Drill: Determine the intervals of increase and decrease by using the first derivative method.

1. Drill: Determine the intervals of increase and decrease by using the first derivative method. • F(x) = x2 + 1 • F(x) = x2 – 2x + 3 • F(x) = 2x3 - 3x2 + 1 F’(x) = 2x 0 = 2x, x = 0 F’(x) = 2x – 2 0 = 2x – 2 ; x = 1 F’(x) = 6x2 – 6x = 6x (x – 1) 0 = 6x (x – 1); x = 0, x = 1 - + 1 + - + 0 1 - + 0

2. 7.1: Integral at Net Change Day #1 Homework, p 386:1-8 Day #2 Homework, p. 386: 9-29 (odd)

3. What we’ll learn about… • Linear motion revisited • General strategy • Consumption over time • Net change from data • Work • WHY???? The integral can be used to calculate net change and total accumulation.

4. Interpreting a Velocity Function. • Describe the motion of which shows the velocity of a particle moving along a horizontal s-axis for 0 < t < 5. • Solve graphically. • V(0) = -8 • Initial velocity of 8 cm/sec, to the left. • V(1.25) = 0 • Slows 0 cm/sec at 1.25 seconds • V(t)>0 • it moves to the right with increasing speed. • V(5) = 24.8 • Reaches a velocity of 24.8 cm/sec at the end.

5. Finding Position from DisplacementSuppose the initial position of the particle in the previous example is s(0) = 9. What is the particle’s position at t = 1 second? Let u = (t+1), du = dt • The position at t = 1 is the initial position s(0) + the displacement that the position changed from t = 0 to t = 1. (Δs) • When velocity is constant, displacement = rate X time • However, in our case, the velocity varies. Final position = initial position + displacement: 9 – 11/3 = 16/3

6. Same problem, but what if t = 5 seconds? • We can model the displacement in the same way. Final position = initial position +displacement: 9 +35= 44

7. To Check Graphically You could use parametric equations to show the previous answers you got through analytic methods are correct. You can then use trace to determine the position at time t.

8. Calculating Total Distance Traveled • Find the total distance traveled by the particle in the example. (remember: 0 < t < 5) • We could partition the time interval to indicate the shift from negative to positive and evaluate, taking the absolute value of the negative integral.

9. Using the Calculator

10. DrillEvaluate the following integrals using integration by parts, or substitution Let u = x du = dx dv = sinxdx v = -cosx Let u = x2 du = 2x dx du/2x = dx • x5ln(x)dx • xsin(x)dx • xsin(x2)dx Let u = ln x du = (1/x)dx dv = x5dx v = x6/6

11. Modeling the Effects of Acceleration • A car moving with initial velocity of 5 mph accelerates at the rate of a(t) = 2.4t mph per second for 8 seconds. How fast is the car going when the 8 seconds are up? • Step 1: Approximate the net change in velocity as a Riemann sum. When acceleration is constant, • Velocity change = acceleration X time applied • Step 2: Write a definite integral • Step 3: Evaluate your definite integral: • If the initial velocity is 5 mph then 5 + 76.8 = 81.8 mph

12. Modeling the Effects of Acceleration • A car moving with initial velocity of 5 mph accelerates at the rate of a(t) = 2.4t mph per second for 8 seconds. How far did the car travel in those 8 seconds? distance = 244.8 mph X seconds Convert to miles by multiplying by 1/3600 = .068 miles

13. Consumption over time • From 1970 to 1980, the rate of potato consumption in a particular country was C(t) = 2.2 + 1.1t millions of bushels per year, with t being years since the beginning of 1972 to the end of 1973. Solution: we seek the cumulative effective of the consumption rate for 2 < t < 4.

14. STEP 1: Riemann Sum • STEP 2: Definite integral • STEP 3: Evaluate using the calculator

15. Finding Gallons Pumped from Rate Data • A pump connected to a generator operates at a varying rate, depending on how much power is being drawn from the generator to operate other machinery. The rate (gallons per minute) at which the pump operates is recorded at 5 minute intervals for one hour. How many gallons were pumped during that hour?

16. Solution: Let R(t), 0 < t < 60 be the pumping rate as a continuous function of time for the hour. Partition the hour into short subintervals of length Δt on which the rate is nearly constant. Because we have no formula, we cannot evaluate an indefinite integral, so we will use the trapezoidal formula. a = 0, b = 60, n = 12 (b-a)/12 = 5 X2 5/2(1433) = 3582.5 gallons

17. Work • When a body moves a distance d along a straight line as a result of the action of a force of constant magnitude F in the direction of motion, the work done by the force is W = Fd • W = Fd is the constant-force formula for work. • The units of work are force X distance • Metric: newton-meter • English: foot-pound

18. Work • Hooke’s Law for springs: the force that it takes to stretch or compress a spring x units from its natural (unstressed) length is a constant times x: F = kx • k is measured in force units per length and is a characteristic of the spring called the force constant.

19. A Bit of Work • It takes a force of 10 N to stretch a spring 2 m beyond its natural length. How much work is done in stretching the spring 4 m from its natural length? • Let F(x) = the force in newtons required to stretch the spring x meters from it’s natural length. • Hooke’s law: F(x) = kx, for some constant k. • We know that F(2) = 10 = k(2), so k = 5 N/m • Therefore, F(x) = 5x

20. Solution • Construct an integral for the work done in applying F over the interval x = 0 to x = 4. 5(4)2 /2 – 5(0)2 /2 = 40 N·m