Lecture 7 Derivatives as Rates

Average Rate of Change If f is a function of t on the interval [a,b] then the average rate of change of f on the interval [a,b] with respect to t is When the units of f(t) are distance and those of t are time the average rate of change is called the average velocity

Need only the values a,b, f(a), f(b) A particle moves along a line so that its distance from the origin at time 4 seconds is 9 feet and its distance from the origin at time t = 10 seconds is 2 feet. What is its average velocity over the interval [4,10]? Here a = 4 sec., f(4) = 9 ft., b = 10 sec, f(10) = 2 ft. Average velocity = (2 – 9)/(10-4) ft/sec = -7/6 ft/sec

Average Velocity is Slope of Secant Line

Motion on a Line x(-1) = 0 Suppose a particle is moving along the x- axis and its x-coordinate at time t seconds is given by feet. x(0) = -1x(1) = 0x(2) = 3x(3) = 8

Previous Example t in seconds, x in feet over the interval [-1,3] ft/sec = = 2 ft/sec The average velocity over the interval [-1,3] is 2 ft per second

What is the velocity at t = 2? Can’t calculate average velocity over interval [2,2] (dividing by 0). Can calculate average velocity over [2,2+h] The limiting average velocity as h -> o must be the “instantaneous” velocity at time t = 2.

Instantaneous Velocity If f(t) is the position on a line of a particle at time t then the instantaneous velocity of the particle at time t =a is Which is f ‘ (a)

Example A particle moves along the x-axis so that its position at time hours is miles What is its instantaneous velocity at time t = 3 hours? At what times is it moving to the left of the origin and at what times is it moving to the right?

solution miles/hr x ‘ is positive for t < 0 so for t < 0 the particle is moving in the positive direction which is to the right. It is negative for time t > 0 which means that for time t >0 it Is moving to the left

Projectile Motion After t seconds the height, in feet of a projectile fired straight upward from the ground is a. What is the (upward) velocity 3 seconds after it is fired? b. What is the (upward) velocity when it hits the ground? c. At what time does the projectile stop ascending and begin to return to Earth?

Solution: Let v(t) = velocity at time t. v(t) = h’ (t) = ft/sec (a)At time t = 3 the velocity is v(3) = -32* = 4 ft/sec (b) The projectile stops rising when its velocity is reduced to 0. That is when -32 t = 0 or when t = 100/32 sec. (c) The projectile returns to Earth when its height is 0 i.e. when 0 = 0 =( -16 t + 100) t has solutions t = 0 and t = 100/16 The time t = 0 is when it was launched, t = 100/16 is when it returned. The velocity then is v(100/16) = -100 ft/sec

Two roads meet at an angle of 90 degrees, one running North- South and the other East-West. Car A is north of the intersection traveling south at 50 miles per hour and car B is east of the intersection traveling west at 80 miles per hour. At what rate (correct to within.001) in miles per hour is the distance between them changing at the time when car A is 5 miles from the intersection and car B is seven miles from the intersection

Let to = time A is at (0,5) and B is at (7,0)

May assume to = 0

Problem on Average Velocity A driver travels the first 40 miles of a trip to Louisville at 80 miles per hour and the remaining 30 miles at 60 miles per hour. What was her average velocity for the trip?