Drill 80(5) = 400 miles 48 (3) = 144 miles a(t) = 10 A train travels at 80 mph for 5 hours. How far does it travel? A truck travels for an average of 48 mph for 3 hours. How far does the truck travel? Beginning at a standstill, a car maintains a constant acceleration of 10 ft/s2 for 10 seconds. What is its velocity after 10 seconds? Give your answer in ft/sec and then convert to mi/hr. 80(5) = 400 miles 48 (3) = 144 miles a(t) = 10 v(t) = 10t + C, v(0) = 0, so v(t) = 10t v(10) = 100 ft/s

Estimating Finite Sums Lesson 5.1

Objectives Students will be able to approximate the area under the graph of a nonnegative continuous function by using rectangle approximation methods. interpret the area under a graph as a net accumulation of a rate of change.

Distance Traveled Figure 1: A train moves along a track at a steady rate of 75 mph from 7 AM to 9 AM. What is the total distance traveled by the train? Can be represented by determining the area of the rectangle above. The distance traveled would be 75 mph (2 hours ) = 150 miles Looking at the rectangle, A = length x width = 150 miles 75 mph Velocity (mph) Time 7 9

Solution: Graph and find the area of the shape. Area = ½ base (height) A car accelerates from 0 to 88 feet in 11 seconds. If the acceleration is CONSTANT, how far will the car have traveled in that time? Solution: Graph and find the area of the shape. Area = ½ base (height) = ½ (11)(88) = 484 feet

v(t) = 800 – 32t. Find the distance traveled between 8 and 20 seconds. Solution: Graph and find the area of the shape. The shape is a trapezoid. A = ½ h (b1 + b1 ) h = 20 -8 = 12 b1 = v(8) =544 b2 = v(20) = 160 A = ½ (12)(544 + 160) = 4224 units

Distance Traveled If the velocity varies over the timer interval [a,b], does the shaded region give the distance traveled? Yes, but because it is an irregular region, we would need to partition it into narrow rectangles and add the areas of each rectangle. Please note: the smaller the width of the rectangle, the more accurate the area. a b

Example 1 Finding Distance Traveled when Velocity Varies A particle starts at x = 0 and moves along the x-axis with velocity v(t) = 3t2 for time t > 0. Where is the particle at t = 6? Solution: Graph the function of v and partition the time interval into subintervals of length Δt. We will define Δt to be 1, which means the base of each smaller rectangle will be 1. The height of each rectangle will be v(t), where t is represented by some point within each interval along the x-axis. Our intervals will be broken down as follows: 0 1 2 3 4 5 6

Choosing values of t: When we use the left-handed endpoints, the estimate is usually less than the actual area under the curve In this case, the left-hand endpoints would be 0, 1, 2, 3, 4, 5, and to find the height, we would find v(0), v(1), v(2), etc. When we use the right-handed endpoints, the estimate is usually more than the actual area under the curve In this case, the left-hand endpoints would be 1, 2, 3, 4, 5, 6 and to find the height, we would find v(1), v(2), v(3), etc. When we use the mid-points of the intervals, the estimate is usually most accurate. In this case, the mid-points would be .5, 1.5, 2.5, 3.5, 4.5 and 5.5, and to find the height, we would find v(.5), v(1.5), v(2.5), etc

Example 1 Finding Distance Traveled when Velocity Varies A particle starts at x = 0 and moves along the x-axis with velocity v(t) = 3t2 for time t > 0. Where is the particle at t = 6? Left-Side Rectangles Velocity Time

Example 1 Finding Distance Traveled when Velocity Varies A particle starts at x = 0 and moves along the x-axis with velocity v(t) = 3t2 for time t > 0. Where is the particle at t = 6? Right-Side Rectangles Velocity Time

Example 1 Finding Distance Traveled when Velocity Varies A particle starts at x = 0 and moves along the x-axis with velocity v(t) = 3t2 for time t > 0. Where is the particle at t = 6? Midpoint Rectangles Velocity Time

Homework, day 1 Page 270: 1-6 Please note the following for 5 and 6 LRAM: use the left endpoints RRAM: use the right endpoints MRAM: use the mid-points

Drill: Find the area under the curve listed below, [0, 4] Drill: Find the area under the curve listed below, [0, 4]. Δx = 1 Determine using LRAM, RRAM and MRAM

We could estimate the area under the curve by drawing rectangles touching at their left corners. Approximate area:

We could also use a Right-hand Rectangular Approximation Method (RRAM). Approximate area:

Another approach would be to use rectangles that touch at the midpoint Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM). In this example there are four subintervals. As the number of subintervals increases, so does the accuracy. Approximate area:

Example 2 Finding Information Given Data People who sail ships at sea use dead reckoning to calculate the distance a ship has gone. (The term dead reckoning comes from “ded-reckoning,” which is short for “deduced reckoning.”) Suppose that a ship is maneuvering by changing speed rapidly. The table shows its speeds at 2-min intervals. (A knot – abbreviated kn – is a nautical mile per hour. A nautical mile is about 2000 yd.) Use this information to find the distance the ship traveled in the 12-min time interval. Time (min) 2 4 6 8 10 12 Speed (kn) 33 25 27 13 21 5 9

Example 2 Finding Information Given Data Time (min) 2 4 6 8 10 12 Speed (kn) 33 25 27 13 21 5 9 Left-Side Rectangles

Example 2 Finding Information Given Data Time (min) 2 4 6 8 10 12 Speed (kn) 33 25 27 13 21 5 9 Right-Side Rectangles

Seconds after injection5 Example 3 Dye Concentration Data (graph on p. 269) Estimate the cardiac output of the patient whose data appears above. Cardiac output = amount of dye/area under the curve Assume our patient’s amount of dye = 5.6 mg Seconds after injection5 5 41 9 11 13 15 17 19 21 23 25 27 29 31 Dye concentration 3.8 8 6.1 3.6 2.3 1.45 .91 .57 .36 .23 .14 .09

Cardiac Output To compute MRAM, we would actually need to draw the rectangles and determine the height of each rectangle. LRAM RRAM 2( 0 + 3.8 + 8 + 6.1 + 3.6 + 2.3 + 1.45 + .91 + .57 + .36 + .23 + .14 + .09) 2(27.55) = 55.1 mg/L · sec 2(3.8 + 8 + 6.1 + 3.6 + 2.3 + 1.45 + .91 + .57 + .36 + .23 + .14 + .09 + 0) 2(27.55) = 55.1 mg/L · sec 5.6 mg/ 55.1 mg/L · sec = .102 L/sec .102 L/sec X 60 sec/1 min = 6.10 L/minute

Homework Page 270: 15-19 On #15, use MRAM, meaning you will have to graph the rectangles and determine heights.