# Click to see more. Rate of Change And Limits What is Calculus?

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Click to see more. Rate of Change And Limits What is Calculus?

Two Basic Problems of Calculus
1. Find the slope of the curve y = f (x) at the point (x, f (x)) (x, f(x)) (x, f(x)) The slope problem is the first major problem of calculus. You already know how to find the slope of a line. But how can you find the slope at a point on a curve. As you can see, the slope of a curve can change from one point to another. There are many applications to the slope problem, since a slope indicates a rate of change. (x, f(x))

2. Find the area of the region bounded above by the curve y = f(x), below by the x-axis and by the vertical lines x = a and x = b Area y = f(x) a b x The area problem is the second major problem of calculus. There are many applications.

From BC (before calculus)
We can calculate the slope of a line given two points Calculate the slope of the line between the given point P (.5, .5) and another point on the curve, say Q(.1, .99). The line is called a secant line. .

Slope of Secant line PQ x f(x) .1 .99 .2 .98 .3 .92 .4 .76 P(0.5, 0.5)
Point Q x f(x) .1 .99 .2 .98 .3 .92 .4 .76 Let x values get closer and closer to .5. Determine f(x) values.

Slope of Secant line PQ As Q gets closer to P, the
Slope of the secant line PQ Gets closer and closer to the slope Of the line tangent to the Curve at P.

Slope of a curve at a point
Figure 1.4: The tangent line at point P has the same steepness (slope) that the curve has at P. The slope of the curve at a point P is defined to be the slope of the line that is tangent to the curve at point P. In the figure the point is P(0.5, 0.5) The slope of the curve at P(0.5, 0.5) is defined to be the slope of the line that is tangent to the curve at point P.

Slope formula In calculus we learn how to calculate the slope at a given point P. The strategy is to take use secant lines with a second point Q. and find the slope of the secant line. Continue by choosing second points Q that are closer and closer to the given point P and see if the difference quotient gets closer to some fixed value. .

Slope Find the slope of y = x2 at the point (1,1) Find the equation of the tangent line. A

Find slope of tangent line on f(x) =x2 at the point (1,1)
Approaching x = 1 from the right x f(x) Slope of secant between (1,1) and (x, f(x)) 2 4 3 1.5 2.25 2.5 1.1 1.21 2.1 1.01 1.021 2.01 1.001 2.001 Slope appears to be getting close to 2.

Find slope of tangent line on f(x) =x2 at the point (1,1)
Approaching x = 1 from the left x f(x) Slope of secant between (1,1) and (x, f(x)) 1 .5 .25 1.5 .9 .81 1.9 .99 .9801 1.99 .999 1.999 Slope appears to be getting close to 2.

Write the equation of tangent line
As the x value of the second point gets closer and closer to 1, the slope gets closer and closer to 2. We say the limit of the slopes of the secant is 2. This is the slope of the tangent line. To write the equation of the tangent line use the point-slope formula

Average rate of change (from bc)
If f(t) represents the position of an object as a function of time, then the rate of change is the velocity of the object. Find the average velocity if f (t) = 2 + cost on [0, ] 1. Calculate the function value (position) at each endpoint of the interval f() = 2 + cos () = 2 – 1 = 1 f(0) = 2 + cos (0) = = 3 2. Use the slope formula The average velocity on on [0, ] is

Instantaneous rate of change
To calculate the instantaneous rate of change of we could not use the slope formula since we do not have two points. To approximate instantaneous calculate the average rates of change in shorter and shorter intervals to approximate the instantaneous rate of change.

2.2 To understand the instantaneous rate of change (slope) problem and the area problem, you will need to learn about limits This important concept ties the ideas of calculus together.

The answer can be found graphically, numerically and analytically.
Limits What happens to the value of f (x) when the value of x gets closer and closer and closer (but not necessarily equal) to 2? We write this as: Read as the limit of f(x) as x approaches 2. The answer can be found graphically, numerically and analytically.

What happens to f(x) as x gets closer to 2?
Graphical Analysis 5 4 3 2 1 6 8 10 12 14 16 18 20 f (x) x Input the function to Y1 in your graphing calculator. The “hole” is greatly exaggerated in this slide but can not be seen with the calculator. What happens to f(x) as x gets closer to 2?

Numerical Analysis Use one sided limits
Start to the left of 2 and choose x values getting closer and closer (but not equal) to 2 x 1.5 1.9 1.99 1.999 1.9999 f (x) 9.25 You will have to approach 2 from the left (numbers less than 2) and from the right (numbers greater than 2). Use your graphing calculator table function. Set the table so the independent variable is on ASK. That way you can select x and the calculator will return a value for f(x). See what happens if you try to ask for the value at x = 2. 11.41 11.941 Could x get closer to 2? Does f(x) appear to get closer to a fixed number?

Numerical Analysis Start to the right of 2 and choose x values getting closer and closer (but not equal ) to 2 x 2.5 2.1 2.01 2.001 2.0001 f (x) 15.3 12.61 Use your graphing calculator table function. Set the table so the independent variable is on ASK. That way you can select x and the calculator will return a value for f(x). See what happens if you try to ask for the value at x = 2. If the limit exists, f(x) must approach the same value from both directions. Does the limit exist? Guess what it is.

Limits that do not exist
Figure 1.8: The functions in Example 7. Limits that do not exist In order for a limit to exist, the function must approach the same value From the left and from the right.

Infinite Limits What happens to the function value as x gets closer and closer to 3 from the right? x 3.5 3.1 3.01 3.001 3.0001 y 3 11 101 1001 10001 100001 The function increases without bound so we say There is a vertical asymptote at x = 3.

The line x=a is a Vertical Asymptote if at least one is true.
Identify any vertical asymptotes:

Graph of f(x) True or false x = 2 is in the domain of f (b) (c)

2.3 Functions That Agree at All But One Point
If f(x) = g(x) for all x in an open interval except x = c then: Example then A long as the limit in the denominator is not equal to 0, you can use direct substitution here. Evaluate by direct substitution 2-5 = -3 As x gets closer and closer and closer to 2, the function value gets closer and closer to -3.

Analytic = = As x gets closer and closer to 2 (but not equal to 2)
Using direct substitution, What are the two functions that agree in value with the exception of at a =2? What about the limits as x approaches 2? As x gets closer and closer to 2 (but not equal to 2) f(x) gets closer and closer to 12

Compute some limits

Basic Limits If b and c are real numbers and is n a positive integer 1. Guess an answer and click to check. Guess an answer and click to check. Guess an answer and click to check. = -2 Ex: 2. = 5 Ex: Try to see how to apply basic limits. 3. = 9 Ex:

Properties of Limits Multiplication by a constant b
Limit of a sum or difference Limit of a product Limit of a power These properties look complicated but in practice are easy to use. Don’t spend too much time analyzing them. They are presented here to justify the use of direct substitution in many functions. Limit of a quotient when denominator is not 0.

Using Properties of Limits
Properties allow evaluation of limits by direct substitution for many functions. Ex.: Since the limit in the denominator is not equal to 0 (it is 3), you can use direct substitution to evaluate the limit. As x gets closer and closer to 3, the function value gets closer and closer to 9.

Direct substitution Analytic Techniques
First substitute the value of x being approached into the function f(x). If this is a real number then the limit is that number. If the function is piecewise defined, you must perform the substitution from both sides of x. The limit exists if both sides yield the same value. If different values are produced, we say the limit does not exist.

Rationalize a numerator or denominator Simplify a complex fraction
Analytic Techniques Rewrite algebraically if direct substitution produces an indeterminate form such as 0/0 Factor and reduce Rationalize a numerator or denominator Simplify a complex fraction For analytic techniques you will need a good algebraic background. When you rewrite you are often producing another function that agrees with the original in all but one point. When this happens the limits at that point are equal.

Find the indicated limit
direct substitution fails Rewrite and cancel = - 5 now use direct sub.

Find the indicated limit
direct substitution fails Rewrite and cancel now use direct sub.

Find the indicated limit
5 calculate one sided limits 7 Since the one-sided limits are not equal, we say the limit does not exist. There will be a jump in the graph at x =2

Determine the limit on y = sin θ/θ as θ approaches 0.
Figure 1.24: The graph of f () = (sin )/. Determine the limit on y = sin θ/θ as θ approaches 0. Although the function is not defined at θ =0, the limit as θ is 1.

A one-sided limit Figure 1.37: The graph of y = e1/x for x < 0 shows limx0– e1/x = 0. (Example 11)

Limits that are infinite (y increases without bound)
An infinite limit will exist as x approaches a finite value when direct substitution produces If an infinite limit occurs at x = c we have a vertical asymptote with the equation x = c.

2.5 Continuity in (a) at x = 0 but not in other graphs.
Figure 1.50: The function in (a) is continuous at x = 0; the functions in (b) through ( f ) are not.

Conditions for continuity
A function y = f(x) is continuous at x = c if and only if: The function is defined at x = c The limit as x approaches c exists The value of the function and the value of the limit are equal.

Find the reasons for discontinuity in b, c, d, e and f.
(a) Is continuous. (b) f(0) is not defined c) the value of the function does not equal the value of the limit (d) the limit as x approaches 0 does not exist (e) and (f) f(0) is not defined

Figure 1.53: Composites of continuous functions are continuous.
Composite Functions If two functions are continuous at x = c then their composition will be continuous. is continuous for all reals. Example:

Exploring Continuity Are there values of c and m that make the function continuous At x = 1? Find c and m or tell why they do not exist.

Exploring Continuity

2.6 Slope of secant line and slope of tangent line

s(t) = 8(t3 – 6 t2 +12t) t s 56 64 72 Position of a car at t hours. 1. Draw a graph. 2. Does the car ever stop? 3. What is the average velocity for the following intervals a. [0, 2], b. [.5, 1.5] c. [.9,1.1] 4. Estimate the instantaneous velocity at t = 1

s(t) = 8(t3 – 6 t2 +12t) 2. Appears to stop at t =2. (Velocity= 0)
3. What is the average velocity for [0, 2], [.5, 1.5] [.9,1.1] t s(t) 2 62 .5 37 1.5 63 .9 53.352 1.1 58.168 a) 31 mph b) 26 mph c) mph

Find an equation of the tangent line to y = 2x3 – 4 at the point P(2, 12)
So, m = 24. Use the point slope form to write the equation

Slope of the tangent line at x= a
Figure 1.62: The tangent slope is f (x0 + h) – f (x0) h Slope of the tangent line at x= a lim h0 Q(a + h, f (a + h)) f(a+h) – f(a) The letter h represents the change in x which is sometimes denoted as Δx. Notice this difference quotient is the change in y over the change in x. P(a, f(a)) a + h a

Other form for Slope of secant line of tangent line
Let h = x - a Then x = a + h

Find an equation of the tangent line at (3, ½) to
At a = 3, m = - 1/8 Using the point-slope formula:

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