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2.6 Related Rates

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**After this lesson, you should be able to:**

Find a related rate. To determine how variables change with respect to time. Use related rates to solve real-life problems.

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**Related Rates In real-life, we meet those kind of questions often:**

1. How fast is the ladder slipping down the wall? 2. How fast is the shadow moving? 3. How quickly is the angle decreasing? 4. How fast is the height changing in the water tank? 5. How fast is the area/volume changing?

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When water is drained out of a conical tank, the volume V, the radius r, and the height h of the water level are all functions of time t. We know that all the variables are related by the volume formula: The rate of change of V is related to the rates of change both r and h

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Procedure: Sketch a picture and label constants and all values that vary List all given rates and note if they are increasing (+) or decreasing (-). Write an equation relating the quantities with the unknown rate of change with the given rates of change. CR – Differentiate with respect to time & solve for the unknown rate Substitute the given values in and simplify Don’t forget to use units!!! Also note: If the quantity increases: + answer If the quantity decreases: - answer Change in distance? m/s Change in area? m2/s Change in volume? m3/s

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Related Rates Example 1 Suppose x and y are both differentiable functions of t and are related by the equation y = x Find dy/dt, given that dx/dt = 2 when x = 1 Solution When x = 1 and dx/dt = 2, then

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Related Rates Example 2 Suppose air is being pumped into a spherical balloon at the rate of 10 cubic centimeters per minute. How fast is the radius of the balloon increasing when the radius is 5 cm? Solution Let V be the volume of the balloon and r be its radius. We know that all the variables are related by the volume formula: or,

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Related Rates Example 2 Suppose air is being pumped into a spherical balloon at the rate of 10 cubic centimeters per minute. How fast is the radius of the balloon increasing when the radius is 5 cm? Solution Since the volume is increasing at a rate of 10 cm3/min, the rate of change of the volume is When r = 5, the rate of change of the radius is

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Related Rates Example 3 A pebble is dripped into a calm pond, causing in the form of concentric circles. The radius r of the outer ripple is increasing at a rate of 1 feet per second. When the radius is 4 feet, at what rate is the total area A of the disturbed water changing? Solution Note that the disturbed water area A and radius r are related by the formula:

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Related Rates Example 3 A pebble is dripped into a calm pond, causing in the form of concentric circles. The radius r of the outer ripple is increasing at a rate of 1 feet per second. When the radius is 4 feet, at what rate is the total area A of the disturbed water changing? Solution Since the radius of outer ripple is increasing at a rate of 1 ft3/sec, the rate of change of the radius is Then,

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Related Rates Example 4 A patrol car is parked 50 feet from a long warehouse. The revolving light on top of the car turns at a rate of 30 revolutions per minute. How fast is the light beam moving along the wall when the beam makes angles of (a) = 30o (b) = 60o and (c) = 70o with the line perpendicular from the light to the wall? 50 x

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**Related Rates Solution**

The information that “30 revolution per minute” means 50 x The relationship among the variables is Therefore, or,

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Related Rates Solution (a) = 30o 50 x (b) = 60o (c) = 70o

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Related Rates Read and understand the Example 6 on P 153

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Related Rates Example 5 A winch at the tip of a 12-meter building pulls pipe of the same length to a vertical position. The winch pulls in a rope at a rate of – 0.2 meter/sec. Find the rate of vertical change and the rate of horizontal change at the end of the pipe when y = 6 s (x, y) 12 12 Solution The relationships among the three variables are: and When y = 6 =12/2, and

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**Related Rates and Solution and s (x, y) 12**

Taking the derivative with respect of t to the above 2 equations, we have 12 (1) (2)

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**Related Rates and Solution Equation (1) – (2), we have s (x, y) or 12**

m/s From Equation (1), we have m/s

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Homework Section 2.6 page 154 #xxxx

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**Related Rates and Solution and s (x, y) 12**

Taking the derivative with respect of t to the above 2 equations, we have 12

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and Related Rates Solution s (x, y) 12 12

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Differentiation: Related Rates – Day 1

Differentiation: Related Rates – Day 1

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