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in terms of that of another quantity.

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1 in terms of that of another quantity.
Related Rates If we are pumping air into a balloon, both the volume and the radius of the balloon are increasing and their rates of increase are related to each other. However, it is much easier to measure directly the rate of increase of the volume than the rate of increase of the radius. In this section, we will learn: How to compute the rate of change of one quantity in terms of that of another quantity. In a related-rates problem, the idea is to compute the rate of change of one quantity in terms of the rate of change of another quantityβ€”which may be more easily measured. The procedure is to find an equation that relates the two quantities and then use the Chain Rule to differentiate both sides with respect to time.

2 Example 1: Consider a sphere of radius 10cm. If the radius changes 0.1cm (a very small amount) how much does the volume change? π‘‘π‘Ÿ=0.1 π‘π‘š 𝑑𝑉=? The volume would change by approximately

3 Example 2: Now, suppose that the radius is changing at an instantaneous rate of 0.1 cm/sec. At what rate is the sphere growing when r = 10 cm? π‘‘π‘Ÿ 𝑑𝑑 =0.1π‘π‘š/ sec 𝑑𝑉 𝑑𝑑 =? The sphere is growing at a rate of

4 Air is being blown into a sphere at the rate of 6 cubic inches per minute.
How fast is the radius changing when the radius of the sphere is 2 inches?

5 Example: Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm3/s. How fast is the radius of the balloon increasing when the diameter is 50 cm? 𝑑𝑉 𝑑𝑑 =100 π‘π‘š 3 𝑠 π‘‘π‘Ÿ 𝑑𝑑 =? π‘€β„Žπ‘’π‘› π‘Ÿ=25 π‘π‘š 𝑑𝑉 𝑑𝑑 = 𝑑𝑉 π‘‘π‘Ÿ π‘‘π‘Ÿ 𝑑𝑑 =4πœ‹ π‘Ÿ 2 π‘‘π‘Ÿ 𝑑𝑑 π‘‘π‘Ÿ 𝑑𝑑 = 1 4πœ‹ π‘Ÿ 2 𝑑𝑉 𝑑𝑑 β†’ π‘‘π‘Ÿ 𝑑𝑑 = 1 4πœ‹ = 1 25πœ‹ The radius of the balloon is increasing at the rate of 1/(25Ο€) β‰ˆ cm/s.

6 How fast is the surface dropping?
Example 4: Water is draining from a cylindrical tank at 3 liters/second. How fast is the surface dropping? = ? r is constant

7 How fast is the balloon rising?
Example 4: Hot Air Balloon Problem @ How fast is the balloon rising?

8 𝑑=𝑣𝑑 2𝑧 𝑑𝑧 𝑑𝑑 =2π‘₯ 𝑑π‘₯ 𝑑𝑑 +2𝑦 𝑑𝑦 𝑑𝑑 5 𝑑𝑧 𝑑𝑑 =4βˆ™40+3βˆ™30 𝑑𝑧 𝑑𝑑 =50 π‘šπ‘–/β„Žπ‘Ÿ
Example 5: Truck A travels east at 40 mi/hr. Truck B travels north at 30 mi/hr. How fast is the distance between the trucks changing 6 minutes later? @ 6 min x= 4 mi & y = 3 mi from initial position 𝑑=𝑣𝑑 z2 = x2 + y @ 6 min z = 5 mi 2𝑧 𝑑𝑧 𝑑𝑑 =2π‘₯ 𝑑π‘₯ 𝑑𝑑 +2𝑦 𝑑𝑦 𝑑𝑑 5 𝑑𝑧 𝑑𝑑 =4βˆ™40+3βˆ™30 𝑑𝑧 𝑑𝑑 =50 π‘šπ‘–/β„Žπ‘Ÿ

9 𝑑𝑧 𝑑𝑑 =? 2𝑧 𝑑𝑧 𝑑𝑑 =2π‘₯ 𝑑π‘₯ 𝑑𝑑 +2𝑦 𝑑𝑦 𝑑𝑑
Example 6: Car A is traveling west at 50 mi/h and car B is traveling north at 60 mi/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection? dx / dt = –50 mi/h & dy / dt = –60 mi/h. both x & y are decreasing 𝑑𝑧 𝑑𝑑 =? 2𝑧 𝑑𝑧 𝑑𝑑 =2π‘₯ 𝑑π‘₯ 𝑑𝑑 +2𝑦 𝑑𝑦 𝑑𝑑 z2 = x2 + y2 β‡’ When x = 0.3 mi & y = 0.4 mi, z = 0.5 mi.


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