Appendix A:  Optional Learning Objectives

Objectives

Instructional Notes

To determine function values for, and draw the graphs of step fuctions

A piecewise function, whose graph looks like a series of steps, is called a step function . The function , called the greatest integer function , names the greatest integer that is less than or equal to x .

Thus , and . This function appears on the TI-83 graphing calculator as int ( X ).

Another example of a step function is the floor function (the function rounds non-integer values down), and when this name is chosen, the notation is often used. Thus .

The graph of this function appears below.

Another notation used for the greatest integer function is .

Students should also be exposed to the least integer function or ceiling function , , which returns the smallest integer that is greater than or equal to x . Thus , , and . This function rounds non-integer values up.

Examples/Activites

Suggestions/Extensions

Evaluate each of the following.

1.                       2.

3.                       4.

5.                          6.

7.                          8.

9.                       10.

11.                    12.

13.                         14.

15.                16.

Draw the graph of the greatest integer function using a graphing calculator. What are the limitations of the calculator's portrayal of the graph?

Draw the graph of the iPart function, which removes the decimal portion of any number and only returns the integer part. How does this graph differ from the graph of the greatest integer function?

Draw the graphs of each of the following functions.

1.

2.

3.

4.

5.

Solve the following equations or inequalities.

1.

2.

Have the students discuss where the floor and ceiling functions would be used.

Have the students research the Heaviside function (IL).

Objectives

Instructional Notes

To identify and use limit properties in evaluating limits.

To make and justify decisions based upon understanding of calculus concepts (CCT).

In C.2 and C.3, we have been evaluating limits using limit properties intuitively. The purpose of this section is to raise these properties to a conscious level, to formalize them, and to recognize them as limits are evaluated. The (epsilon)- (delta) definition of limits is not intended.

The limit properties follow without proof.

If c and k are real numbers, n is an integer, and and are real numbers, then:

1. If is the constant function , then .

2. If is the identity function , then .

3. Sum : .

4. Difference : .

5. Product : .

6. Quotient: provided .

7. Constant Multiple: ( k is a real number).

8. Power: , n is a positive integer.

9. Root: , provided the root exists.

Use words to express numbers 3 to 9:

•  The limit of a sum is the sum of the limits.

•  The limit of a difference is the difference of the limits.

•  The limit of a product is the product of the limits.

•  The limit of a quotient is the quotient of the limits, provided the denominator is not 0.

•  The limit of a constant multiple of a function is the constant times the limit of the function.

•  The limit of a positive integral power of a function is the power of the limit of the function.

•  The limit of a root of a function is the root of the limit provided the root exists.

Examples/Activities

Suggestions/Extensions

The examples that follow show how the limit properties can be used to justify the steps involved in finding a limit. Guide the students through exercises like the ones that follow and then provide a few similar ones and have students carry out and justify their steps. The point is not to memorize how to do exercises like these but with the assistance of the list of properties to break down limit questions step-by-step.

The numbers in parentheses below are the property numbers listed on the previous page.

a) Find .

b) Find .

c) Find .

Students could also examine the rigorous approach to limits.

If and , then try to determine whether each of the following statements is true or false. For those you think are false, provide the counter-example.

1.

2.

3.

4.

Objectives

Instructional Notes

To understand the Intermediate Value Theorem and apply it to finding zeros of a function.

To find alternate examples of relationships that demonstrate continuous functions or not (CCT).

To demonstrate trust in own feelings, judgement, and abilities to be self-reliant (PSVS).

Introduce this theorem by asking students to identify if the relationships described are continuous functions:

•  your height and time

•  the amount of money in your wallet and time

•  the amount of gas in the gas tank and the distance travelled

•  the cost of a long distance phone call and the number of minutes you've been talking

•  the temperature outside the car and the distance travelled

•  the depth of the water in a lake and your distance from the shore

•  the cost of a city bus ride and the distance travelled

•  the cost of a taxi ride and the distance traveled.

Ask the students what knowledge is conveyed about each of these functions if students know that it is continuous.

In pairs, have students create two examples of relationships that demonstrate continuous functions and two examples of relationships that do not demonstrate continuous functions. Have pairs form small groups of four and list the relationships by groups (e.g., continuous functions or not) on large chart paper for display. (CCT, PSVS)

If the function is continuous on the closed interval , and L is a number that lies between and , then there exists a number such that .

The theorem states that as x goes through values from a to b , will take on every value between and . Geometrically, every horizontal line you draw, whose height lies between and , will cross the graph of at least once. This theorem is known as an existence theorem. It does not tell you how to find c , or how many values of c might be found, it just asserts that it is possible to find at least one value of c .

For the value of L in this figure, there are three values of c between a and b such that .

For the value of L in this figure, there is one value of c between a and b such that .

Examples/Activities

Suggestions/Extensions

Explain why must have a zero between 5 and 7.

Verify that the Intermediate Value Theorem applies in the indicated interval. In addition, find the value of c that the theorem guarantees.

a) ; ;

b) ; ;

c) ; ;

d) ; ;

If , prove that there is a number c such that . Do not find the value of c .

Objectives

Instructional Notes

To understand the Intermediate Value Theorem and apply it to finding zeros of a function.

Ask the students why it is important that the function be continuous in order to find c . The figures below illustrate the difficulty that could arise if it is not.

Of what use is the Intermediate Value Theorem? Suppose you need to solve the equation . If we let , we observe that (negative) and (positive). Since is continuous (remember all polynomial functions are), then there must be a value of c between 1 and 2 for which . Thus, we know that there is at least one root to the equation that lies in the interval . We can repeat the use of the Intermediate Value Theorem as many times as we wish to zoom in on the solution. The following method, known as the bisection method , could be used.

•  Since is negative and is positive, find the value of the function halfway between 1 and 2, namely find . Now which is positive. Thus the root lies in .

•  Try – 1.25 lies halfway between 1 and 1.5. which is negative. Thus, the root lies in .

•  Try – 1.375 is halfway between 1.25 and 1.5. which is positive. Thus, the root lies between 1.25 and 1.375.

•  This process could be continued until the desired degree of accuracy is obtained.

With the aid of a calculator, you may be able to find the root more quickly by guessing intelligently rather than using the bisection method. For example, if and , then clearly there is a root between 1

and 2. However, is so close to zero that to guess the root as being 1.5 is not reasonable. Instead you might try or even .

The bisection method lends itself well to a computer algorithm that can be written to find a root. Of course, graphing calculators will find the x -intercept of a function.

Examples/Activities

Suggestions/Extensions

With the aid of a calculator (not using the graphing features), use the bisection method together with intelligent guessing to determine the root of the equation , that lies between 0 and 1, correct to two decimal places.

Sketch the graph of a function for which , , and for any number L that lies between 3 and 10 you can find a number c such that , even though is not continuous on .

Micah stepped into the elevator, closed the door, punched his stopwatch, and ascended 90 metres to the 10 th floor of his apartment building without stopping. The climb took 28 seconds. He set his stopwatch back to 0 and took the elevator back down, once again without stopping. The descent took 20 seconds. Use the Intermediate Value Theorem to prove that there was a time on his stopwatch at which he was the same height above ground on his way up as on his way down.

(Hint: Let represent his height above ground on the way up after riding the elevator for t seconds. Similarly let be his height above ground on his way down after riding the elevator for t seconds. Let . Now and are continuous. Would be continuous? Is positive or negative? Is positive or negative? What then is true about at some time between 0 and 20 seconds? What then is true about and at that time?)

Can you provide another argument, not based on the Intermediate Value Theorem, as to why this is true?

Objectives

Instructional Notes

To identify the location of an oscillating discontinuity.

An oscillating discontinuity occurs at a value of x near to which a function refuses to settle down. Ask the students to use calculators to draw the graph of over small intervals symmetric about 0 to see if students can predict . Consider the graph drawn over three different intervals.

As , the function oscillates more rapidly. Ask the students to explain why that might be. Recall that the graph of goes through one period for . In other words, takes units to complete one cycle. How many cycles does go through as changes from to .01? If , then, taking reciprocals, . Thus, changes by 900 units. must pass through or almost 143 cycles as changes from to .01. Thus, as , will never settle down does not exist.

Oscillating discontinuities will occur for trigonometric functions at if is undefined at .  

Examples/Activities

Suggestions/Extensions

Determine if the function is continuous or not. If it is not continuous, determine the value(s) of x at which any discontinuity occurs, and classify the discontinuity.

a)

b)

c)

d)

e)

f)

Define so that the function is continuous at .

Create a function (not just the graph) that has a jump discontinuity at , a removable discontinuity at , an oscillating discontinuity at , but is continuous everywhere else.

On a very warm day, the water in a cooler gets used up quickly and must be replenished. The number of litres of water in the cooler after t hours is given by the function . Sketch the graph of the function. Where are the points of discontinuity? At what rate is the water disappearing?

Objectives

Instructional Notes

(Continued)

To identify the location of an oscillating discontinuity.

Students should be reminded of the continuity principles mentioned in C.1.

Trigonometric functions of the form , , and have oscillating discontinuities at x values for which , provided .

Objectives

Instructional Notes

To understand and apply Rolle's Theorem and the Mean Value Theorem.

To consider various points of view and alternative perspectives (CCT).

Ask the students to draw the graph of a function over the closed interval that is both continuous and differentiable over the open interval , has an absolute maximum and/or an absolute minimum, and has . Have students compare their graphs. What common theme emerges? (CCT)

The Extreme Value Theorem in E.2 guarantees that if a function is continuous on a closed interval , then it has an absolute maximum and absolute minimum on that interval. These extrema may occur at the endpoints of the interval.

Rolle's Theorem, named after the French mathematician Michel Rolle (1652-1719), describes the conditions under which an absolute maximum or minimum is reached at somewhere between the endpoints of a closed interval.

Students should be shown Rolle's Theorem, but do not need to be able to prove it.

Rolle's Theorem

If a function

•  is continuous on the closed interval

•  is differentiable on the open interval

•  has

then there is at least one number for which .

Rolle's Theorem is illustrated below. Note that is both continuous on and differentiable on . Furthermore, . Note that there are (in this case) two points and in at which the tangent line has a slope of zero.

Have the students check the graph they sketched against the theorem. Remind the students of the statement in C.2 indicating that the sums, differences, products, and quotients of constant, polynomial, rational, power, root, trigonometric, exponential, and logarithmic functions are continuous for all values in their domain and for values that do not make the denominator zero.

Lead students through two typical examples such as the ones that follow.

Example: Find the x -intercepts of the function and show that somewhere between the x -intercepts.

Solution: At an x -intercept, . Thus, . Solving by factoring, and . Thus, . Of course, all polynomial functions are continuous and differentiable: . If , then so . Note that .

Examples/Activities

Suggestions/Extensions

Explain why Rolle's Theorem does not apply to the function even though .

If the condition of differentiability were removed from Rolle's Theorem, how would the conclusion of the theorem have to be modified?

Consider a golf ball's height above ground t seconds after it is hit. At the moment it is hit ( ), its height above ground is zero. When the ball comes back to the fairway, the height above ground is also zero. Now the rate of change of its height relative to time, in other words the derivative of height function, would give the vertical velocity of the golf ball. What does Rolle's Theorem guarantee about the vertical velocity of the ball? Why is that common sense?

Objectives

Instructional Notes

To understand and apply Rolle's Theorem and the Mean Value Theorem.

Example: Show that satisfies the conditions of Rolle's Theorem on the interval . Then find all values of c that satisfy the conclusion of Rolle's Theorem.

Solution: Since is the sum of a constant function with the product of a polynomial and a root function, it is continuous and differentiable on the interval . . Thus, the conditions of Rolle's Theorem are satisfied. Now . will be zero if the numerator or if . Note that , and the conclusion of Rolle's Theorem is satisfied.

Rolle's Theorem is often used to introduce the Mean Value Theorem.

Mean Value Theorem. If is a continuous function on the closed interval and differentiable on the open interval , then there exists a number c such that .

The word “mean” in the title of the theorem implies average. Geometrically, the theorem implies that somewhere between and , we find a number c such that the tangent line drawn at is parallel to the slope of the secant line that connects to . That is, the average rate of change of the function between and is equal to the instantaneous rate of change of the function at some point between a and b . Therefore, the average rate of change using the endpoints of the interval equals the instantaneous rate of change at some point in the interval.

Examples/Activities

Suggestions/Extensions

Determine if Rolle's Theorem can be applied to the function on the given interval. If it can be applied, find all values of c in the interval for which .

1. ;

2. ;

3. ;

4. ;

5. ;

6. ;

7. ;

8. ;

Objectives

Instructional Notes

E.2B (OPTIONAL)

(Continued)

To understand and apply Rolle's Theorem and the Mean Value Theorem.

Example: Verify that the function satisfies the Mean Value Theorem on the interval and then find all values of c that satisfy the conclusion of the theorem.

Solution: is a polynomial function and is thus continuous and differentiable. Here and . and . , thus . The Mean Value Theorem states that . Thus, . Solving for c , , giving . Note that .

Example: Suppose that is continuous and differentiable everywhere. On the interval , . If , what is the largest possible value for ?

Solution: Since the function is continuous and differentiable, the Mean Value Theorem assures us that there is a point for which . Multiplying both sides by 4 gives , and solving for , we have . Since for all , and since , then and thus . Thus, . Thus, the largest possible value for is 88.

Examples/Activities

Suggestions/Extensions

Apply the Mean Value Theorem to the given function on the given interval and find all values of c within the interval for which .

1. ;

2. ;

3. ;

4. ;

5. ;

6. ;

7. ;

Suppose that is continuous and differentiable everywhere. On the interval , . If , what is the smallest possible value for ?

The following problem is typical of those that illustrate an application of the Mean Value Theorem.

Problem: A car, travelling 100 km/h, passed a parked RCMP officer. Exactly 3 minutes later the car, travelling 90 km/h at this time, passed another parked RCMP officer 6 km down the road from the first officer. Was the car exceeding the speed limit at some point during those 3 minutes?

Make sure the students are provided with an opportunity to try to solve the speeding example without a specified use of the Mean Value Theorem. It is important that the students realize that despite working with calculus concepts, there are still options for solutions and solution strategies. The purpose of this type of problem is to make sense of the theorem in a context familiar to the students and not to force the students to solve such problems using the theorem.

Solution: Let be the function describing the distance that the car has travelled t minutes after passing the first officer. Then km, and km. According to the Mean Value Theorem, there is a point at which

The mean or average speed of the car was 120 km/h and somewhere between 0 and 3 minutes, the rate of change of position versus time, that is the instantaneous velocity of the car, was the same. Thus, the car was speeding.

Objectives

Instructional Notes

To find the x -intercept(s) of a function using Newton 's Method.

A day or two before tackling this objective, show the students the graph of and ask them to see if they can find a way to determine the x -intercepts of the function to two decimal places without using the quadratic formula or a graphing calculator. Remind the students that if the graph of a function had to be replaced by a line near any point P , then the tangent line drawn at P is the best replacement.

Suppose that is continuous on the interval and differentiable on the interval . If and differ in sign, the Intermediate Value Theorem guarantees that has a zero somewhere in the interval . Suppose that zero occurs at (as in the figure below). We wish to approximate the value of c as we do not know this value in advance. Newton 's method assumes that a tangent line drawn close to c will cross the x -axis very close to c .

We know that the slope of the tangent line drawn at is given by  . We see that the tangent line passes through the points and . Thus, the slope of the tangent line is or ‚ . Equating the two expressions for the tangent line slope, we have . Solving for yields . You can find a third estimate for c by repeating the procedure. Draw a tangent at (not illustrated above) and let it cross the x -axis at . Then, it can be shown that . This process can be repeated until the required degree of accuracy for estimating c is reached.

Examples/Activities

Suggestions/Extensions

Without the use of a calculator, use Newton 's Method to find the value of for the given value of .

1. ;

2. ;

3. ;

4. ;

5. ;

Use a calculator (but not its graphical features) to determine at least one x -intercept for each of the following functions. Give the answer to four decimal places. Find your first estimate by locating values of a and b for which and have opposite signs.

1.

2.

3.

4.

Use Newton 's Method to find the following approximations to four decimal places. You may use a calculator (but not its graphical features).

1. (Hint: Let . Then and . Consider the graph of and find the x -intercept.)

2.

3.

Suppose that during the first 10 seconds of lift-off, a space probe's height above the earth is given by the function . At what time is the probe 30 m above the ground? Use Newton 's Method to approximate the roots to two decimal places.

Newton 's Method is an iterative process. Each time you cycle through Newton 's Method, you are performing an iteration. The number of iterations you perform depends on the degree of accuracy required and how quickly the estimates converge (get closer to the actual x -intercept).

Objectives

Instructional Notes

E.7 (OPTIONAL)

(Continued)

To find the x -intercept(s) of a function using Newton 's Method.

Use the challenge problem you gave the students a few days earlier as your first example. Suppose that you estimate the tangent line to cross the x -axis at 1. Then . Now ƒ . Since , . Thus, and .

Examples/Activities

Suggestions/Extensions

The TI-83 calculator can handle Newton 's Method efficiently. Consider the example shown in this activity. In the window, enter the functions as shown below.

Note that the original function is placed in Y 1 , the derivative is placed in Y 2 , and the formula for the next estimate is placed in Y 3 .

Objectives

Instructional Notes

To use differentials to approximate values of functions.

If was a curve in a highway, and if at you left the curve following the tangent line, then after travelling horizontal units, you would be off the highway by units.

Recall that the tangent line is the line that most closely follows a curve at the point of contact. We can use this property to approximate the values of functions ( , , ) without using a calculator. This topic, like Newton 's Method, is significant for its historical appreciation rather than for its current practical value. Clearly, if we wanted the value of , we would simply use a calculator. To appreciate the complexity of this objective, imagine the 1600s when calculus was developing and calculators were not available.

Differentials do, however, set a visual stage for the development of integrals and integration.

It is unrealistic to expect students to come up with the ideas related to this objective without assistance.

Begin by making two definitions. As you move from point P to point Q along the graph of , the value of x will change by a certain amount, say , and the value of y will change by a certain amount, say . The differential of x , abbreviated by the symbol , is the actual change in x as you move from point P to point Q . In other words, . The differential of y , abbreviated by the symbol , is an approximation to and is given by the relationship . If for horizontal units, you follow the tangent line instead of the function, y will change by units. By examining the figure and explanation below, you can see the rationale for the definition.

The tangent line drawn at P follows the curve closely for a while. If you begin at P and follow the tangent line for horizontal units instead of the graph of , you will be at point R instead of at point Q . The y value at R will be very close to the y value at Q if is small.

The slope of the tangent line at P is . The slope of the tangent line is also the ratio of the rise to the run in right triangle RPS . That is . As these two slopes are the same, we have . Solving for RS , which is our definition for . Thus, .

Examples/Activities

Suggestions/Extensions

In the first part of the assignment, students should practise determining the differential, dy , for different functions.

Find the differential, dy , for each of the following functions.

1.

2.

3.

4.

5.

6.

Use differentials to find an approximation for each of the following. Express your answer in fractional form. Do not use a calculator.

1.

2.

3.

4.

5.

6.

7.

8.

Have the students relate their work with differentials to where differentiating started .

Objectives

Instructional Notes

(Continued)

To use differentials to approximate values of functions.

Note that this is the first time in the course where and have individual identities. To this point, students have only seen these identities together in the form . The definition of the differential is easy for students to remember, since in the past we have used . Now if and have individual values, then we can solve for to obtain . Students will be more successful if they have a mental image of and as shown earlier.

Example 1: Consider the function . If x changes from 4 to 4.1, find , , , .

Solution: the exact change in y . Thus, . is the exact change in y as x changes from 4 to 4.1. Thus, . Note that to find without a calculator would involve a bit of time.

Now . Since , then . Thus, . If the tangent line is drawn at , then . Thus, . It can be seen that is a close approximation to , and required no complicated calculation. Thus, if someone asked you to find the value of without a calculator, you could simply find and add on or 4.8, obtaining 68.8. You would differ from the exact answer of 68.921 by only 0.121, a very small percentage difference.

Example 2: Use differentials to find an approximation to .

Solution: We know that . Have the students organize their work in a table format. This will help them see the problem in the context of the figure used to explain differentials.

The first difficulty students will have is to determine the function . Ask them what is being done to 101. We want to find its square root. Thus, the function is . We know that , so we will follow the graph of the tangent line drawn to the graph of at and find the approximate change in y as x changes from 100 to 101. Since , in this case we have . Thus, . So if x changes from 100 to 101, the square root of x will change by approximately . A good statement for students to use at the conclusion of the problem is: . Thus or . The calculator value for is 10.04987562. Our approximation is very close.

Examples/Activities

Suggestions/Extensions

The side of a square is measured to be 20 cm with a possible error of 1 mm. Use differentials to approximate the error in calculating the area of the square.

The edge of a cube is measured to be 10 cm with a possible error of 1 mm. Use differentials to approximate the error in calculating the volume and the surface area of the cube.

A layer of gold, .0001 cm in thickness, is to be applied to the rounded surface of a hemispherical dome. If the dome has a radius of 6 m, use differentials to approximate the volume of gold that will be required. Express your answer in .

The profit for a business selling x units is given by the function . Use differentials to find the approximate change in profit as the number of units produced changes from 200 to 210 units.

The concentration of a pollutant, t hours after being tossed into a swimming pool, is given by the function . Use differentials to approximate the change in concentration when t changes from to .

Objectives

Instructional Notes

(Continued)

To use differentials to approximate values of functions.

Example 3: The radius of a sphere is measured to be 10 cm. The measurement is accurate to within 1 mm. Use differentials to approximate the resulting error in calculating the volume of the sphere.

Solution: We know that . Thus, . The degree of accuracy in r is mm or cm. If we follow the graph of the tangent line drawn to the function at , moving horizontal units in either direction, then the approximate change in V is . Thus, the reported volume will be accurate to within . Is this a large error? This question is best answered by calculating the relative error , a ratio of dV to V . Now or .

Objectives

Instructional Notes

To differentiate cosecant, secant, tangent, and cotangent functions and solve related problems.

To develop the ability to plan, monitor, and evaluate their own learning (IL).

Knowing how to differentiate sine and cosine functions leads easily to the derivatives of the remaining four trigonometric functions. This is one section that students could probably handle as an independent study activity. Students could develop the formulas if you provide the “bottom line” so students know in what form they should leave their derivative. Another possibility is to assign one function to each of four groups and ask them to develop the formulas for the class playing the role of teacher. (IL)

Students should know the graphs of these trigonometric functions in order to have a better understanding of the results they are obtaining.

1. The Derivative Of Cosecant Functions

Suppose . Using the quotient rule, .

If y is a cosecant function of u and u is a function of x , we have:

2. The Derivative Of Secant Functions

Rather than use a quotient rule again, use the power rule and the chain rule. Suppose . Then, .

If y is a secant function of u and u is a function of x , we have:

3. The Derivative Of Tangent Functions

Suppose that . Using the quotient rule, .

If y is a tangent function of u and u is a function of x , we have:

4. The Derivative Of Cotangent Functions

Develop this result using the product rule.

If , then

If y is a cotangent function of u and u is a function of x , we have:

Examples/Activities

Suggestions/Extensions

Differentiate each of the following functions.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

Find the equation of the tangent line drawn to the given curve at the given point. Leave and radicals in the answer where they arise.

1. ,

2. ,

3. ,

4. ,

Find two critical numbers for the function in the interval . Find all local extrema in that interval.

Students could research the other hyperbolic functions, namely , , , and , and find their derivatives (IL).

Objectives

Instructional Notes

To differentiate cosecant, secant, tangent, and cotangent functions and solve related problems.

To help students remember the formulas for the derivatives of the trigonometric functions, point out that the derivatives of all the functions starting with “co” have a negative coefficient.

Ask the students to suggest some functions that could be used as the differentiation examples for this objective. Apart from remembering the formulas, students will have little trouble with the differentiation if they were successful in differentiating sine and cosine functions.

The tangent function is often needed to solve related rate problems involving angles. Consider the following example. There is more than one way to solve this problem, but the method shown is tidy because it uses the identity , thereby saving the need to solve the right triangle in order to find .

Examples/Activities

Suggestions/Extensions

1. A car, travelling east on a highway at a speed of 100 km/h, passes through an intersection. A policeman is parked 50 m south of the intersection. At what rate is angle changing when the car is 200 m past the intersection?

2. As the sun sets, its angle of elevation is decreasing at a rate of radians per hour. At what rate is the shadow of a 12 m tall vertical pole lengthening when the angle of elevation of the sun is radians? Assume the pole sits on level ground.

3. A spotlight on a fire truck is rotating at a rate of radians per second and is located 30 m from an apartment building. Find the speed at which the spot of light is moving along the side of the apartment building when the spot of light is at point P , which is 10 metres from Q – see the figure below.

4. Jill is jogging east at a rate of 3m/s and is moving away from an intersection. Jack is jogging north at a rate of 2m/s and is approaching the same intersection. At what rate is changing at the moment when Jill is 20 m from the intersection and Jack is 10 m from the intersection?

5. A steel I-beam must be moved around a right-angled corner formed by two hallways, one of which is 3 m wide and the other is 4 m wide. If the beam cannot be tilted, what is the longest beam that can get around the corner? Ignore the width of the beam. (Hint: Show that the length of the beam, , can be given by the expression .)

Objectives

Instructional Notes

To find the derivative of the inverse of a function, and in particular the derivatives of the inverse trigonometric functions, namely: , , and .

Students should be reminded that they have already met several functions and their inverses, and in particular they studied inverse functions in Math B30.

                 The Function                                  The Inverse of the function

1.

multiplying a number by 5

dividing a number by 5

2.

adding 3 to a number

subtracting 3 from a number

3.

squaring a positive number

finding the square root of a positive number

4.

raising 2 to a power

finding the logarithm of a number in base 2

5.

finding the sine of an angle

finding an angle having a given sine value

Examples/Activities

Suggestions/Extensions

Complete the following tables. Based on the values in the tables, are the functions inverses of one another?

1.

2.

3.

Objectives

Instructional Notes

G.6 (OPTIONAL)

(Continued)

To find the derivative of the inverse of a function, and in particular the derivatives of the inverse trigonometric functions, namely: , , and .

With the assistance of appropriate examples remind students that:

• you can find the inverse of a function f by interchanging the x and y coordinates of f .
• the domain of a function is the range of its inverse and the range of a function is the domain of its inverse.
• the inverse of a function f may not necessarily be a function unless some restrictions are placed on the domain of f . One-to-one functions always have inverses that are functions. A function is one-to-one if it passes the horizontal line test.
• if a function has an inverse that is also a function, then the inverse of f is denoted by the symbol , not to be confused with the reciprocal of f .
• the graph of a function and the graph of its inverse are always mirror images about one another, the mirror being the line . (That is because if is on the graph of f , then is on the graph of the inverse of f .)

Shown below is the graph of . Clearly it does not pass the horizontal line test – it is not one-to-one. Thus, its inverse is not a function unless some restriction is placed on the domain of .

Examples/Activities

Suggestions/Extensions

1. Find the inverse of each of the following functions. Note any necessary restrictions.

(a)

(b) ,

(c) ,

(d)

(e)

(f)

Objectives

Instructional Notes

G.6 (OPTIONAL)

(Continued)

To find the derivative of the inverse of a function, and in particular the derivatives of the inverse trigonometric functions, namely: , , and .

The reflection of about the line is not a function since it fails the vertical line test. By restricting the domain of to be , the range will still be and the reflection will pass the vertical line test. See the boxed-in region below.

Note that the domain of is , and the range is . Note that if you drew a tangent line to the graph of , its slope would never be negative.

Examples/Activities

Suggestions/Extensions

Find the derivative of each of the following functions. You may leave radicals in the denominator.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

It is common to write as to avoid confusing with .

Note that is the same as .

Students could research and draw the graphs of , , and . Students could find the domain and range of each function and determine the derivatives of these functions. (IL)

Show that is equivalent to .

Objectives

Instructional Notes

G.6 (OPTIONAL)

(Continued)

To find the derivative of the inverse of a function, and in particular the derivatives of the inverse trigonometric functions, namely: , , and .

Students need to know the following table of exact values in order to readily work with inverse trigonometric functions.

After establishing the meaning of the inverse sine function, its graph, and its domain and range, it is important to do some oral exercises such as the following to solidify understanding.

1. Give the value of each of the following if such a value exists.

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

(j) (k) (l)

The derivative of .

The function:

The meaning of the function:

Differentiate implicitly with respect to x .

Solve for .  .

and since , we have . Thus . Substituting in  yields:

Recall from the graph of the inverse sine function, that the slope of the tangent line is never negative. Thus, .

Examples/Activites

Suggestions/Extensions

1. Find the slope of the tangent line drawn to each function at the given point.

(a) ,

(b) ,

(c) ,

2. Find the equation of the tangent line to the given function at the given point. You may leave radicals and in your answer.

(a) , .

(b) , .

(c) , .

3.  Find the critical numbers of and then determine all local extrema.

4.  The bottoms of two vertical poles, one 2 m tall and the other 5 m tall, are 10 m apart. A wire goes from the top of each pole to a stake in the horizontal ground so that the poles and stake are collinear. What is the size of the maximum angle that can be formed by the two wires? (Maximize where .)

5.  A hockey player is skating along a straight line that is perpendicular to the red goal line and 3 m to the left of the nearest goal post (point B ) – see the figure. Find the distance, x , from the red line that maximizes the angle , the angle within which the player can shoot the puck to hit the net and score. Assume that the goal is 2 m wide.

(Hint: In right , .  . In right , . ‚ . Subtracting ‚ from  yields ƒ . Now differentiate ƒ relative to x and continue.)

It can be shown that graphs of inverse functions have reciprocal slopes. In particular, if f and g are inverse functions, then the slope of the tangent line at on function f is the reciprocal of the slope of the tangent line drawn at on function g . For example, consider the inverse functions , and . Now , and thus at the point , . , and thus at the point , .

In general, if and are inverse functions, then provided .