Calculus I Notes on Rates of Change and the Derivative

Calculus I Notes on Rates of Change and the Derivative. Rates of Change: We live in a constantly changing world, and we do have some quantities to tal...
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Calculus I Notes on Rates of Change and the Derivative. Rates of Change: We live in a constantly changing world, and we do have some quantities to talk about how fast something is changing, its rate of change. Rate of change in : Respect to: Is: position time speed or velocity speed time acceleration or deceleration cost number of items produced marginal cost revenue number of items produced marginal revenue profit number of items produced marginal profit radioactivity time decay rate Since you are probably most familiar with velocity and acceleration from daily activities like driving, I will focus on them for applications. If we are going to make predictions of quantities related to a rate of change we have to be able to quantify it mathematically. Example #1: Let’s imagine a simple example of a spring without friction. Thus, it will continue to bounce up and down without stoping. The bottom of the spring will move in simple harmonic motion. Look at the figure to the right. As the spring bounces, the bottom of the spring will move according to a sinusoidal wave. For simplicity, let’s assume the position of the bottom of the spring, p, is given by p  sin(t) where t  time . Looking at the figure to answer the following questions. a) Where is the spring moving the slowest? b) Where is it moving the fastest? c) What are you using to answer a) and b)? ► a) It would be moving the slowest to either the top or the bottom of the curve when it is turning around. b) It would be moving the fastest when it is halfway between the top and the bottom. c) The more horizontal the curve, the slower it’s moving. While, the steeper the curve the faster it’s moving. □ So the steepness of the graph of a function determines the rate of change of that function. The steeper the curve, the faster it is changing.

Tangent Lines: So we need to quantify how fast a graph is increasing or decreasing. This leads to the following definition. Tangent lines An line is the tangent line to the graph of y  f (x) at x  c if and only if, 1) the line passes though the point on the graph of y  f (x) at x  c and 2) the line is increasing/decreasing at the same rate as the graph of y  f (x) at x  c . Note, this definition of a tangent line might be different than ones you may have gotten in a geometry class for the tangent line of a circle. Though, if you think about it, the definition above is a generalization of the one from geometry. Seminole State:Rickman

Notes on Rates of Change and the Derivative.

Page #1 of 8

Now remember that the slope of a line is the measurement of the steepness of a line. Therefore, if we want to find the rate of change of a function we need to find the slope of its tangent line at that point. Example #2: Using the provided graph, find the rate of change of the graph at the given points. Assume the shown lines are tangent lines. ► Since the rate of change of a function is the slope of the tangent line at that point we just need to find the slopes of the lines using the slope formula. y  y1 m 2 x 2  x1 For the tangent line at (2,8): The green line passes though (2,8) and (1,6). Thus, m  1628  -2-1  2 and the rate of change at x  2 is 2. For the tangent line at (5,5): The blue line passes though (5,5) and (6,1). Thus, m  1655  -14  - 4 and the rate of change at x  5 is -4. □ Note, in the previous example we had a rate of change that was negative. What would that mean? Look back at the graph. The rate of change was negative when the graph was decreasing, going down as you moved from left to right. While the graph was increasing, going up as you moved from left to right, at the point when it had a positive rate of change. Example #3: The provided graph is the graph of a position function, y = P(t), giving the position, y in cm, of an item attached to a spring at a time, t in sec. A position of y=0 corresponds to the position that the spring would stay at rest if it wasn’t bouncing, its equilibrium position. A positive position means the item is above the equilibrium position, and a negative position means it is below. Estimate the velocity of the item based on the graph at the given points. Assume the shown lines are tangent lines. ► First, note that since specific points on each line aren’t given, we will have to make guesses at what points are on each line. Thus, your answers may vary a little from my answers. Also, remember that velocity is the rate of change of position with respect to time. So to find the velocities, we will have to get the slopes of the lines. a) The point of tangency looks like (1sec, 3cm) , and it also look like it’s going through

(3sec, 8.1cm) . Thus, 3cm cm velocity  8.1cm  5.1cm  2.55 sec 3sec 1sec 2sec cm Therefore, at this point it’s going up at about 2.55 sec .

b) The point of tangency looks like (5sec, 0cm) , and it also look like it’s going through

( 7sec, -6.2cm) . Hence,

velocity 

-6.2cm  0cm 7sec 5sec



-6.2cm 2sec

cm  -3.1 sec

cm Therefore, at this point it’s going down at about 3.1 sec .

c) The point of tangency looks like ( 7.5sec,-5cm) and it also look like it’s going through (8sec, -5cm) . Hence, -5cm  5cm 0cm cm velocity  8sec  0.5sec  0 sec  7.5sec Therefore, at this point it’s at rest, but it’s only at rest for a moment before it begins to go back up. □

So if we have the graph of the tangent line, we can at least estimate the rate of change of the function at a point, but how do we get the exact slope of the tangent line or the exact rate of change if all we have is the function?

Seminole State:Rickman

Notes on Rates of Change and the Derivative.

Page #2 of 8

Average Rate of Change and Secant Lines: The problem with finding the exact tangent line from just the function is that I need 2 points to get the slope and I only have 1 point on the function that I know is also on the tangent line. Thus, before we can talk about how to get to the tangent line, we need to talk about secant lines and what they tell us about rate of change. Secant lines An line is the secant line to the graph of y  f (x) at x  a and x  b if and only if, the line passes though the points on the graph of y  f (x) at both x  a and x  b . So note the difference between the secant line and the tangent line. See figure to the right. While the slope of the tangent line gives how fast the curve is increasing at that single point, the slope of the secant line gives how fast the function would have to change on the interval, [a,b], if it changed at a constant rate so that it would end up at the same point. The slope of the secant line gives on average the rate of change of the function on an interval. Average Rate of Change The average rate of change, ARC, of f (x) on [a, b] = the slope of the secant line passing through the points on y  f (x) at x  a and x  b =

f (b)  f (a) ba

Note that the formula for ARC is just a variation of the slope formula, m 

y2  y1 x 2  x1

. So when

you are finding ARC, you are finding the slope of the line through the points  a, f (a)  and  b, f (b)  . Example #4: Find the exact average rate of change, ARC, of f (x)  6 tan(x) on 0, 3  . ► f (0)  6 tan(0)  0

f  3   6 tan  3   6 ARC 

6 3 0  0 3

 3  6

3

 18 3

□ Example #5: Find the exact average acceleration of v(t)  8t  3t 2 on 1hr,5hr  . v(t) is measured in mi . hr ►

v(1)  8(1)  3(1)2  5 mi hr v(5)  8(5)  3(5)2  -35 mi hr Averageacceleration 

-35 mi  5 mi hr hr 5hr 1hr



-40 mi hr 4hr

 -10 hrmi2

Note since v(1)>0 it was travelling forward at t = 1hr, v(5)

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