1. Related Rates
Chapter 5.5

2. What is Changing?

3. Related Rates
In the previous animation, the following parameters were changing
The radius of the circle
The area of the circle
Time
Note that we can consider the rate of change of, say, the area of the circle with respect to its radius: 𝑑𝐴 𝑑𝑟
But the area is also changing with time; is there a way to find the rate of change of area with respect to time: 𝑑𝐴 𝑑𝑡 ?

4. Related Rates
The same can be said of the radius: it is changing with respect to time
Can we find the rate of change of the radius with respect to time: 𝑑𝑟 𝑑𝑡 ?
Although it was imperceptible in the animation, the radius was related to time by the function 𝑟(𝑡)= 𝑡 2
Hence, given 𝐴=𝜋 𝑟 2 , we have:
𝑑𝐴 𝑑𝑟 =2𝜋𝑟
𝑑𝑟 𝑑𝑡 =2𝑡

5. Related Rates Hence, given 𝐴=𝜋 𝑟 2 , we have:
𝑑𝐴 𝑑𝑟 =2𝜋𝑟
𝑑𝑟 𝑑𝑡 =2𝑡
Can we use these two results to find 𝑑𝐴 𝑑𝑡 ?

6. Related Rates Note that multiplying these results gives 𝑑𝐴 𝑑𝑟 ⋅ 𝑑𝑟 𝑑𝑡
Now recall that, by the Chain Rule 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑢 ⋅ 𝑑𝑢 𝑑𝑥
Applying the Chain Rule 𝑑𝐴 𝑑𝑡 = 𝑑𝐴 𝑑𝑟 ⋅ 𝑑𝑟 𝑑𝑡

7. Related Rates
However, we may not have a function that explicitly relates radius to time; it is enough to know that the radius is a function of time (and, therefore, so is area a function of time)
In that case, we apply the Chain Rule “implicitly”
𝐴=𝜋 𝑟 2
𝑑 𝑑𝑡 𝐴 = 𝑑 𝑑𝑡 [𝜋 𝑟 2 ]

8. Related Rates In that case, we apply the Chain Rule “implicitly”
𝐴=𝜋 𝑟 2
𝑑 𝑑𝑡 𝐴 = 𝑑 𝑑𝑡 [𝜋 𝑟 2 ]
𝑑𝐴 𝑑𝑡 =2𝜋𝑟⋅ 𝑑𝑟 𝑑𝑡
We must think of 𝑟 as 𝑟(𝑡), an inner function of 𝑡

9. Example 1: Finding Related Rate Equations
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 𝑑𝑉 𝑑𝑡 and 𝑑𝑟 𝑑𝑡 .
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 𝑑𝑉 𝑑𝑡 , 𝑑ℎ 𝑑𝑡 , and 𝑑𝑟 𝑑𝑡

10. Example 1: Finding Related Rate Equations
Since 𝑉= 4𝜋 3 𝑟 3 and 𝑟(𝑡) is a function of time, then we can think of the volume function (also of time) as
𝑉 𝑡 = 4𝜋 3 𝑟 𝑡 3
Note that taking the derivative requires that we take the derivative of the “inner” function, 𝑟
𝑑𝑉 𝑑𝑡 = 4𝜋 3 ⋅3⋅ 𝑑𝑟 𝑑𝑡 𝑟 𝑡 2 =4𝜋 𝑟 2 ⋅ 𝑑𝑟 𝑑𝑡

11. Example 1: Finding Related Rate Equations
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 𝑑𝑉 𝑑𝑡 , 𝑑ℎ 𝑑𝑡 , and 𝑑𝑟 𝑑𝑡 .
Both 𝑟 and ℎ are functions of time
𝑉 𝑡 = 𝜋 3 𝑟 𝑡 2 [ℎ 𝑡 ]
Note that we must use the Product Rule

12. Example 1: Finding Related Rate Equations
𝑉 𝑡 = 𝜋 3 𝑟 𝑡 2 [ℎ 𝑡 ] 𝑑𝑉 𝑑𝑡 = 𝜋 3 𝑟 𝑡 2 ⋅ 𝑑ℎ 𝑑𝑡 +2⋅ 𝜋 3 ⋅ 𝑑𝑟 𝑑𝑡 𝑟 𝑡 ℎ 𝑡 𝑑𝑉 𝑑𝑡 = 𝜋 3 𝑟 2 ⋅ 𝑑ℎ 𝑑𝑡 + 2𝜋 3 𝑟ℎ⋅ 𝑑𝑟 𝑑𝑡

13. Solution Strategy for Related Rates Problems
Understand the problem: in particular, identify the variable whose rate of change you seek and the variable (or variables) whose rate of change you know.
Develop a mathematical model of the problem: draw a picture and label the parts that are important to the problem. Be sure to distinguish constant quantities from variables that change over time. Only constant quantities can be assigned numerical values at the start.

14. Solution Strategy for Related Rates Problems
Write an equation relating the variable whose rate of change you seek with the variable(s) whose rate of change you know: the formula is often geometric, but it could come from a scientific application
Differentiate both sides of the equation implicitly with respect to time t: be sure to follow the differentiation rules. The Chain Rule will be especially critical, as you will be differentiating with respect to the parameter t

15. Solution Strategy for Related Rates Problems
Substitute values for any quantities that depend on time: notice that it is only safe to do this after the differentiation step. Substituting too soon “freezes the picture” and makes changeable variables behave like constants, with zero derivatives.
Interpret the solution: translate your mathematical result into the problem setting (with appropriate units) and decide whether the result makes sense.

16. Example 2: A Rising Balloon
A hot-air balloon rising straight up from a level field is tracked by a range finder 500 feet from the lift-off point. At the moment the range finder’s elevation angle is 𝜋 4 , the angle is increasing at the rate of radians per minute. How fast is the balloon rising at that moment?

17. Example 2: A Rising Balloon
Understand the problem
Let 𝜃 be the angle of elevation and 𝑦 be the height of the balloon.
We are given that 𝑑𝜃 𝑑𝑡 =0.14 radians per minute
We are asked to find 𝑑𝑦 𝑑𝑡 , with units of feet per minute

18. Example 2: A Rising Balloon
Develop a mathematical model of the problem.

19. Example 2: A Rising Balloon
Write an equation relating the variable whose rate of change you seek with the variable(s) whose rate of change you know.
Since we need to incorporate the rate of change of the angle and find the rate of change of the height, we can use the tangent function
tan 𝜃 = 𝑦 500
𝑦=500 tan 𝜃

20. Example 2: A Rising Balloon
Differentiate both sides of the equation implicitly with respect to time, t.
𝑦=500 tan 𝜃
𝑑𝑦 𝑑𝑡 =500 sec 2 𝜃 ⋅ 𝑑𝜃 𝑑𝑡

21. Example 2: A Rising Balloon
Substitute values for any quantities that depend on time.
At this point you many have to find one or more variables that may not have been provided explicitly. We were given that 𝑑𝜃 𝑑𝑡 =0.14 when 𝜃= 𝜋 4 , so we can now substitute
𝑑𝑦 𝑑𝑡 =500 sec 2 𝜋 =140

22. Example 2: A Rising Balloon
Interpret the solution
At the instant when the angle of elevation of the range finder is 𝜋 4 radians, the balloon is rising at a rate of 140 feet per minute.
Note that we do not know the time when this occurs (nor do we need to know it)

23. Example 3: A Highway Chase
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 straight east. When the cruiser is 0.6 miles north of the intersection and the car is 0.8 miles 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?

24. Example 3: A Highway Chase
If x and y represent the distance from the intersection of the cruiser and speeder, respectively, and if D represents the distance between them, then 𝑑𝐷 𝑑𝑡 =20, 𝑑𝑥 𝑑𝑡 =−60, 𝑑𝑦 𝑑𝑡 =? We are also given that 𝑥=0.6, 𝑦=0.8 We relate these with the Pythagorean Theorem 𝑥 2 + 𝑦 2 = 𝐷 2

25. Example 3: A Highway Chase
𝑥 2 + 𝑦 2 = 𝐷 2 Differentiate implicitly with respect to time 2𝑥⋅ 𝑑𝑥 𝑑𝑡 +2𝑦⋅ 𝑑𝑦 𝑑𝑡 =2𝐷⋅ 𝑑𝐷 𝑑𝑡 𝑑𝑦 𝑑𝑡 = 1 𝑦 𝐷⋅ 𝑑𝐷 𝑑𝑡 −𝑥⋅ 𝑑𝑥 𝑑𝑡 Note that D was not given in the problem, but we can use the Pythagorean theorem to find its value at this instant in time: 𝐷=1

26. Example 3: A Highway Chase
𝑑𝑦 𝑑𝑡 = 1 𝑦 𝐷⋅ 𝑑𝐷 𝑑𝑡 −𝑥⋅ 𝑑𝑥 𝑑𝑡 Substitute: 𝑑𝑦 𝑑𝑡 = ⋅20− 0.6 −60 𝑑𝑦 𝑑𝑡 =70 The car was travelling at 70 mph at that instant in time

27. Example 4: Filling a Conical Tank
Water runs into a conical tank at the rate of 9 cubic feet per minute. 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?

28. Example 4: Filling a Conical Tank
We are given that 𝑑𝑉 𝑑𝑡 =9 and are asked to find 𝑑ℎ 𝑑𝑡 , where h is the height of the water in the tank at time t. The volume is 𝑉= 𝜋 3 𝑟 2 ℎ Since no information is provided concerning the rate of change of the radius, we must find an equation relating radius and height.

29. Example 4: Filling a Conical Tank
Note that the ratio of height to radius is always the same (2:1)

30. Example 4: Filling a Conical Tank
We have ℎ 𝑟 = 2 1 so 𝑟= 1 2 ℎ Our volume formula becomes 𝑉= 𝜋 ℎ 2 ℎ= 𝜋 12 ℎ 3 Now differentiate implicitly with respect to t 𝑑𝑉 𝑑𝑡 = 𝜋 12 ⋅3 ℎ 2 ⋅ 𝑑ℎ 𝑑𝑡 𝑑ℎ 𝑑𝑡 = 4 𝜋 ℎ 2 ⋅ 𝑑𝑉 𝑑𝑡

31. Example 4: Filling a Conical Tank
𝑑ℎ 𝑑𝑡 = 4 𝜋 ℎ 2 ⋅ 𝑑𝑉 𝑑𝑡 Substitute: 𝑑ℎ 𝑑𝑡 = 4 36𝜋 ⋅9= 1 𝜋 ≈0.318 The height is rising at the rate of approximately feet per minute at the instant when the height is 6 feet.

32. Example 5: Filling a Water Trough
A trough is 12 feet long and 3 feet across the top. Its ends are isosceles triangles with altitudes of 3 feet.
If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is one foot deep?
If the water is rising at a rate of 3/8-inch per minute when ℎ=2, determine the rate at which water is being pumped into the trough.

33. Example 5: Filling a Water Trough
A trough is 12 feet long and 3 feet across the top. Its ends are isosceles triangles with altitudes of 3 feet.
If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is one foot deep? foot per minute
If the water is rising at a rate of 3/8-inch per minute when ℎ=2, determine the rate at which water is being pumped into the trough. 9 cubic inches per minute

34. Example 6: Baseball
A baseball diamond has the shape of a square with sides 90 feet long. A player is running from first to second base at a speed of 28 feet per second. Find the rate at which the distance from home plate is changing when the player is 30 feet from second base.

35. Example 6: Baseball
A baseball diamond has the shape of a square with sides 90 feet long. A player is running from first to second base at a speed of 28 feet per second. Find the rate at which the distance from home plate is changing when the player is 30 feet from second base. ≈ feet per second

36. Example 7: Air Traffic Control
An air traffic controller spots two planes at the same altitude converging on a point as they fly at right angles to each other. One plane is 150 miles from the point moving at 450 miles per hour. The other plane is 200 miles from the point moving at 600 miles per hour.
At what rate is the distance between the planes decreasing?
How much time does the air traffic controller have to get one of the planes on a different flight path?

37. Example 7: Air Traffic Control
An air traffic controller spots two planes at the same altitude converging on a point as they fly at right angles to each other. One plane is 150 miles from the point moving at 450 miles per hour. The other plane is 200 miles form the point moving at 600 miles per hour.
At what rate is the distance between the planes decreasing? −750 mph
How much time does the air traffic controller have to get one of the planes on a different flight path? 20 minutes

38. Example 8: Shadow Length
A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground. When he is 10 feet from the base of the light
at what rate is the tip of his shadow moving?
at what rate is the length of his shadow changing?

39. Example 8: Shadow Length
A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground. When he is 10 feet from the base of the light
at what rate is the tip of his shadow moving? feet per second
at what rate is the length of his shadow changing? feet per second

40. Exercise 5.6
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