Net Change as the Integral of a Rate Section 5.5 Mr. Peltier

Net Change as the Integral of a Rate Consider the following problem: Water flows into an empty bucket at a rate of r(t) liters per second. How much water is in the bucket after 4 seconds? – Constant flow – Variable flow Amount of water is area under curve from 0 to 4

Net Change as the Integral of a Rate Therefore, the net change in s(t) over an interval [t 1, t 2 ] is given by the integral Integral of the rate of changeNet change over [t 1, t 2 ]

Net Change as the Integral of a Rate EX: Water leaks from a tank at a rate of 2 + 5t liters/hour, where t is the number of hours after 7 a.m. How much water is lost between 9 a.m. and 11 a.m.?

Net Change as the Integral of a Rate Integrating Velocity – Velocity is derivative of position function – V(t) = s’(t) – Integral of velocity equals net change in position, or displacement over time interval [t 1, t 2 ] – Displacement does not equal distance traveled! If I drive to Littlerock and back, my displacement is zero, but I traveled 16 miles…

Net Change as the Integral of a Rate Therefore, for an object in linear motion with velocity v(t), then Displacement during [t 1, t 2 ] = Distance traveled during [t 1, t 2 ] =

Net Change as the Integral of a Rate EX: A particle has velocity v(t) = t 3 – 10t t meters/second. Compute the displacement and total distance traveled over [0, 6] Displacement

Net Change as the Integral of a Rate Total Distance Traveled By the graph we can see the velocity changes from positive to negative at t = 4, so to compute the total distance, we need to subtract the negative area from the positive area

Assignment Page 283 Problems 1-4, 9-12