127)(0.15)=(2(5)(0) 5(da/dt) + 1.3 = 0 - 1.3 - 1.3 5(da/dt)= -1.3 5 5 (da/dt)= -0.26 m/sec Step3:Use the Pythagorean Theorem to find side B. Step4:Use the derivative of the Pythagorean Theorem to find the missing information. Chain rule (first derivative time the derivative of what is on the inside) is used in this process. The C part of the Pythagorean Theorem when solving for a derivative cancels out. C is a constant and therefore does not have a derivative. Step 5:Solve the equation for dA/dt. The end of the plank sliding along the ground at a decreasing rate of 0.26 meters per second when it is 2.5 meters from the wall of the building.