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Related rates problems involve either increasing or decreasing rates of change. They always involve two or more quantities that change with time. They may also involve constraints. Either a rate of change that is true for an interval of time or a rate of change that is true at a particular instant of time is given. The rate of change of one of the variables is given and the other rate of change is unknown. Analyze carefully the problem to determine which of these rates of change is given. Determine the relationship between the variables and find out which one can be eliminated. Eliminate the variable using your knowledge of geometry or trigonometry. The following are the suggested steps in solving related rate problems. Step 1: Draw the picture if applicable. Step 2: a. Write the given data including the known rate of change. b. Identify what rate of change is asked for in the problem. If the rate of change is decreasing, make the quantity negative. Step 3: Use your knowledge of geometry or trigonometry to be able to write one variable in terms of the other variable. Step 4: Write an equation that related to the variables. Step 5: Differentiate both sides of the equation with respect to time. The Chain Rule and Implicit Differentiation are needed in this step. Do not substitute values for variables that are changing before taking the derivative. Step 6: Substitute the known rate of change and the known quantity in the equation. Step 7: Solve for the required rate of change. Example 1: A ladder 12 meters long is leaning against the wall of building. The bottom of the ladder is sliding away from the wall at the rate of 1.5 m/s. How fast is the top of the ladder sliding down when it is 3 meters above the ground?Solution: 1 - 12 m 4 = 3m 1.5 m a. y = 3 meters z = 12 meters - = 1.5meters dt dy dt - =? b. x2 + y2 = 22 x2 + (3 ) 2 = (12) 2 x2+9= 144 x2 = 144 -9 x2 = 135 x = V135 Differentiate: C. x2+ y2 = 22 2x -+ 2y- dy = 2z at at dz dx + y at at dy = 2 It Substitute: 135(1.5) + 3 dt = 12(0) 135 (1.5) 3 dy dt) 3 dy -=-5.809 m/s dt It is negative because the ladder is moving down towards zero height.Example 5: A plastic container in the shape of a cylinder is being filled with coconut juice at the rate of 450T cubic centimeters per second. If the radius of the plastic container is 20 cm, how fast is the height of the coconut juice rising? Solution: 20 cm 20 cm r = 20 cm dv = 450m cm3/s dt dh =? V = Trzh V = 1 (20) 2 h V = 400Th dt Differentiate: Substitute: dV dh dh = 400n- 450m = 400m- dt dt dt dh 450T dt 400TC 9 =-cm3/sSolve the following problems. 1. A ladder 10 meters long is leaning against the wall of a building. The base of the ladder is sliding away from the wall at a rate of 2 m/s. How fast is the top of the ladder moving down the wall when the base of the ladder is 6 meters away from the wall? E 2. A right cylindrical tank filled with water has a radius of 2 meters. If the water is being drained at the rate of 10 cubic meters per minute, how fast is the height of the water decreasing