Section 3.5 – Related Rates ADV Calculus Section 3.5 – Related Rates

Related Rates An equation that relates two or more variables and their corresponding rates of change over time. The Chain Rule is VERY IMPORTANT for this!!!

Example 1 Assume that the radius r of a sphere is a differentiable function of t, and let V be the volume of the sphere. Find an equation that relates and . Remember, 𝑑𝑉 𝑑𝑡 =4𝜋 𝑟 2 𝑑𝑟 𝑑𝑡

Example 2 Assume that the radius r and height h of a cone are differentiable functions of t, and let V be the volume of the cone. Find an equation that relates , and . Remember, Hint: Must use product rule! 𝑑𝑉 𝑑𝑡 = 2 3 𝜋𝑟ℎ 𝑑𝑟 𝑑𝑡 + 1 3 𝜋 𝑟 2 𝑑ℎ 𝑑𝑡

Strategy for Solving Related Rate Problems 1. Understand the problem. Identify the variable whose rate of change you seek and the variables whose rate of change you know. 2. Develop a mathematical model of the problem. Draw a picture and label the parts. Make sure to distinguish between constant quantities and variables that change over time. 3. Write an equation relating the variable whose rate of change you seek with the variables you know. 4. Differentiate both sides of the equation implicitly with respect to time t. 5. Substitute values for any quantities that depend on time. (only do this after you differentiate!) 6. Interpret the solution.

Example 3 A police cruiser, approaching a right-angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving due east. When the cruiser is 0.6 mi north of the intersection and the car is 0.8 mi to the east, the police determine with radar that the distance between them and the car is increasing at 20 mph. If the cruiser is moving at 60 mph at the instant of measurement, what is the speed of the car? x = position of car at time t y = position of cruiser at time t s = distance between car and cruiser at time t

Example 3 𝑥=0.8 𝑚𝑖, 𝑦=0.6 𝑚𝑖, 𝑑𝑦 𝑑𝑡 =−60 𝑚𝑝ℎ, 𝑑𝑦 𝑑𝑡 =20 𝑚𝑝ℎ 𝑠= 𝑥 2 + 𝑦 2 = 0.8 2 + 0.6 2 =1 𝑠 2 = 𝑥 2 + 𝑦 2 2𝑠 𝑑𝑠 𝑑𝑡 =2𝑥 𝑑𝑥 𝑑𝑡 +2𝑦 𝑑𝑦 𝑑𝑡 𝑠 𝑑𝑠 𝑑𝑡 =𝑥 𝑑𝑥 𝑑𝑡 +𝑦 𝑑𝑦 𝑑𝑡 1 20 = 0.8 𝑑𝑥 𝑑𝑡 + 0.6 −60 𝑑𝑥 𝑑𝑡 =70𝑚𝑝ℎ

Example 4 Water runs into a conical tank at the rate of 9 ft3/min. The tank stands point down and has a height of 10 feet and a base radius of 5 feet. How fast is the water level rising when the water is 6 feet deep? ℎ 2 =𝑟 𝑉= 1 3 𝜋 𝑟 2 ℎ= 1 3 𝜋 ℎ 2 2 ℎ= 𝜋 ℎ 3 12 𝑑𝑉 𝑑𝑡 = 𝜋 ℎ 2 4 𝑑ℎ 𝑑𝑡 9= 𝜋 6 2 4 𝑑ℎ 𝑑𝑡 𝑑ℎ 𝑑𝑡 ≈0.318𝑓𝑡/𝑚𝑖𝑛