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Week 3

Study Sections: 1.6, 3.5 [3.6], 3.6 [3.8] in the textbook

· Course Unit: More Derivative Techniques and Applications · Inverse Trigonometric Functions · Implicit Differentiation · Derivatives of Inverse Trigonometric Functions · Logarithmic Differentiation

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WEEK 3

Course Unit: More About Derivatives

In the next two weeks, we return to functions of a single variable and expand our knowledge of differential calculus. This is a continuation of the material that you should already know from high school, and which is covered in the Quick and Easy Differential Calculus Notes. Here we will learn more derivative techniques and see how derivatives are used to model many physical processes. We will also learn how derivatives can be used to approximate functions; such approximations can be used to estimate experimental measurement errors. Let's begin with a question to motivate a discussion of inverse trigonometric functions, the first topic in this section: Example 16. A projectile is fired from the origin, with angle of elevation (radians) and speed 60 meters per second.

Assuming that air resistance is negligible and that the only force on the projectile is gravity (producing a downward acceleration of g = 9.8 meters per second per second), answer the following questions: 1. Find a formula for the position of the projectile at time t. 2. How far down range will the projectile land? 3. At what angle should the projectile be launched so that it will land 300 meters down range? 1. To solve the first problem, suppose we let r(t) = (x(t), y(t)) denote the position of the particle at time t (in seconds), where the first coordinate

COURSE UNIT: MORE ABOUT DERIVATIVES

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measures the distance down range and the second coordinate measures the height of the projectile, both in meters. To solve the problem we must first find expressions for x(t) and y(t). Collecting information given in the problem, we see that r(0) = (0, 0) , and r (0) = 60 cos(), 60 sin() We also know that at all time the acceleration of the projectile is entirely downward, produced by gravity: r (t) = 0, -g . This is the derivative of the velocity, so the velocity can be calculated by "anti-differentiating" (finding functions of which these two components are derivatives): r (t) = A, -gt + B Where A and B are unknown constants. To find A and B we use the information we have about r (0): 60 cos(), 60 sin() = r (0) = A, B , from which it follows that A = 60 cos() and B = 60 sin(). We can now plug this information into the formula we found for r (t): r (t) = 60 cos(), -gt + 60 sin() . We can now repeat the same anti-differentiation procedure to get r(t) from r (t): g r(t) = (60 cos()t + C, - t2 + 60 sin()t + D) , 2 where C and D are unknown constants. To find these constants we use the initial information r(0): g (0, 0) = r(0) = (60 cos()0 + C, - 02 + 60 sin()0 + D) = (C, D) , 2 so C = 0 and D = 0. Thus the path of the projectile is described by g r(t) = (60 cos()t, - t2 + 60 sin()t) . 2

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2. To see where the projectile lands, we put the second component equal to zero: so g - t2 + 60 sin()t = 0 , 2

g t - t + 60 sin() = 0 . 2 t = 0 corresponds to the fact that the projectile is fired from the origin. The g other value of t is the one we are interested in:- t + 60 sin() = 0; that is, 2 2 t = 60 sin() is the time at which the projectile lands. g To find how far down range the projectile will land, we simply substitute this time value it into x(t) = 60 cos()t to get 2 2 2 x(60 sin() ) = 60 cos()60 sin() = 3600 cos() sin() . g g g Thus the projectile will land 3600 cos() sin() 2 meters down range. g

3. To solve the third part of the problem we have to find the value of for which 2 3600 cos() sin() = 300 . g Isolating in this question does not seem easy at first. Going part-way we have g 2 cos() sin() = 300 , 3600 but what is the next step? In order to go one we first have to recall Corollary 1.1 at the end of the first week, where we learned that sin(2) = 2 cos() sin() . If we plug this into the preceding equation we get sin(2) = 300 g . 3600

At least this way appears only once on the left side of the equation. Using the value g = 9.8 we get sin(2) = 0.817 .

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Now we could experiment with inputs to the sine function (making sure the calculator is in radian mode!) until we find one that produces .817 as output. Fortunately we do not have to do that, for your calculator is equipped with an inverse function for the sine function. This inverse is variously known as arcsin(y) or sin-1 (y). The former is the better name to give it, because the latter makes it look as if we are talking about the reciprocal 1/ sin(y) which is not at all the same thing. Arcsin, or the inverse sine function is the function that undoes what the sine function does. We will discuss this function in more detail after this problem. For now let us use the sin-1 button on the calculator to complete it: sin-1 (0.817) = 0.956 That is, the angle whose sine is equal to 0.817 measures 0.956 radians. If you do not believe it, see what you get if you take calculate sin(0.956). Thus, 2 = 0.956, and so 0.478 radians (about 27.4 degrees).

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1.6 Inverse Trigonometric Functions

We will now apply our understanding of inverse functions to discuss inverse trigonometric functions more carefully. We would like to be able to talk about the inverse of the sine function, for example, as we did when we solved Example 16. As you will have learned in high school, and as reviewed in the Quick and Easy Calculus notes available through the course website, a function must be one-to-one in order for it to have an inverse function. In the case of the function f (x) = sin x, one can clearly observe by graphing the function that it does not possess this property (does not pass horizontal line test) and that it therefore does not have an inverse. How then can we find an inverse for f (x) = sin x? What did the calculator think it was doing when we used the sin-1 button? We get around this problem by analyzing f (x) = sin x only from - to . The graph of the sin function from - to 2 2 2 is shown in the following graph: 2

f (x) = Sin x 1

- 2 -1

2

For convenience we could call this new function Sin (x), where Sin (x) = sin(x) provided - x . 2 2

1.6

INVERSE TRIGONOMETRIC FUNCTIONS

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Sin (x) satisfies the horizontal line test and therefore has an inverse function which we call the inverse sine function and denote it as sin-1 (x). Note that sin-1 (x) = 1 . sin(x)

In fact the practice of writing the inverse of the sine function as sin-1 is a bad habit that we have to get used to. It is too well-established to change; in particular, that is how the function is indicated on calculators. As mentioned earlier, the inverse of sin is also often written as arcsin(x) , and its graph is the following one:

2

f (x) = arcsin x

-1

1

- 2

The domain of arcsin is: [-1, 1] . The range of arcsin is: - , . 2 2

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Note that, as always, the graph can be produced using Maple. Simply enter the command > plot(arcsin(x), x=-2..2, scaling=constrained);

Since arcsin undoes what sin does, and vice-versa, the following equations are true, but only for the specified values of x: sin-1 (sin x) = x, for - x 2 2 sin(sin-1 x) = x, for -1 x 1. Example 17. What is the value of arcsin(0.5)? sin(/6) = 0.5, so arcsin(0.5) = /6.

Example 18. sin(-7/5) = 0.951, so what is the value of sin-1 (0.951)? Careful here! The temptation is to give -7/5 as the answer. The problem is that this is not in the range of the function sin-1 , which is [-, ]. Instead we need to observe that if sin(-7/5) = 0.951 then also sin(-7/5+2) = 0.951; that is, sin(3/5) = 0.951. Since 3/5 is in the range of sin-1 , this is the correct answer.

1.6

INVERSE TRIGONOMETRIC FUNCTIONS

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The inverse of cosine is obtained by a calculation similar to the way the inverse of sine was determined. We analyze cosine from 0 to ; this is shown in the following graph:

1

f (x) = Cos x

0

-1

For convenience, we could call this new function Cos (x) where Cos (x) = cos(x) provided 0 x . Cos (x) satisfies the horizontal line test and therefore has an inverse function which we call the inverse cosine function and denote it as cos-1 (x), 1 . cos x The inverse cosine function is also often written as cos-1 x = arccos x, and it is graphed on the following page. noting that

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f (x) = arccos x

-1 The domain of arccos is: [-1, 1]. The range of arccos is: [0, ].

1

Concept Question 7. When you enter cos-1 2 on your calculator, it objects. Why is that? A. The numbers involved are too large for the calculator to handle B. The calculator does not understand this business of taking the inverse using only part of the cosine function C. The cosine function does not really have an inverse D. The number 2 is outside the domain of the function arccos The answer is D. The number 2 is not in the domain of the function arccos.

1.6

INVERSE TRIGONOMETRIC FUNCTIONS

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Concept Question 8. What is the value of arccos(cos(7/6))? A. 7/6 B. 210 C. 5/6 You might think that since arccos is also referred to as "inverse cosine", therefore arccos(cos(7/6)) = 7/6. However, 7/6 is not in the range of the function arccos, which is the interval [0, ], so we cannot get that as the answer. arccos(cos(7/6)) is the angle (in radians) which lies between 0 and and whose cosine is the same as cos(7/6). We can determine that by looking at the graph of the cosine function:

1

0.5

0 0 2 4 x -0.5 6

-1

Thus, the correct answer is - /6 = 5/6, which is C.

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The inverse of tangent is determined in the same way, only analyzing it from - to , this is shown in the following graph: 2 2

f (x) = T an x

- 2

2

As done before, we name this portion of the tan function T an (x), where T an(x) = tan x provided - < x < . 2 2 T an (x) satisfies the horizontal line test and therefore has an inverse, which we call the inverse tangent function and denote it as tan-1 x, once again noting that tan-1 x = 1 . tan x

1.6

INVERSE TRIGONOMETRIC FUNCTIONS

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The inverse tangent function is also often written as arctan x, and its graph is the following one:

2

f (x) = tan-1 x

- 2

This graph can be produced in Maple, using the command > plot(arctan(x), x=-5..5, scaling=constrained); If you want the horizontal asymptotes as well, use the command > plot([arctan(x), Pi/2, -Pi/2], x=-5..5, scaling=constrained); The domain of arctan is: (-, ). The range of arctan is: (- , ). 2 2

Example 19. What is the value of arctan(1)? tan(/4) = 1, so arctan(1) = /4

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3.5 [3.6] Implicit Differentiation

The relation between an "input" x and "output" y may be given explicitly. Here is an example: y = 1 - x2 . This same relation can be given implicitly: x2 + y 2 = 1. When the relation is given implicitly, there are often several y's for a given x, unless you add some information, e.g. y 0. Sometimes an implicit relation can be turned into an explicit relation by algebra, as in the above example, although usually this cannot be done! For example, consider the relation: cos(x - y) = y sin x.

dy You might think that to calculate dx you must first write y explicitly in terms of x. In this section we learn that this is not necessary. Hence the term "Implicit Differentiation".

Consider the following example: x 1 + y + y x - 1 = x + 5. In this relation, think of y as somehow dependent on x. That is y = y(x); in other words y is a function of x. To help you think this way, you can write the expression in this way: x 1 + y(x) + y(x) x - 1 = x + 5. Now differentiate both sides: 1 1 + y(x) + x · · 2

dy dx

1 + y(x)

dy dx

+

dy 1 1 · x - 1 + y(x) · · = 1. dx 2 x-1

Next, collect terms involving

and go back to writing y instead of y(x): dy =1- dx y 1+y- . 2 x-1

x + x-1 2 1+y

3.5 [3.6] IMPLICIT DIFFERENTIATION Solving for

dy dx

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gives:

1 - 1 + y - 2y dy x-1 = . x dx + x-1 2 1+y

As in the example just demonstrated, it is typical of implicit differentiation that the answer involves both x and y. You can think of x 1+y +y x -1 = x +5;

or, to express it more completely in set notation, {(x, y) R2 | x 1 + y + y x - 1 = x + 5} , as some sort of curve in the (x, y)-plane.

(2,3)

Of course, this curve is not presented to us as a parametric curve (as a vector valued function). It may be possible to find a way to parametrize this particular set of points, but it is probably difficult. Fortunately we do not need to. Note that (x, y) = (2, 3) satisfies the equation, therefore (2, 3) lies on the curve. We can calculate the slope at the point (2, 3) by substituting dy for x = 2 and y = 3 into the formula obtained earlier for dx . This works out dy to dx = - 5 . This tells us that the slope of the curve (that is, the tangent 3 5 line to the curve) at (2,3) is equal to - 3 . Note that we are not speaking of a velocity here, for to have a velocity you first need a parametrization. If we did have a parametrization for this curve - that is, if we had a particle moving along this curve - then its velocity as it passes through the point (2, 3) would lie along the tangent line whose slope we just calculated.

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Example 20. Find the points where the ellipse x2 - xy + y 2 = 3, crosses the x-axis and show that the tangent lines at these points are parallel. When the ellipse crosses the x-axis, y = 0. If we substitute this value for y into the given equation of the ellipse we obtain the expression x = ± 3. 3, 0 and P2 - 3, 0 . So the ellipse crosses the the x-axis at the points P1

To show that the tangent lines are parallel at these points, we must show dy that their slopes are equal. So we need to find the value of dx at P1 and P2 . Lets differentiate the equation of the ellipse: 2x - y + x dy dx + 2y dy =0 dx

2x - y - x

dy dy + 2y =0 dx dx dy (2y - x) = y - 2x dx dy y - 2x = dx 2y - x -2 3 = - 3 =2 2 3 = 3 =2

At P1 , dy dx At P2 ,

P1

dy dx

P2

So the slopes of the tangent lines to the ellipse x2 - xy + y 2 = 3 at the points P1 3, 0 and P2 - 3, 0 are equal. So the tangent lines are parallel at these points.

3.5 [3.6] IMPLICIT DIFFERENTIATION

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Example 21. Let y = sin-1 x, and differentiate the equation sin(sin-1 x) = d x, or sin(y) = x, to get a formula for sin-1 (x). dx Recall that when y = sin-1 x then - y . Differentiating sin y = x 2 2 implicitly with respect to x, we get cos y Therefore, dy = 1. dx

1 dy = . dx cos y

Since -

y we know that cos y 0, and so 2 2 cos y = 1 - sin2 y = 1 - y2 .

Thus

d 1 sin-1 x = dx 1 - x2

MEMORIZE

Note: This solution tacitly assumes that we already know that the function sin-1 (x) is differentiable, for only then are we permitted to differentiate y the way we did. Notice that Maple calculates derivatives for you. In this case you can do it in two ways: You can work with the function arcsin directly and enter > D(arcsin); or you can start with the expression arcsin(x) (that is; put the variable in), and enter > diff(arcsin(x), x);

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Here is a list of the derivatives of the inverse trigonometric functions. Memorize the first three of these. They will be used a lot in the course.

d sin-1 x dx

=

d tan-1 x = dx d cos-1 x dx d sec-1 x dx = =

1 ; 1 - x2 1 ; 1 + x2 - 1 ; 1 - x2 1 . x x2 - 1

Example 22. Differentiate y = tan-1 (sin x).

dy 1 d = · (sin x) dx 1 + (sin x)2 dx = = 1 · cos x 1 + (sin x)2 cos x 1 + sin2 x

3.6 [3.8] LOGARITHMIC DIFFERENTIATION

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3.6 [3.8] Logarithmic Differentiation

Example 23. Find dy when y = (x2 + 1)x . dx

Logarithmic differentiation is mainly used when the input variable appears in the base and in the exponent. You may incorrectly think that d 2 (x + 1)x = x(x2 + 1)x-1 · (2x). dx

d This is incorrect for the power rule dx xr = rxr-1 is valid only when r is a constant! We need to take the natural logarithm of both sides of the given equation and then implicitly differentiate:

y = (x2 + 1)x ln y = x ln(x2 + 1) 1 dy 1 = 1 · ln(x2 + 1) + x · 2 · 2x y dx x +1 = ln(x2 + 1) + Now solve for

dy : dx

2x2 x2 + 1

2x2 dy = y ln(x2 + 1) + 2 dx x +1 = (x2 + 1)x ln(x2 + 1) + 2x2 x2 + 1

Maple is not fazed by logarithmic derivatives. Try > diff((x2+1)x, x);

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