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Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 63

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Figure 19.9. Use a straightedge and the calibrated unit circle drawn below to estimate each of the following values. (a) cos(1.1) (b) sin(1.1) (c) sin(3.5) (d) cos(4.2) (e) cos(5.9) (f) sin(2.2) (g) sin(5.7). Find the following.. w) (e) sin(w + 6π ) (f ) cos(w − 2π. t ) (e) sin(t − π ) (f ) sin(t − 10π. (g)...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 64

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Graph the following.. Find the period, amplitude, and balance value of each of the following functions.. sin(0.2x) π + π (d) j (x. Graph the height of a point on the Ferris wheel versus time, assuming that at t = 0 the point is at height 0. Give an equation whose graph is the picture you’ve drawn.. (a) What is...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 65

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the line drawn from the origin to (0, 1) makes an angle of π 2 radians with the positive horizontal axis.. Figure 19.22. When we write sin(1.1) and want to think of 1.1 as an angle, then it is an angle in radians. sin(1.1 radians).. or π radians.. π radians π radians = 180 degrees 1. EXERCISE 19.12 Convert from...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 66

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Figure 20.6. From times of antiquity, scientists and engineers have used triangle trigonometry to con- struct edifices and to estimate distances and heights—from estimating the height of the great Egyptian pyramids in Giza to estimating the sizes and distances of the sun and the moon.. In the case of a right triangle, if we know the measure of one of...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 67

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EXAMPLE 20.5 You’re interested in knowing the height of a very tall tree. You position yourself so that your line of sight to the top of the tree makes a 60 ◦ angle with the horizontal. You measure the distance from where you stand to the base of the tree to be 45 feet. How tall is the tree?. It’s...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 68

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(a) sin −1 (sin. 20.4 SOLVING TRIGONOMETRIC EQUATIONS. If we can get the trigonometric equation into the form sin u = k, cos u = k, or tan u = k, where k is a constant, then we can use the inverse trigonometric functions to help solve for u.. EXAMPLE 20.9 Solve for x.. But there are actually two solutions...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 69

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C an obtuse angle (ii) Figure 20.39. Check that regardless of whether we use part (i) or part (ii) of Figure 20.39, we have. a , or, equivalently, x = a cos C sin C = y. c 2 = a 2 cos 2 C − 2ab cos C + b 2 + a 2 sin 2 C c 2...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 70

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20.7 A BRIEF INTRODUCTION TO VECTORS. In fact, we can have the arrow answer both the amount and direction questions by having the length of the arrow indicate “how far” or “how fast” and the direction the arrow is pointing indicate the direction of the displacement or motion. A velocity vector is a vector whose length represents speed and whose...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 71

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words, what is the horizontal component of its displacement vector?). Force A has a horizontal component of 3 pounds and a vertical component of 4 pounds.. Force B has a horizontal component of 5 pounds and a vertical component of 12 pounds.. (a) What is the strength of force A? What angle does this force vector make with the horizontal?...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 72

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EXERCISE 21.2 Show that d. EXERCISE 21.3 Show that d. EXAMPLE 21.1 Differentiate the following.. sin(π/180. (d) What is the maximum value of for e −0.3x sin x for x ≥ 0? At what x -value is this maximum attained? Your answers must be exact, not numerical approximations from a calculator. Is the derivative. 21.3 APPLICATIONS. where g is the...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 73

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If the origin is placed at the center of the wheel a point on the rim has a horizontal position of (7, 0) at time t = 0. What is the horizontal component of the point’s velocity at t = 2?. In this problem you will show that y = C 1 sin kx + C 2 cos kx is...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 74

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Net Change in Amount and Area: Introducing the. 22.1 FINDING NET CHANGE IN AMOUNT: PHYSICAL AND GRAPHICAL INTERPLAY. How do we calculate the instantaneous rate of change of a quantity?. How do we calculate the slope of the line tangent to a curve at a point?. the slope of the graph of an “amount”. function can be interpreted as the...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 75

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Let’s return for a moment to the problem of calculating the cheetah’s net change in position and approach it with a different mindset. Then the change in position from t = 1 to t = 4 is given by s(4. In Chapter 24 we will explore the relationship between the two mindsets presented in Example 22.5b and arrive at a...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 76

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(a) The lead the chicken has when the goat begins to move (b) The distance between the goat and the chicken at time t = 1.5 (c) The distance between the goat and the chicken at time t = 3 (d) The time at which the goat is farthest ahead of the chicken. At this time, what is the relationship...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 77

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22.4 Properties of the Definite Integral 741. (Hint: Look at the sign of the integrand.) (b) Put in ascending order:. For each of the following, sketch a graph of the indicated region and write a definite integral (or, if you prefer, the sum and/or differences of definite integrals) that gives the area of the region.. (a) The area between the...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 78

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Figure 23.9. SOLUTION Let’s begin with − 2 A f (x) because none of the domain is to the left of the anchor point.. graph of –2 A f sign of f. 2 A f (x) is the area of the trapezoid bounded by f (t. (See Figure 23.10.) Its area is. x is negative Figure 23.10. For x on...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 79

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The Fundamental Theorem of Calculus. 24.1 DEFINITE INTEGRALS AND THE FUNDAMENTAL THEOREM. We concluded the previous chapter with the Fundamental Theorem of Calculus, version 1.. One reason this result is so exciting is that we can use it to obtain a simple and beautiful method for computing definite integrals. A function F is an antiderivative of f if its derivative...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 80

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net change in. When we interpret the integrand as a rate function the Fundamental Theorem of Calculus becomes transparent.. EXAMPLE 24.9 If f (t ) is the rate at which water is flowing into or out of a reservoir and we let W (t ) be the amount of water in the reservoir at time t, then dW dt =...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 81

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24.2 The Average Value of a Function: An Application of the Definite Integral 781 6. 5e − t is the number of grams of the chemical at time t, what is the average number of grams of the chemical in the mixture on the time interval [0, 1]?. (d) What is the object’s average velocity on [0, 2]?. Let I...

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 82

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du = 2x dx so x dx = 1 2 du. 2 sin u + C Now return to the original variable.. Why Is the Substitution du = u (x) dx Legitimate?. In the example above we went from du dx = u (x) to du = u (x) dx. Replacing du dx dx by du is legitimate. This is...