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Do Now. A car moving with initial velocity of 5 mph accelerates at. mph per second for 8 seconds. (a) How fast is the car going when the 8 seconds are up?. (b) How far did the car travel during these 8 seconds?. Section 7.1b. Modeling with Integrals as Net Change.
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Do Now A car moving with initial velocity of 5 mph accelerates at mph per second for 8 seconds. (a) How fast is the car going when the 8 seconds are up? (b) How far did the car travel during these 8 seconds?
Section 7.1b Modeling with Integrals as Net Change
Right in with Practice Problems The integral is a natural tool to calculate net change and total accumulation of more quantities than just distance and velocity (ex: growth, decay, consumption, acceleration). Whenever you want to find the cumulative effect of a varying rate of change, integrate it!!!
Right in with Practice Problems A car moving with initial velocity of 5 mph accelerates at mph per second for 8 seconds. (a) How fast is the car going when the 8 seconds are up? Net velocity change: mph Final velocity after 8 seconds: Initial velocity + Net velocity change mph
Right in with Practice Problems A car moving with initial velocity of 5 mph accelerates at mph per second for 8 seconds. (b) How far did the car travel during these 8 seconds? Velocity function: mph Distance traveled: mph x sec 244.8 mi x sec 1 hr mi x 1 hr 3600 sec
Right in with Practice Problems The graph of the velocity of a particle moving on the x-axis is given below. The particle starts at x = 2 when t = 0. (a) Find where the particle is at the end of the trip. (b) Find the total distance traveled by the particle. v (m/sec) Final position: 3 t (sec) 10 meters –3
Right in with Practice Problems The graph of the velocity of a particle moving on the x-axis is given below. The particle starts at x = 2 when t = 0. (a) Find where the particle is at the end of the trip. (b) Find the total distance traveled by the particle. v (m/sec) Total distance: 3 t (sec) 10 meters –3
Right in with Practice Problems From 1970 to 1980, the rate of potato consumption in a particular country was millions of bushels per year, with t being years since the beginning of 1970. How many bushels were consumed from the beginning of 1972 to the end of 1973? We need the cumulative effect of the consumption rate for Consumption: Evaluate numerically: million bushels
Right in with Practice Problems Suppose that the average rate of electricity consumption for a certain home is modeled by the given function, where C(t) is measured in kilowatts and t is the number of hours past midnight. Find the average daily consumption for this home, measured in kilowatt-hours. Consumption: kilowatt-hours
Right in with Practice Problems Oil flows through a cylindrical pipe of radius 3 in, but friction from the pipe slows the flow toward the outer edge. The speed at which the oil flows at a distance r in from the center is inches per second. In a plane cross section of the pipe, a thin ring with thickness at a distance r inches from the center approximates a rectangular strip when you straighten it out. What is the area of the strip (and hence the approximate area of the ring)? Width Length Area
Right in with Practice Problems Oil flows through a cylindrical pipe of radius 3 in, but friction from the pipe slows the flow toward the outer edge. The speed at which the oil flows at a distance r in from the center is inches per second. (b) Explain how we can find the rate at which oil passes through this ring. Volume per sec. = Inches per sec. x Cross section area = flow in
Right in with Practice Problems Oil flows through a cylindrical pipe of radius 3 in, but friction from the pipe slows the flow toward the outer edge. The speed at which the oil flows at a distance r in from the center is inches per second. (c) Set up and evaluate a definite integral that will give the rate (in cubic inches per sec) at which oil is flowing through the pipe.
WORK In science, the term work refers to a force acting on a body and the body’s subsequent displacement. 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 the motion, the work done by the force is This is called the constant-force formulafor work. Units of work are force x distance. In the metric system, the unit is Newton-meter (called a Joule); in the U.S. customary system, the most common unit is the foot-pound.
Hooke’s Law Hooke’s Law for springs says that the force it takes to stretch or compress a spring xunits from its natural (unstressed) length is a constant times x. In symbols: where k, measured in force units per unit length, is a characteristic of the spring called the force constant. Another important note: Numerically, work is area under the force graph!!!
Right in with Practice Problems 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? First, we need to find the “force equation” for this particular spring: for this particular spring
Right in with Practice Problems 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? Now, we construct an integral for the work done in applying F over the interval from x= 0 to x= 4:
