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Differentials ∆ When we first started to talk about derivatives, we said that ∆ becomes when the change in x and change in y become very small. dy can be considered a very small change in y. dx can be considered a very small change in x. Let y f x be a differentiable function. The differential d x is an independent variable. The differential dy is: dy f x dx dy = 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 quantity—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. Example 1: Consider a sphere of radius 10cm. If the radius changes 0.1cm (a very small amount) how much does the volume change? = 0.1 V 4 r =? 3 3 dV 4 r dr 2 dV 4 10cm 0.1cm 2 dV 40 cm 3 The volume would change by approximately 40 cm 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? = 0.1/ sec V 4 r =? 3 3 dV 4 r 2 dt dV dt dV dt dr dt 4 10cm 40 cm 2 cm 0.1 sec 3 sec The sphere is growing at a rate of 40 cm 3 / sec Example 3: 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? V 4 r 3 3 dr dt 1 4 (25) 2 100 1 25 The radius of the balloon is increasing at the rate of 1/(25π) ≈ 0.0127 cm/s. Example 4: Water is draining from a cylindrical tank at 3 liters/second. How fast is the surface dropping? dV L 3 dt 3000 V r h r dt dh dt 2 r is constant dh dt 3000 3 sec sec 2 dV cm cm 3 sec r 2 dh dt =? Steps for Related Rates Problems: 1. Draw a picture (sketch). 2. Write down known information. 3. Write down what you are looking for. 4. Write an equation to relate the variables. 5. Differentiate both sides with respect to t. 6. Evaluate. Example 4: Hot Air Balloon Problem @ d 4 dt rad 0.14 m in h How fast is the balloon rising? h tan 500ft 500 d sec 2 dt sec 4 2 2 1 dh 500 dt 2 0.14 1 dh 500 dt 0.14 500 dh dt 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 z2 = x2 + y2 @ 6 min z = 5 mi 2 = 2 + 2 5 = 4 ∙ 40 + 3 ∙ 30 = 50 /ℎ = 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 =? z2 = x2 + y2 ⇒ 2 = 2 + 2 When x = 0.3 mi & y = 0.4 mi, z = 0.5 mi.