Derivatives and Rates of Change Section 2.7

Example 1, equation of tangent line Given parabola y = x2 at the point P(1, 1). Solution: a = 1 and f (x) = x2, so the slope is Using the point-slope form of the equation the equation of the tangent line at (1, 1) y – 1 = 2(x – 1) y = 2x – 1 = 1 + 1 = 2

Example 1, equation of tangent line Solution: a = 3 and f (x) = 3/x, so the slope is

The average velocity over this time interval is

Example 3, velocity of ball Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground. (a) What is the velocity of the ball after 5 seconds? Velocity after 5 sec v(5) = (9.8)(5) = 49 m/s.

Example 3, velocity of ball (b) How fast is the ball traveling when it hits the ground? Ball must travel 450 m the ball will hit the ground at the time t s(t) = 450 4.9t2 = 450 Velocity of the ball as it hits the ground is v(t) = 9.8t t2 = v( ) = 9.8 t =  9.6 sec = 94 m/s

Example 4, find the derivative f (x) = x2 – 8x + 9 Solution:

Example 5, find tangent line f (x) = x2 – 8x + 9 at point (3, -6) Solution:

Rates of Change In other words, f (a) is the velocity of the particle at time t = a. The speed of the particle is the absolute value of the velocity, that is, | f (a) |.

Example 5 A manufacturer produces bolts of a fabric with a fixed width. The cost of producing x yards of this fabric is C = f (x) dollars. (a) What is the meaning of the derivative f (x)? What are its units? Instantaneous rate of change of the production cost with respect to the number of yards produced. Since C is measured in dollars and x in yards, it follows that the units for f (x) are dollars per yard.

Example 6(b) – Solution (b) In practical terms, what does it mean to say f (1000) = 9? After 1000 yards of fabric have been manufactured, the rate production cost is increasing is $9/yard.

Example 6 f(t) = number of centimeters of rainfall since midnight t = time in hours (a) What is the meaning of f -1(x)? Domain? Range? - Inverse is time where a certain #of cm has fallen - Domain; 0 to max. total rainfall - Range; Midnight to the end of the storm (b) Interpret, f(5)=2 - After 5 hours 2 cm of rain has fallen

f(t) = number of centimeters of rainfall since midnight t = time in hours (c) Interpret, f -1(5)= 2 - 5 cm of rain has fallen after 2 hours (d) Interpret, f ‘(5)= ½ - After 5 hours the rain is falling at a rate of ½ cm/hr (e) Interpret, (f -1) ‘ (5)= ½ - After 5 cm of rain has fallen, time is passing at a rate of ½ hr/cm

2.7 Derivatives and Rates of Change Summarize Notes Read Section 2.7 Homework Pg. 150 #5,7,9,13,23,27,29,31,39,43,47