Section 4.6: Related Rates Overview: Learning how to relate the rate of change of some aspect of a system to the rate of change of another. Explanation: Some examples of systems that can be analyzed Examples: Example 3: A Highway Chase A police cruiser, approaching a rightangled intersection from the north chasing a speeding car that has turned the corner and is now moving straight 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? Solution: Step 1: Picture and variables . We picture the car and cruiser in the coordinate plane, using the positive x axis as the eastbound highway and the positive y axis as the southbound highway. We let t represent time and set ● x = position of car at time t , ● y = position of cruiser at time t , and ● s = distance between car and cruiser at time t . We assume x, y, and s are differentiable functions of t . Step 2: Numerical information : At the instant in question, x = 0.8 mi, y = 0.6 mi, dy =60mph, ds =20 mph. dt dt Note dx/dt is negative because y is decreasing. Step 3: To find: dx/dt Step 4: How the variables are related: s2 = x2 + y 2 (We could also use s = √x2 + y 2 ) Step 5: Differentiate with respect to t. 2 s ds 2 x dx +2 y dy dt dt dt ds dt = = 1 s (x + y ) (x + y ) 1 x √ 2 +y2 dy dt dx dt dx dt dy dt Step 6: Evaluate . Use x = 0.8, y = 0.6, dy/dt =60, ds/dt =20, and solve for dx/dt . Interpret: At the moment in question, the car’s speed is 70mph. Example 4: Filling A Conical Tank 3 Water runs into a conical tank at the rate of 9 ft /min. The tank stands point down and has a height of 10ft and a base radius of 5ft. How fast is the water level rising when the water is 6ft deep? Solution: Step 1: Picture and variables: The variables in the problem are 3 ● V =volume (ft ) of the water in the tank at time t (min), ● x =radius (ft) of the surface of the water at time t , and ● y =depth (ft) of water in tank at time t . We assume V, x , and y are differentiable functions of t . The constants are the dimensions of the tank. Step 2: Numerical information. At the time in question, the volume of water increasing 9 3 ft /min (thus we say dV/dt=9) and that the height y = 6. The question asks us to find how fast the water level is increasing, in other words how fast the height is increasing → dy/dt. We also know that the dimensions of the cone holding the water are that the height = 10ft and the radius is 5 feet. Step 3: To find: dy/dt We need to find a way to relate the variables we were giving, the height, the radius and the volume and rearrange Step 4: How the variables are related: The water forms a cone with volume V = ⅓πx2 y This equation involves x as well as V and y . Because no information is given about x and dx/dt at the time in question, we need to eliminate x . The similar triangles give us a way to express x in terms of y : 5/10 = x/y → 5y /10 = x 2 Therefore, V = ⅓π(½y) y → V = ⅓π(¼y 3 ) Step 5: Differentiate with respect to t. dV /dt = ¼πy 2 dy /dt where the third power drops from the y and cancels with the ⅓ and we pop out dy/dt due to the chain rule. Step 6: Evaluate . Use y =6 and dV/dt =9 to solve for dy/dt . (dV /dt)/¼πy 2 = dy /dt → 9/(¼π62 ) = dy /dt → π = dy /dt Interpret: At the moment in question, the water level is rising at about ft/min when the water level is at 6 feet Tips: 1. Make sure to review your geometry and Trig, YES you need to know the volumes of cones and spheres and cylinders as well as the Pythagorean theorem. These things will continue to appear in later chapters. 2. These problems are scary looking, but you can do them. Just take it one step at a time and remember that a derivative is just the rate of change of something. If you lay out your knowns and unknowns and give them common variables you’ll begin to recognize what equations to use. 3. These are some of the hardest problems that are based purely on the Derivative and are not terribly common on the AP Exam. They do, however, like to throw in ones in the MC about particles moving along slopes and how far said particles are from the origin. Try to do a couple of those problems from the book, example 2 in the chapter is exactly this problem. HINT: it is just a right triangle…. relate the hypotenuse ( the distance from the origin) to the side length data they give you. 4. Triangles are terribly common so you may wish to rush up on your common special triangles, such as the 345 or the 454590. Exercises: Hauling in a Dinghy. A dinghy is pulled toward a dock by a rope from a bow though a ring on the dock 6 ft above the bow. THe rope is hauled in at the rate of 2 ft/sec. a. How fast is the boat approaching the dock when 10 ft of rope are out? b. At what rate is angle between the rope and straight down to the water changing changing at that moment? Section 5.5 Trapezoidal Rule Overview: How to more accurately estimate the area under a curve using trapezoids, as opposed to LRAM, MRAM, MRAM. Explaination: The Trapezoidal rule uses trapezoids instead of rectangles in order to estimate the area under a curve in order to more accurately map the area under a curve. When using rectangles we tend to over or underestimate the area under a curve more extremely than we do with trapezoids which have angles tops. Examples: Example 1: Applying The Trapezoidal Rule 2 Ues the Trapezoidal Rule with n =4 to estimate ∫ x2 dx . Compare the estimate 1 2 with the value of NINT ( x , x ,1,2) and with the exact value. Solution: 2 Partition [1,2] into four subintervals of equal length. Then evaluate Y=x at each partition point. Using these y values, n=4, and h= (21)/4=¼ in the Trapezoidal Rule, we have T = h2 (y 0 + 2y 1 + 2y 2 + 2y 3 + y 4 ) 36 49 T = 18 (1 + 2( 25 ) + 2( 16 ) + 2( 16 ) + 4) 16 75 T = 32 = 2.333 2 The value of NINT (x ,x,1,2) is 2. 333 . The exact value of the integral is 2 3 ∫ xb dx = x3 21 = 83 − 13 = 73 1 The T approximation overestimates the integral by about half a percent of its true value of 7/3. The percentage error is (2.343757/3)/(7/3) 0.00446. Example 2: Averaging Temperatures An observer measures the outside temperature every hour from noon until midnight, recording the temperatures in the following table. ≈ Time N Temp 63 1 2 3 4 65 66 68 70 5 6 7 8 9 10 11 M 69 68 68 65 64 62 58 55 What was the average temperature for the 12hour period? Solution: We are looking for the average value of a continuous function (temperature) from which we know values at discrete times that are one unit apart. We need to find av(f ) = 1 b f (x) dx a∫ b− a without having a formula for f(x). the integral, however, can be approximated by the Trapezoidal Rule, using the temperatures in the table as function values at the points of a 12subinterval partition of the 12hour interval (making h=1). T = h2 (y 0 + 2y 1 + 2y 2 + ∙∙∙ + 2y 11 + y 12 ) T = 12 (63 + (2 ∙ 62) + (2 ∙ 66) + ... + 2 ∙ 58 + 55) =782 b Using T to approximate ∫ f (x) dx , we have a av(f ) ≈ 1 ∙ b−a T = 1 ∙ 12 782 ≈ 65.17 Rounding to be consistent with the data given, we estimate the average temperature as 65 degrees. Tips: 1. Remember h is divided 2 2. Remember the first and last y values are not multiplied by two 3. “Memorize to the point of stabbing out your own eyeballs and then you will do well.” Szostakovik Exercises: Volume of water in a swimming pool: A rectangular swimming pool is 30 ft wide and 50 ft long. The table below shows the depth h(x) of the water at 5 ft intervals from one end of the pool to the other. Estimate the volume of water in the pool using the trapezoidal rule with n = 10, applied to the integral. 50 V = ∫ 30 • h(x) dx 0 Position (ft) Depth (ft) Position (cont.) Depth (cont.) x h(x) x h(x) 0 6.0 30 11.5 5 8.2 35 11.9 10 9.1 40 12.3 15 9.9 45 12.7 20 10.5 50 13.0 25 11
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