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Find the cosine and sine of the following angles.\n\n1. \( \theta = {270}^{ \circ } \) 2. \( \theta = - \pi \) 3. \( \theta = {45}^{ \circ } \) 4. \( \theta = \frac{\pi }{6} \) 5. \( \theta = {60}^{ \circ } \) | Solution.\n\n1. To find \( \cos \left( {270}^{ \circ }\right) \) and \( \sin \left( {270}^{ \circ }\right) \), we plot the angle \( \theta = {270}^{ \circ } \) in standard position and find the point on the terminal side of \( \theta \) which lies on the Unit Circle. Since \( {270}^{ \circ } \) represents \( \frac{3}{4... | Yes |
If \( \theta \) is a Quadrant II angle with \( \sin \left( \theta \right) = \frac{3}{5} \), find \( \cos \left( \theta \right) \). | When we substitute \( \sin \left( \theta \right) = \frac{3}{5} \) into The Pythagorean Identity, \( {\cos }^{2}\left( \theta \right) + {\sin }^{2}\left( \theta \right) = 1 \), we obtain \( {\cos }^{2}\left( \theta \right) + \frac{9}{25} = 1 \). Solving, we find \( \cos \left( \theta \right) = \pm \frac{4}{5} \). Since ... | Yes |
Find the cosine and sine of the following angles.\n\n1. \( \theta = {225}^{ \circ } \) 2. \( \theta = \frac{11\pi }{6} \) 3. \( \theta = - \frac{5\pi }{4} \) 4. \( \theta = \frac{7\pi }{3} \) | Solution.\n\n1. We begin by plotting \( \theta = {225}^{ \circ } \) in standard position and find its terminal side overshoots the negative \( x \) -axis to land in Quadrant III. Hence, we obtain \( \theta \) ’s reference angle \( \alpha \) by subtracting: \( \alpha = \theta - {180}^{ \circ } = {225}^{ \circ } - {180}^... | No |
Suppose \( \alpha \) is an acute angle with \( \cos \left( \alpha \right) = \frac{5}{13} \). 1. Find \( \sin \left( \alpha \right) \) and use this to plot \( \alpha \) in standard position. | 1. Proceeding as in Example 10.2.2, we substitute \( \cos \left( \alpha \right) = \frac{5}{13} \) into \( {\cos }^{2}\left( \alpha \right) + {\sin }^{2}\left( \alpha \right) = 1 \) and find \( \sin \left( \alpha \right) = \pm \frac{12}{13} \). Since \( \alpha \) is an acute (and therefore Quadrant I) angle, \( \sin \le... | Yes |
Find all of the angles which satisfy the given equation.\n\n1. \( \cos \left( \theta \right) = \frac{1}{2} \) 2. \( \sin \left( \theta \right) = - \frac{1}{2} \) 3. \( \cos \left( \theta \right) = 0 \) . | Solution. Since there is no context in the problem to indicate whether to use degrees or radians, we will default to using radian measure in our answers to each of these problems. This choice will be justified later in the text when we study what is known as Analytic Trigonometry. In those sections to come, radian meas... | No |
Theorem 10.3. If \( Q\left( {x, y}\right) \) is the point on the terminal side of an angle \( \theta \), plotted in standard position, which lies on the circle \( {x}^{2} + {y}^{2} = {r}^{2} \) then \( x = r\cos \left( \theta \right) \) and \( y = r\sin \left( \theta \right) \) . | Moreover, \[ \cos \left( \theta \right) = \frac{x}{r} = \frac{x}{\sqrt{{x}^{2} + {y}^{2}}}\text{ and }\sin \left( \theta \right) = \frac{y}{r} = \frac{y}{\sqrt{{x}^{2} + {y}^{2}}} \] | Yes |
1. Suppose that the terminal side of an angle \( \theta \), when plotted in standard position, contains the point \( Q\left( {4, - 2}\right) \) . Find \( \sin \left( \theta \right) \) and \( \cos \left( \theta \right) \) . | 1. Using Theorem 10.3 with \( x = 4 \) and \( y = - 2 \), we find \( r = \sqrt{{\left( 4\right) }^{2} + {\left( -2\right) }^{2}} = \sqrt{20} = 2\sqrt{5} \) so that \( \cos \left( \theta \right) = \frac{x}{r} = \frac{4}{2\sqrt{5}} = \frac{2\sqrt{5}}{5} \) and \( \sin \left( \theta \right) = \frac{y}{r} = \frac{-2}{2\sqr... | Yes |
Suppose we are in the situation of Example 10.1.5. Find the equations of motion of Lakeland Community College as the earth rotates. | Solution. From Example 10.1.5, we take \( r = {2960} \) miles and and \( \omega = \frac{\pi }{{12}\text{ hours }} \) . Hence, the equations of motion are \( x = r\cos \left( {\omega t}\right) = {2960}\cos \left( {\frac{\pi }{12}t}\right) \) and \( y = r\sin \left( {\omega t}\right) = {2960}\sin \left( {\frac{\pi }{12}t... | Yes |
Find the measure of the missing angle and the lengths of the missing sides of: | Solution. The first and easiest task is to find the measure of the missing angle. Since the sum of angles of a triangle is \( {180}^{ \circ } \), we know that the missing angle has measure \( {180}^{ \circ } - {30}^{ \circ } - {90}^{ \circ } = {60}^{ \circ } \) . We now proceed to find the lengths of the remaining two ... | Yes |
1. \( \sec \left( {60}^{ \circ }\right) \) | 1. According to Theorem 10.6, \( \sec \left( {60}^{ \circ }\right) = \frac{1}{\cos \left( {60}^{ \circ }\right) } \) . Hence, \( \sec \left( {60}^{ \circ }\right) = \frac{1}{\left( 1/2\right) } = 2 \) . | Yes |
Find all angles which satisfy the given equation.\n\n1. \( \sec \left( \theta \right) = 2 \) 2. \( \tan \left( \theta \right) = \sqrt{3} \) 3. \( \cot \left( \theta \right) = - 1 \) . | Solution.\n\n1. To solve \( \sec \left( \theta \right) = 2 \), we convert to cosines and get \( \frac{1}{\cos \left( \theta \right) } = 2 \) or \( \cos \left( \theta \right) = \frac{1}{2} \) . This is the exact same equation we solved in Example 10.2.5, number 1, so we know the answer is: \( \theta = \frac{\pi }{3} + {... | Yes |
1. \( \frac{1}{\csc \left( \theta \right) } = \sin \left( \theta \right) \) | To verify \( \frac{1}{\csc \left( \theta \right) } = \sin \left( \theta \right) \), we start with the left side. Using \( \csc \left( \theta \right) = \frac{1}{\sin \left( \theta \right) } \), we get:\n\n\[ \frac{1}{\csc \left( \theta \right) } = \frac{1}{\frac{1}{\sin \left( \theta \right) }} = \sin \left( \theta \rig... | Yes |
Theorem 10.11. Domains and Ranges of the Circular Functions | - The function \( f\left( t\right) = \cos \left( t\right) \; \bullet \) The function \( g\left( t\right) = \sin \left( t\right) \)\n\n- has domain \( \left( {-\infty ,\infty }\right) \; - \) has domain \( \left( {-\infty ,\infty }\right) \)\n\n- has range \( \left\lbrack {-1,1}\right\rbrack \; - \) has range \( \left\l... | Yes |
1. Find the exact value of \( \cos \left( {15}^{ \circ }\right) \) . | In order to use Theorem 10.13 to find \( \cos \left( {15}^{ \circ }\right) \), we need to write \( {15}^{ \circ } \) as a sum or difference of angles whose cosines and sines we know. One way to do so is to write \( {15}^{ \circ } = {45}^{ \circ } - {30}^{ \circ } \). \n\n\[ \cos \left( {15}^{ \circ }\right) = \cos \lef... | Yes |
Rewrite \( {\sin }^{2}\left( \theta \right) {\cos }^{2}\left( \theta \right) \) as a sum and difference of cosines to the first power. | Solution. We begin with a straightforward application of Theorem 10.18\n\n\[ \n{\sin }^{2}\left( \theta \right) {\cos }^{2}\left( \theta \right) = \left( \frac{1 - \cos \left( {2\theta }\right) }{2}\right) \left( \frac{1 + \cos \left( {2\theta }\right) }{2}\right) \n\]\n\n\[ \n= \frac{1}{4}\left( {1 - {\cos }^{2}\left(... | Yes |
1. Write \( \cos \left( {2\theta }\right) \cos \left( {6\theta }\right) \) as a sum. | Identifying \( \alpha = {2\theta } \) and \( \beta = {6\theta } \), we find\n\n\[ \cos \left( {2\theta }\right) \cos \left( {6\theta }\right) = \frac{1}{2}\left\lbrack {\cos \left( {{2\theta } - {6\theta }}\right) + \cos \left( {{2\theta } + {6\theta }}\right) }\right\rbrack \]\n\n\[ = \frac{1}{2}\cos \left( {-{4\theta... | Yes |
1. Find a cosine function whose graph matches the graph of \( y = f\left( x\right) \) . | We fit the data to a function of the form \( C\left( x\right) = A\cos \left( {{\omega x} + \phi }\right) + B \) . Since one cycle is graphed over the interval \( \left\lbrack {-1,5}\right\rbrack \), its period is \( 5 - \left( {-1}\right) = 6 \) . According to Theorem 10.23, \( 6 = \frac{2\pi }{\omega } \), so that \( ... | Yes |
Consider the function \( f\left( x\right) = \cos \left( {2x}\right) - \sqrt{3}\sin \left( {2x}\right) \). Find a formula for \( f\left( x\right) \): 1. in the form \( C\left( x\right) = A\cos \left( {{\omega x} + \phi }\right) + B \) for \( \omega > 0 \) | The key to this problem is to use the expanded forms of the sinusoid formulas and match up corresponding coefficients. Equating \( f\left( x\right) = \cos \left( {2x}\right) - \sqrt{3}\sin \left( {2x}\right) \) with the expanded form of \( C\left( x\right) = A\cos \left( {{\omega x} + \phi }\right) + B \), we get \[ \c... | Yes |
Graph one cycle of the following functions. State the period of each.\n\n1. \( f\left( x\right) = 1 - 2\sec \left( {2x}\right) \) | To graph \( y = 1 - 2\sec \left( {2x}\right) \), we follow the same procedure as in Example 10.5.1. First, we set the argument of secant, \( {2x} \), equal to the ’quarter marks’ \( 0,\frac{\pi }{2},\pi ,\frac{3\pi }{2} \) and \( {2\pi } \) and solve for \( x \) .\n\n<table><tr><td>\( a \)</td><td>\( {2x} = a \)</td><t... | Yes |
Graph one cycle of the following functions. Find the period.\n\n1. \( f\left( x\right) = 1 - \tan \left( \frac{x}{2}\right) \) | Solution.\n\n1. We proceed as we have in all of the previous graphing examples by setting the argument of tangent in \( f\left( x\right) = 1 - \tan \left( \frac{x}{2}\right) \), namely \( \frac{x}{2} \), equal to each of the ’quarter marks’ \( - \frac{\pi }{2}, - \frac{\pi }{4},0,\frac{\pi }{4} \) and \( \frac{\pi }{2}... | Yes |
Properties of the Arccosine and Arcsine Functions | Everything in Theorem 10.26 is a direct consequence of the facts that \( f\left( x\right) = \cos \left( x\right) \) for \( 0 \leq x \leq \pi \) and \( F\left( x\right) = \arccos \left( x\right) \) are inverses of each other as are \( g\left( x\right) = \sin \left( x\right) \) for \( - \frac{\pi }{2} \leq x \leq \frac{\... | Yes |
Example 10.6.6. \( {}^{7} \) The roof on the house below has a ’ \( 6/{12} \) pitch’. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to ... | Solution. If we divide the side view of the house down the middle, we find that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using Theorem 10.10, we find the angle of inclination, labeled \( \theta \) below, satisfies \( \tan \left( \theta \right) = \frac{6}{12} = \frac... | Yes |
1. Find all angles \( \theta \) for which \( \sin \left( \theta \right) = \frac{1}{3} \) . | If \( \sin \left( \theta \right) = \frac{1}{3} \), then the terminal side of \( \theta \), when plotted in standard position, intersects the Unit Circle at \( y = \frac{1}{3} \) . Geometrically, we see that this happens at two places: in Quadrant I and Quadrant II. If we let \( \alpha \) denote the acute solution to th... | Yes |
1. \( \cos \left( {2x}\right) = - \frac{\sqrt{3}}{2} \) | The solutions to \( \cos \left( u\right) = - \frac{\sqrt{3}}{2} \) are \( u = \frac{5\pi }{6} + {2\pi k} \) or \( u = \frac{7\pi }{6} + {2\pi k} \) for integers \( k \) . Since the argument of cosine here is \( {2x} \), this means \( {2x} = \frac{5\pi }{6} + {2\pi k} \) or \( {2x} = \frac{7\pi }{6} + {2\pi k} \) for in... | Yes |
1. \( 3{\sin }^{3}\left( x\right) = {\sin }^{2}\left( x\right) \) | We resist the temptation to divide both sides of \( 3{\sin }^{3}\left( x\right) = {\sin }^{2}\left( x\right) \) by \( {\sin }^{2}\left( x\right) \) (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor.\n\n\[ 3{\sin }^{3}\left( x\right) = {\sin }^{2}\left( x\right) \]\n... | Yes |
1. \( 2\sin \left( x\right) \leq 1 \) | We begin solving \( 2\sin \left( x\right) \leq 1 \) by collecting all of the terms on one side of the equation and zero on the other to get \( 2\sin \left( x\right) - 1 \leq 0 \) . Next, we let \( f\left( x\right) = 2\sin \left( x\right) - 1 \) and note that our original inequality is equivalent to solving \( f\left( x... | Yes |
Express the domain of the following functions using extended interval notation. \( f\left( x\right) = \csc \left( {{2x} + \frac{\pi }{3}}\right) \) | To find the domain of \( f\left( x\right) = \csc \left( {{2x} + \frac{\pi }{3}}\right) \), we rewrite \( f \) in terms of sine as \( f\left( x\right) = \frac{1}{\sin \left( {{2x} + \frac{\pi }{3}}\right) } \) . Since the sine function is defined everywhere, our only concern comes from zeros in the denominator. Solving ... | Yes |
1. \( \arcsin \left( {2x}\right) = \frac{\pi }{3} \) | To solve \( \arcsin \left( {2x}\right) = \frac{\pi }{3} \), we first note that \( \frac{\pi }{3} \) is in the range of the arcsine function (so a solution exists!) Next, we exploit the inverse property of sine and arcsine from Theorem 10.26\n\n\[ \arcsin \left( {2x}\right) = \frac{\pi }{3} \]\n\n\[ \sin \left( {\arcsin... | Yes |
Recall from Exercise 55 in Section 10.1 that The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height 136 feet. It completes two revolutions in 2 minutes and 7 seconds. Assuming that the riders are at the edge of the circle, find a sinusoid which ... | Solution. We sketch the problem situation below and assume a counter-clockwise rotation. \( {}^{3} \)\n\n\n\n\( {}^{3} \) Otherwise, we could just observe the motion of the wheel from the other side.\n\n\n\nWe know f... | Yes |
1. Find a sinusoid which models these data and use a graphing utility to graph your answer along with the data. | To get a feel for the data, we plot it below.\n\n\n\nThe data certainly appear sinusoidal, \( {}^{5} \) but when it comes down to it, fitting a sinusoid to data manually is not an exact science. We do our best to fin... | Yes |
Theorem 11.1. Equation for Free Undamped Harmonic Motion: Suppose an object of mass \( m \) is suspended from a spring with spring constant \( k \) . If the initial displacement from the equilibrium position is \( {x}_{0} \) and the initial velocity of the object is \( {v}_{0} \), then the displacement \( x \) from the... | \[ \text{-}\omega = \sqrt{\frac{k}{m}}\text{and}A = \sqrt{{x}_{0}^{2} + {\left( \frac{{v}_{0}}{\omega }\right) }^{2}} \] \[ \text{-}A\sin \left( \phi \right) = {x}_{0}\text{and}{A\omega }\cos \left( \phi \right) = {v}_{0}\text{.} \] | Yes |
Suppose an object weighing 64 pounds stretches a spring 8 feet. 1. If the object is attached to the spring and released 3 feet below the equilibrium position from rest, find the equation of motion of the object, \( x\left( t\right) \) . When does the object first pass through the equilibrium position? Is the object hea... | Solution. In order to use the formulas in Theorem 11.1, we first need to determine the spring constant \( k \) and the mass of the object \( m \) . To find \( k \), we use Hooke’s Law \( F = {kd} \) . We know the object weighs 64 lbs. and stretches the spring \( 8 \) ft.. Using \( F = {64} \) and \( d = 8 \), we get \(... | Yes |
Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, find the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundreth of a degree. | Solution. For definitiveness, we label the triangle below.  \( b = 4 \) To find the length of the missing side \( a \), we use the Pythagorean Theorem to get \( {a}^{2} + {4}^{2} = {7}^{2} \) which then yields \( a =... | Yes |
Theorem 11.2. The Law of Sines: Given a triangle with angle-side opposite pairs \( \left( {\alpha, a}\right) \) , \( \left( {\beta, b}\right) \) and \( \left( {\gamma, c}\right) \), the following ratios hold\n\n\[ \frac{\sin \left( \alpha \right) }{a} = \frac{\sin \left( \beta \right) }{b} = \frac{\sin \left( \gamma \r... | The proof of the Law of Sines can be broken into three cases. For our first case, consider the triangle \( \bigtriangleup {ABC} \) below, all of whose angles are acute, with angle-side opposite pairs \( \left( {\alpha, a}\right) ,\left( {\beta, b}\right) \) and \( \left( {\gamma, c}\right) \) . If we drop an altitude f... | Yes |
Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle.\n\n1. \( \alpha = {120}^{ \circ }, a = 7 \) units, \( \beta = {45}^{ \circ } \) | Knowing an angle-side opposite pair, namely \( \alpha \) and \( a \), we may proceed in using the Law of Sines. Since \( \beta = {45}^{ \circ } \), we use \( \frac{b}{\sin \left( {45}^{ \circ }\right) } = \frac{7}{\sin \left( {120}^{ \circ }\right) } \) so \( b = \frac{7\sin \left( {45}^{ \circ }\right) }{\sin \left( {... | Yes |
Theorem 11.3. Suppose \( \left( {\alpha, a}\right) \) and \( \left( {\gamma, c}\right) \) are intended to be angle-side pairs in a triangle where \( \alpha, a \) and \( c \) are given. Let \( h = c\sin \left( \alpha \right) \n- If \( a < h \), then no triangle exists which satisfies the given criteria.\n- If \( a = h \... | Theorem 11.3 is proved on a case-by-case basis. If \( a < h \), then \( a < c\sin \left( \alpha \right) \) . If a triangle were to exist, the Law of Sines would have \( \frac{\sin \left( \gamma \right) }{c} = \frac{\sin \left( \alpha \right) }{a} \) so that \( \sin \left( \gamma \right) = \frac{c\sin \left( \alpha \rig... | Yes |
Sasquatch Island lies off the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the first observation point is \( {30}^{ \circ } \) and at the second point the angle is \( {45}^{ \circ } \). Assuming a straight coastline, find the distance ... | Solution. We sketch the problem below with the first observation point labeled as \( P \) and the second as \( Q \). In order to use the Law of Sines to find the distance \( d \) from \( Q \) to the island, we first need to find the measure of \( \beta \) which is the angle opposite the side of length \( 5\mathrm{{mile... | Yes |
Find the area of the triangle in Example 11.2.2 number 1. | Solution. From our work in Example 11.2.2 number 1, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose \( A = \frac{1}{2}{ac}\sin \left( \beta \right) \) from Theorem 11.4 because it uses the most pieces of given information. We are given \( a = 7 \) and \( \beta... | Yes |
Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle.\n\n1. \( \beta = {50}^{ \circ }, a = 7 \) units, \( c = 2 \) units | 1. We are given the lengths of two sides, \( a = 7 \) and \( c = 2 \), and the measure of the included angle, \( \beta = {50}^{ \circ } \) . With no angle-side opposite pair to use, we apply the Law of Cosines. We get \( {b}^{2} = {7}^{2} + {2}^{2} - 2\left( 7\right) \left( 2\right) \cos \left( {50}^{ \circ }\right) \)... | Yes |
A researcher wishes to determine the width of a vernal pond as drawn below. From a point \( P \), he finds the distance to the eastern-most point of the pond to be 950 feet, while the distance to the western-most point of the pond from \( P \) is 1000 feet. If the angle between the two lines of sight is \( {60}^{ \circ... | Solution. We are given the lengths of two sides and the measure of an included angle, so we may apply the Law of Cosines to find the length of the missing side opposite the given angle. Calling this length \( w \) (for width), we get \( {w}^{2} = {950}^{2} + {1000}^{2} - 2\left( {950}\right) \left( {1000}\right) \cos \... | Yes |
Find the area enclosed of the triangle in Example 11.3.1 number 2. | Solution. We are given \( a = 4, b = 7 \) and \( c = 5 \) . Using these values, we find \( s = \frac{1}{2}\left( {4 + 7 + 5}\right) = 8 \) , \( \left( {s - a}\right) = 8 - 4 = 4,\left( {s - b}\right) = 8 - 7 = 1 \) and \( \left( {s - c}\right) = 8 - 5 = 3 \) . Using Heron’s Formula, we get \( A = \sqrt{s\left( {s - a}\... | Yes |
For each point in polar coordinates given below plot the point and then give two additional expressions for the point, one of which has \( r > 0 \) and the other with \( r < 0 \). 1. \( P\left( {2,{240}^{ \circ }}\right) \) | Whether we move 2 units along the polar axis and then rotate \( {240}^{ \circ } \) or rotate \( {240}^{ \circ } \) then move out 2 units from the pole, we plot \( P\left( {2,{240}^{ \circ }}\right) \) below. We now set about finding alternate descriptions \( \left( {r,\theta }\right) \) for the point \( P \) . Since \(... | Yes |
Theorem 11.7. Conversion Between Rectangular and Polar Coordinates: Suppose \( P \) is represented in rectangular coordinates as \( \left( {x, y}\right) \) and in polar coordinates as \( \left( {r,\theta }\right) \). Then\n\n\[ \text{-}x = r\cos \left( \theta \right) \text{and}y = r\sin \left( \theta \right) \]\n\n\[ \... | In the case \( r > 0 \), Theorem 11.7 is an immediate consequence of Theorem 10.3 along with the quotient identity \( \tan \left( \theta \right) = \frac{\sin \left( \theta \right) }{\cos \left( \theta \right) } \). If \( r < 0 \), then we know an alternate representation for \( \left( {r,\theta }\right) \) is \( \left(... | Yes |
Convert each point in rectangular coordinates given below into polar coordinates with \( r \geq 0 \) and \( 0 \leq \theta < {2\pi } \) . Use exact values if possible and round any approximate values to two decimal places. Check your answer by converting them back to rectangular coordinates. | 1. Even though we are not explicitly told to do so, we can avoid many common mistakes by taking the time to plot the points before we do any calculations. Plotting \( P\left( {2, - 2\sqrt{3}}\right) \) shows that it lies in Quadrant IV. With \( x = 2 \) and \( y = - 2\sqrt{3} \), we get \( {r}^{2} = {x}^{2} + {y}^{2} =... | Yes |
Graph the following polar equations.\n\n1. \( r = 4 \) 2. \( r = - 3\sqrt{2} \) 3. \( \theta = \frac{5\pi }{4} \) 4. \( \theta = - \frac{3\pi }{2} \) | Solution. In each of these equations, only one of the variables \( r \) and \( \theta \) is present making the other variable free. \( {}^{1} \) This makes these graphs easier to visualize than others.\n\n1. In the equation \( r = 4,\theta \) is free. The graph of this equation is, therefore, all points which have a po... | No |
Graph the following polar equations.\n\n1. \( r = 4 - 2\sin \left( \theta \right) \) | We first plot the fundamental cycle of \( r = 4 - 2\sin \left( \theta \right) \) on the \( {\theta r} \) -axes. To help us visualize what is going on graphically, we divide up \( \left\lbrack {0,{2\pi }}\right\rbrack \) into the usual four subintervals \( \left\lbrack {0,\frac{\pi }{2}}\right\rbrack ,\left\lbrack {\fra... | No |
Sketch the region in the \( {xy} \) -plane described by the following sets.\n\n1. \( \left\{ {\left( {r,\theta }\right) \mid 0 \leq r \leq 5\sin \left( {2\theta }\right) ,0 \leq \theta \leq \frac{\pi }{2}}\right\} \) | Solution. Our first step in these problems is to sketch the graphs of the polar equations involved to get a sense of the geometric situation. Since all of the equations in this example are found in either Example 11.5.2 or Example 11.5.3, most of the work is done for us.\n\n1. We know from Example 11.5.2 number 3 that ... | Yes |
Theorem 11.9. Rotation of Axes: Suppose the positive \( x \) and \( y \) axes are rotated counterclockwise through an angle \( \theta \) to produce the axes \( {x}^{\prime } \) and \( {y}^{\prime } \), respectively. Then the coordinates \( P\left( {x, y}\right) \) and \( P\left( {{x}^{\prime },{y}^{\prime }}\right) \) ... | \[ \left\{ {\begin{array}{l} x = {x}^{\prime }\cos \left( \theta \right) - {y}^{\prime }\sin \left( \theta \right) \\ y = {x}^{\prime }\sin \left( \theta \right) + {y}^{\prime }\cos \left( \theta \right) \end{array}\;\text{ and }\left\{ \begin{array}{l} {x}^{\prime } = x\cos \left( \theta \right) + y\sin \left( \theta ... | Yes |
Let \( P\left( {x, y}\right) = \left( {2, - 4}\right) \) and find \( P\left( {{x}^{\prime },{y}^{\prime }}\right) \). Check your answer algebraically and graphically. | If \( P\left( {x, y}\right) = \left( {2, - 4}\right) \) then \( x = 2 \) and \( y = - 4 \). Using these values for \( x \) and \( y \) along with \( \theta = \frac{\pi }{3} \), Theorem 11.9 gives \( {x}^{\prime } = x\cos \left( \theta \right) + y\sin \left( \theta \right) = 2\cos \left( \frac{\pi }{3}\right) + \left( {... | Yes |
Graph the following equations.\n\n1. \( 5{x}^{2} + {26xy} + 5{y}^{2} - {16x}\sqrt{2} + {16y}\sqrt{2} - {104} = 0 \) | Since the equation \( 5{x}^{2} + {26xy} + 5{y}^{2} - {16x}\sqrt{2} + {16y}\sqrt{2} - {104} = 0 \) is already given to us in the form required by Theorem 11.10, we identify \( A = 5, B = {26} \) and \( C = 5 \) so that \( \cot \left( {2\theta }\right) = \frac{A - C}{B} = \frac{5 - 5}{26} = 0 \) . This means \( \cot \lef... | Yes |
Example 11.6.3. Use Theorem 11.11 to classify the graphs of the following non-degenerate conics. | Solution. This is a straightforward application of Theorem 11.11.\n\n1. We have \( A = {21}, B = {10}\sqrt{3} \) and \( C = {31} \) so \( {B}^{2} - {4AC} = {\left( {10}\sqrt{3}\right) }^{2} - 4\left( {21}\right) \left( {31}\right) = - {2304} < 0 \) . Theorem 11.11 predicts the graph is an ellipse, which checks with our... | Yes |
Sketch the graphs of the following equations.\n\n1. \( r = \frac{4}{1 - \sin \left( \theta \right) } \) | Solution.\n\n1. From \( r = \frac{4}{1 - \sin \left( \theta \right) } \), we first note \( e = 1 \) which means we have a parabola on our hands. Since \( {ed} = 4 \), we have \( d = 4 \) and considering the form of the equation, this puts the directrix at \( y = - 4 \) . Since the focus is at \( \left( {0,0}\right) \),... | Yes |
Theorem 11.13. Given constants \( \ell > 0, e \geq 0 \) and \( \phi \), the graph of the equation\n\n\[ r = \frac{\ell }{1 - e\cos \left( {\theta - \phi }\right) } \]\n\nis a conic section with eccentricity \( e \) and one focus at \( \left( {0,0}\right) \) . | - If \( e = 0 \), the graph is a circle centered at \( \left( {0,0}\right) \) with radius \( \ell \) .\n\n- If \( e \neq 0 \), then the conic has a focus at \( \left( {0,0}\right) \) and the directrix contains the point with polar coordinates \( \left( {-d,\phi }\right) \) where \( d = \frac{\ell }{e} \) .\n\n- If \( 0... | Yes |
For each of the following complex numbers find \( \operatorname{Re}\left( z\right) ,\operatorname{Im}\left( z\right) ,\left| z\right| ,\arg \left( z\right) \) and \( \operatorname{Arg}\left( z\right) \) . Plot \( z \) in the complex plane. | 1. For \( z = \sqrt{3} - i = \sqrt{3} + \left( {-1}\right) i \), we have \( \operatorname{Re}\left( z\right) = \sqrt{3} \) and \( \operatorname{Im}\left( z\right) = - 1 \) . To find \( \left| z\right| ,\arg \left( z\right) \) and \( \operatorname{Arg}\left( z\right) \), we need to find a polar representation \( \left( ... | Yes |
Theorem 11.14. Properties of the Modulus: Let \( z \) and \( w \) be complex numbers.\n\n- \( \left| z\right| \) is the distance from \( z \) to 0 in the complex plane\n\n- \( \left| z\right| \geq 0 \) and \( \left| z\right| = 0 \) if and only if \( z = 0 \)\n\n- \( \left| z\right| = \sqrt{\operatorname{Re}{\left( z\ri... | To prove the first three properties in Theorem 11.14, suppose \( z = a + {bi} \) where \( a \) and \( b \) are real numbers. To determine \( \left| z\right| \), we find a polar representation \( \left( {r,\theta }\right) \) with \( r \geq 0 \) for the point \( \left( {a, b}\right) \) . From Section 11.4, we know \( {r}... | Yes |
Theorem 11.15. Properties of the Argument: Let \( z \) be a complex number.\n\n- If \( \operatorname{Re}\left( z\right) \neq 0 \) and \( \theta \in \arg \left( z\right) \), then \( \tan \left( \theta \right) = \frac{\operatorname{Im}\left( z\right) }{\operatorname{Re}\left( z\right) } \) .\n\n- If \( \operatorname{Re}\... | To prove Theorem 11.15, suppose \( z = a + {bi} \) for real numbers \( a \) and \( b \) . By definition, \( a = \operatorname{Re}\left( z\right) \) and \( b = \operatorname{Im}\left( z\right) \), so the point associated with \( z \) is \( \left( {a, b}\right) = \left( {\operatorname{Re}\left( z\right) ,\operatorname{Im... | Yes |
Theorem 11.16. Products, Powers and Quotients Complex Numbers in Polar Form:\n\nSuppose \( z \) and \( w \) are complex numbers with polar forms \( z = \left| z\right| \operatorname{cis}\left( \alpha \right) \) and \( w = \left| w\right| \operatorname{cis}\left( \beta \right) \) . Then\n\n- Product Rule: \( {zw} = \lef... | The proof of Theorem 11.16 requires a healthy mix of definition, arithmetic and identities. We first start with the product rule.\n\n\[ {zw} = \left\lbrack {\left| z\right| \operatorname{cis}\left( \alpha \right) }\right\rbrack \left\lbrack {\left| w\right| \operatorname{cis}\left( \beta \right) }\right\rbrack \]\n\n\[... | Yes |
Let \( z = 2\sqrt{3} + {2i} \) and \( w = - 1 + i\sqrt{3} \). Use Theorem 11.16 to find the following\n\n1. \( {zw} \) 2. \( {w}^{5} \) 3. \( \frac{z}{w} \)\n\nWrite your final answers in rectangular form. | Solution. In order to use Theorem 11.16, we need to write \( z \) and \( w \) in polar form. For \( z = 2\sqrt{3} + {2i} \) , we find \( \left| z\right| = \sqrt{{\left( 2\sqrt{3}\right) }^{2} + {\left( 2\right) }^{2}} = \sqrt{16} = 4 \) . If \( \theta \in \arg \left( z\right) \), we know \( \tan \left( \theta \right) =... | Yes |
Theorem 11.17. The \( {n}^{\text{th }} \) roots of a Complex Number: Let \( z \neq 0 \) be a complex number with polar form \( z = r\operatorname{cis}\left( \theta \right) \). For each natural number \( n, z \) has \( n \) distinct \( {n}^{\text{th }} \) roots, which we denote by \( {w}_{0},{w}_{1},\ldots ,{w}_{n - 1} ... | The proof of Theorem 11.17 breaks into to two parts: first, showing that each \( {w}_{k} \) is an \( {n}^{\text{th }} \) root, and second, showing that the set \( \left\{ {{w}_{k} \mid k = 0,1,\ldots ,\left( {n - 1}\right) }\right\} \) consists of \( n \) different complex numbers. To show \( {w}_{k} \) is an \( {n}^{\... | Yes |
Use Theorem 11.17 to find the following:\n\n1. both square roots of \( z = - 2 + {2i}\sqrt{3} \) | ## Solution.\n\n1. We start by writing \( z = - 2 + {2i}\sqrt{3} = 4\operatorname{cis}\left( \frac{2\pi }{3}\right) \) . To use Theorem 11.17, we identify \( r = 4 \) , \( \theta = \frac{2\pi }{3} \) and \( n = 2 \) . We know that \( z \) has two square roots, and in keeping with the notation in Theorem 11.17, we’ll ca... | Yes |
A plane leaves an airport with an airspeed \( {}^{5} \) of 175 miles per hour at a bearing of \( \mathrm{N}{40}^{ \circ }\mathrm{E} \) . A \( {35}\mathrm{{mile}} \) per hour wind is blowing at a bearing of \( \mathrm{S}{60}^{ \circ }\mathrm{E} \) . Find the true speed of the plane, rounded to the nearest mile per hour,... | Solution: For both the plane and the wind, we are given their speeds and their directions. Coupling speed (as a magnitude) with direction is the concept of velocity which we've seen a few times before in this textbook. \( {}^{6} \) We let \( \overrightarrow{v} \) denote the plane’s velocity and \( \overrightarrow{w} \)... | Yes |
Let \( \overrightarrow{v} = \langle 3,4\rangle \) and suppose \( \overrightarrow{w} = \overrightarrow{PQ} \) where \( P\left( {-3,7}\right) \) and \( Q\left( {-2,5}\right) \) . Find \( \overrightarrow{v} + \overrightarrow{w} \) and interpret this sum geometrically. | Solution. Before can add the vectors using Definition 11.6, we need to write \( \overrightarrow{w} \) in component form. Using Definition 11.5, we get \( \overrightarrow{w} = \langle - 2 - \left( {-3}\right) ,5 - 7\rangle = \langle 1, - 2\rangle \) . Thus\n\n\[ \overrightarrow{v} + \overrightarrow{w} = \langle 3,4\rang... | Yes |
Theorem 11.19. Properties of Scalar Multiplication\n\n- Associative Property: For every vector \( \overrightarrow{v} \) and scalars \( k \) and \( r,\left( {kr}\right) \overrightarrow{v} = k\left( {r\overrightarrow{v}}\right) \) . | The proof of Theorem 11.19, like the proof of Theorem 11.18, ultimately boils down to the definition of scalar multiplication and properties of real numbers. For example, to prove the associative property, we let \( \overrightarrow{v} = \left\langle {{v}_{1},{v}_{2}}\right\rangle \) . If \( k \) and \( r \) are scalars... | Yes |
Solve \( 5\overrightarrow{v} - 2\left( {\overrightarrow{v}+\langle 1, - 2\rangle }\right) = \overrightarrow{0} \) for \( \overrightarrow{v} \) . | \[ 5\overrightarrow{v} - 2\left( {\overrightarrow{v}+\langle 1, - 2\rangle }\right) = \overrightarrow{0} \] \[ 5\overrightarrow{v} + \left( {-1}\right) \left\lbrack {2\left( {\overrightarrow{v}+\langle 1, - 2\rangle }\right) }\right\rbrack = \overrightarrow{0} \] \[ 5\overrightarrow{v} + \left\lbrack {\left( {-1}\right... | Yes |
Theorem 11.20. Properties of Magnitude and Direction: Suppose \( \overrightarrow{v} \) is a vector.\n\n- \( \parallel \overrightarrow{v}\parallel \geq 0 \) and \( \parallel \overrightarrow{v}\parallel = 0 \) if and only if \( \overrightarrow{v} = \overrightarrow{0} \) | The proof of the first property in Theorem 11.20 is a direct consequence of the definition of \( \parallel \overrightarrow{v}\parallel \) . If \( \overrightarrow{v} = \left\langle {{v}_{1},{v}_{2}}\right\rangle \), then \( \parallel \overrightarrow{v}\parallel = \sqrt{{v}_{1}^{2} + {v}_{2}^{2}} \) which is by definitio... | Yes |
1. Find the component form of the vector \( \overrightarrow{v} \) with \( \parallel \overrightarrow{v}\parallel = 5 \) so that when \( \overrightarrow{v} \) is plotted in standard position, it lies in Quadrant II and makes a \( {60}^{ \circ } \) angle \( {}^{12} \) with the negative \( x \) -axis. | 1. We are told that \( \parallel \overrightarrow{v}\parallel = 5 \) and are given information about its direction, so we can use the formula \( \overrightarrow{v} = \parallel \overrightarrow{v}\parallel \widehat{v} \) to get the component form of \( \overrightarrow{v} \) . To determine \( \widehat{v} \), we appeal to D... | Yes |
Theorem 11.21. Principal Vector Decomposition Theorem: Let \( \overrightarrow{v} \) be a vector with component form \( \overrightarrow{v} = \left\langle {{v}_{1},{v}_{2}}\right\rangle \) . Then \( \overrightarrow{v} = {v}_{1}\widehat{\imath } + {v}_{2}\widehat{\jmath } \) . | The proof of Theorem 11.21 is straightforward. Since \( \widehat{\imath } = \langle 1,0\rangle \) and \( \widehat{\jmath } = \langle 0,1\rangle \), we have from the definition of scalar multiplication and vector addition that\n\n\[ \n{v}_{1}\widehat{\imath } + {v}_{2}\widehat{\jmath } = {v}_{1}\langle 1,0\rangle + {v}_... | Yes |
Commutative Property: For all vectors \( \overrightarrow{v} \) and \( \overrightarrow{w},\overrightarrow{v} \cdot \overrightarrow{w} = \overrightarrow{w} \cdot \overrightarrow{v} \) . | To show the commutative property for instance, let \( \overrightarrow{v} = \left\langle {{v}_{1},{v}_{2}}\right\rangle \) and \( \overrightarrow{w} = \left\langle {{w}_{1},{w}_{2}}\right\rangle \) . Then\n\n\[ \overrightarrow{v} \cdot \overrightarrow{w} = \left\langle {{v}_{1},{v}_{2}}\right\rangle \cdot \left\langle {... | Yes |
Prove the identity: \( \parallel \overrightarrow{v} - \overrightarrow{w}{\parallel }^{2} = \parallel \overrightarrow{v}{\parallel }^{2} - 2\left( {\overrightarrow{v} \cdot \overrightarrow{w}}\right) + \parallel \overrightarrow{w}{\parallel }^{2} \) . | Solution. We begin by rewriting \( \parallel \overrightarrow{v} - \overrightarrow{w}{\parallel }^{2} \) in terms of the dot product using Theorem 11.22.\n\n\[ \parallel \overrightarrow{v} - \overrightarrow{w}{\parallel }^{2} = \left( {\overrightarrow{v} - \overrightarrow{w}}\right) \cdot \left( {\overrightarrow{v} - \o... | Yes |
Theorem 11.24. Let \( \overrightarrow{v} \) and \( \overrightarrow{w} \) be nonzero vectors and let \( \theta \) the angle between \( \overrightarrow{v} \) and \( \overrightarrow{w} \) . Then\n\n\[ \theta = \arccos \left( \frac{\overrightarrow{v} \cdot \overrightarrow{w}}{\parallel \overrightarrow{v}\parallel \parallel... | We obtain the formula in Theorem 11.24 by solving the equation given in Theorem 11.23 for \( \theta \) . Since \( \overrightarrow{v} \) and \( \overrightarrow{w} \) are nonzero, so are \( \parallel \overrightarrow{v}\parallel \) and \( \parallel \overrightarrow{w}\parallel \) . Hence, we may divide both sides of \( \ov... | Yes |
Find the angle between the following pairs of vectors.\n\n1. \( \overrightarrow{v} = \langle 3, - 3\sqrt{3}\rangle \), and \( \overrightarrow{w} = \langle - \sqrt{3},1\rangle \) | Solution. We use the formula \( \theta = \arccos \left( \frac{\overrightarrow{v} \cdot \overrightarrow{w}}{\parallel \overrightarrow{v}\parallel \parallel \overrightarrow{w}\parallel }\right) \) from Theorem 11.24 in each case below.\n\n1. We have \( \overrightarrow{v} \cdot \overrightarrow{w} = \langle 3, - 3\sqrt{3}\... | Yes |
Theorem 11.25. The Dot Product Detects Orthogonality: Let \( \overrightarrow{v} \) and \( \overrightarrow{w} \) be nonzero vectors. Then \( \overrightarrow{v} \bot \overrightarrow{w} \) if and only if \( \overrightarrow{v} \cdot \overrightarrow{w} = 0 \) . | To prove Theorem 11.25, we first assume \( \overrightarrow{v} \) and \( \overrightarrow{w} \) are nonzero vectors with \( \overrightarrow{v} \bot \overrightarrow{w} \) . By definition, the angle between \( \overrightarrow{v} \) and \( \overrightarrow{w} \) is \( \frac{\pi }{2} \) . By Theorem 11.23, \( \overrightarrow{... | Yes |
Let \( {L}_{1} \) be the line \( y = {m}_{1}x + {b}_{1} \) and let \( {L}_{2} \) be the line \( y = {m}_{2}x + {b}_{2} \) . Prove that \( {L}_{1} \) is perpendicular to \( {L}_{2} \) if and only if \( {m}_{1} \cdot {m}_{2} = - 1 \) . | Solution. Our strategy is to find two vectors: \( \overrightarrow{{v}_{1}} \), which has the same direction as \( {L}_{1} \), and \( \overrightarrow{{v}_{2}} \) , which has the same direction as \( {L}_{2} \) and show \( \overrightarrow{{v}_{1}} \bot \overrightarrow{{v}_{2}} \) if and only if \( {m}_{1}{m}_{2} = - 1 \)... | Yes |
Let \( \overrightarrow{v} = \langle 1,8\rangle \) and \( \overrightarrow{w} = \langle - 1,2\rangle \) . Find \( \overrightarrow{p} = {\operatorname{proj}}_{\overrightarrow{w}}\left( \overrightarrow{v}\right) \), and plot \( \overrightarrow{v},\overrightarrow{w} \) and \( \overrightarrow{p} \) in standard position. | Solution. We find \( \overrightarrow{v} \cdot \overrightarrow{w} = \langle 1,8\rangle \cdot \langle - 1,2\rangle = \left( {-1}\right) + {16} = {15} \) and \( \overrightarrow{w} \cdot \overrightarrow{w} = \langle - 1,2\rangle \cdot \langle - 1,2\rangle = 1 + 4 = 5 \) . Hence, \( \overrightarrow{p} = \frac{\overrightarro... | Yes |
Taylor exerts a force of 10 pounds to pull her wagon a distance of 50 feet over level ground. If the handle of the wagon makes a \( {30}^{ \circ } \) angle with the horizontal, how much work did Taylor do pulling the wagon? Assume Taylor exerts the force of 10 pounds at a \( {30}^{ \circ } \) angle for the duration of ... | Solution. There are two ways to attack this problem. One way is to find the vectors \( \overrightarrow{F} \) and \( \overrightarrow{PQ} \) mentioned in Theorem 11.28 and compute \( W = \overrightarrow{F} \cdot \overrightarrow{PQ} \) . To do this, we assume the origin is at the point where the handle of the wagon meets ... | Yes |
Sketch the curve described by \( \\left\\{ \\begin{array}{l} x = {t}^{2} - 3 \\\\ y = {2t} - 1 \\end{array}\\right. \) for \( t \\geq - 2 \) . | Solution. We follow the same procedure here as we have time and time again when asked to graph anything new - choose friendly values of \( t \), plot the corresponding points and connect the results in a pleasing fashion. Since we are told \( t \\geq - 2 \), we start there and as we plot successive points, we draw an a... | Yes |
Sketch the curves described by the following parametric equations.\n\n1. \( \\left\\{ \\begin{array}{l} x = {t}^{3} \\\\ y = 2{t}^{2} \\end{array}\\right. \) for \( - 1 \\leq t \\leq 1 \) | To get a feel for the curve described by the system \( \\left\\{ {x = {t}^{3}, y = 2{t}^{2}}\\right. \) we first sketch the graphs of \( x = {t}^{3} \) and \( y = 2{t}^{2} \) over the interval \( \\left\\lbrack {-1,1}\\right\\rbrack \) . We note that as \( t \) takes on values in the interval \( \\left\\lbrack {-1,1}\\... | Yes |
Find a parametrization for each of the following curves and check your answers.\n\n1. \( y = {x}^{2} \) from \( x = - 3 \) to \( x = 2 \) | Since \( y = {x}^{2} \) is written in the form \( y = f\left( x\right) \), we let \( x = t \) and \( y = f\left( t\right) = {t}^{2} \) . Since \( x = t \), the bounds on \( t \) match precisely the bounds on \( x \) so we get \( \left\{ {x = t, y = {t}^{2}}\right. \) for \( - 3 \leq t \leq 2 \) . The check is almost tr... | Yes |
Find a parametrization for the following curves.\n\n1. The curve which starts at \( \left( {2,4}\right) \) and follows the parabola \( y = {x}^{2} \) to end at \( \left( {-1,1}\right) \) . Shift the parameter so that the path starts at \( t = 0 \) . | Solution.\n\n1. We can parametrize \( y = {x}^{2} \) from \( x = - 1 \) to \( x = 2 \) using the formula given on Page 1053 as \( \left\{ {x = t, y = {t}^{2}}\right. \) for \( - 1 \leq t \leq 2 \) . This parametrization, however, starts at \( \left( {-1,1}\right) \) and ends at \( \left( {2,4}\right) \) . Hence, we nee... | Yes |
Find the parametric equations of a cycloid which results from a circle of radius 3 rolling down the positive \( x \) -axis as described above. | Solution. We have \( r = 3 \) which gives the equations \( \{ x = 3\left( {t - \sin \left( t\right) }\right), y = 3\left( {1 - \cos \left( t\right) }\right) \) for \( t \geq 0 \) . (Here we have returned to the convention of using \( t \) as the parameter.) Sketching the cycloid by hand is a wonderful exercise in Calcu... | Yes |
Example 1.1.2 (Computing instantaneous velocity for a falling ball). For a falling ball whose position function is given by \( s\left( t\right) = {16} - {16}{t}^{2} \) (where \( s \) is measured in feet and \( t \) in seconds), find an expression for the average velocity of the ball on a time interval of the form \( \l... | Solution. We make the assumptions that \( - {0.5} < h < {0.5} \) and \( h \neq 0 \) because \( h \) cannot be zero (otherwise there is no interval on which to compute average velocity) and because the function only makes sense on the time interval \( 0 \leq t \leq 1 \), as this is the duration of time during which the ... | Yes |
For the function given by \( f\left( x\right) = x - {x}^{2} \), use the limit definition of the derivative to compute \( {f}^{\prime }\left( 2\right) \). | From the limit definition, we know that\n\n\[ \n{f}^{\prime }\left( 2\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {2 + h}\right) - f\left( 2\right) }{h}.\n\]\n\nNow we use the rule for \( f \), and observe that \( f\left( 2\right) = 2 - {2}^{2} = - 2 \) and \( f\left( {2 + h}\right) = \left( {2 + h}... | Yes |
Example 1.5.2. Suppose that \( y = f\left( x\right) \) is a function for which three values are known: \( f\left( 1\right) = \) \( {2.5}, f\left( 2\right) = {3.25} \), and \( f\left( 3\right) = {3.625} \) . Estimate \( {f}^{\prime }\left( 2\right) \) . | Solution. We know that \( {f}^{\prime }\left( 2\right) = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{f\left( {2 + h}\right) - f\left( 2\right) }{h} \) . But since we don’t have a graph for \( y = f\left( x\right) \) nor a formula for the function, we can neither sketch a tangent line nor evaluate the limit exactly. ... | Yes |
Let \( f\left( x\right) = - {4x} + 7 \) and \( g\left( x\right) = {3x} - 5 \) . Determine a formula for \( C\left( x\right) = f\left( {g\left( x\right) }\right) \) and compute \( {C}^{\prime }\left( x\right) \) . How is \( {C}^{\prime } \) related to \( f \) and \( g \) and their derivatives? | Solution. By the rules given for \( f \) and \( g \) ,\n\n\[ C\left( x\right) = f\left( {g\left( x\right) }\right) \]\n\n\[ = f\left( {{3x} - 5}\right) \]\n\n\[ = - 4\left( {{3x} - 5}\right) + 7 \]\n\n\[ = - {12x} + {20} + 7 \]\n\n\[ = - {12x} + {27}\text{.} \]\n\nThus, \( {C}^{\prime }\left( x\right) = - {12} \) . Not... | Yes |
Let \( C\left( x\right) = \sin \left( {2x}\right) \) . Use the double angle identity to rewrite \( C \) as a product of basic functions, and use the product rule to find \( {C}^{\prime } \) . Rewrite \( {C}^{\prime } \) in the simplest form possible. | Solution. By the double angle identity for the sine function,\n\n\[ C\left( x\right) = \sin \left( {2x}\right) = 2\sin \left( x\right) \cos \left( x\right) .\n\]\n\nApplying the product rule and simplifying,\n\n\[ {C}^{\prime }\left( x\right) = 2\sin \left( x\right) \left( {-\sin \left( x\right) }\right) + \cos \left( ... | Yes |
Find a formula for the derivative of \( h\left( t\right) = {3}^{{t}^{2} + {2t}}{\sec }^{4}\left( t\right) \) . | We first observe that the most basic structure of \( h \) is that it is the product of two functions: \( h\left( t\right) = a\left( t\right) \cdot b\left( t\right) \) where \( a\left( t\right) = {3}^{{t}^{2} + {2t}} \) and \( b\left( t\right) = {\sec }^{4}\left( t\right) \) . Therefore, we see that we will need to use ... | Yes |
For the curve given implicitly by \( {x}^{3} + {y}^{2} - {2xy} = 2 \), shown in Figure 2.7.4, find the slope of the tangent line at \( \left( {-1,1}\right) \) . | Solution. We begin by differentiating the curve's equation implicitly. Taking the derivative of each side with respect to \( x \) ,\n\n\[ \frac{d}{dx}\left\lbrack {{x}^{3} + {y}^{2} - {2xy}}\right\rbrack = \frac{d}{dx}\left\lbrack 2\right\rbrack \]\n\nby the sum rule and the fact that the derivative of a constant is ze... | Yes |
Let \( f \) be a function whose derivative is given by the formula \( {f}^{\prime }\left( x\right) = {e}^{-{2x}}(3 - x){\left( x + 1\right) }^{2} \) . Determine all critical numbers of \( f \) and decide whether a relative maximum, relative minimum, or neither occurs at each. | Solution. Since we already have \( {f}^{\prime }\left( x\right) \) written in factored form, it is straightforward to find the critical numbers of \( f \) . Since \( {f}^{\prime }\left( x\right) \) is defined for all values of \( x \), we need only determine where \( {f}^{\prime }\left( x\right) = 0 \) . From the equat... | Yes |
Suppose that pollutants are leaking out of an underground storage tank at a rate of \( r\left( t\right) \) gallons/day, where \( t \) is measured in days. It is conjectured that \( r\left( t\right) \) is given by the formula \( r\left( t\right) = {0.0069}{t}^{3} - {0.125}{t}^{2} + {11.079} \) over a certain 12-day peri... | Solution. We know that since \( r\left( t\right) \geq 0 \), the value of \( {\int }_{4}^{10}r\left( t\right) {dt} \) is the area under the curve on the interval \( \left\lbrack {4,{10}}\right\rbrack \) . If we think about this area from the perspective of a Riemann sum, the rectangles will have heights measured in gall... | Yes |
Evaluate the indefinite integral\n\n\[ \int {x}^{3} \cdot \sin \left( {7{x}^{4} + 3}\right) {dx} \] | Solution. We can make two key algebraic observations regarding the integrand, \( {x}^{3} \cdot \sin \left( {7{x}^{4} + }\right. \) 3). First, \( \sin \left( {7{x}^{4} + 3}\right) \) is a composite function; as such, we know we’ll need a more sophisticated approach to antidifferentiating. Second, \( {x}^{3} \) is almost... | Yes |
Evaluate the indefinite integral\n\n\[ \int x\cos \left( x\right) {dx} \] \n\nusing Integration by Parts. | Solution. Whenever we are trying to integrate a product of basic functions through Integration by Parts, we are presented with a choice for \( u \) and \( {dv} \). In the current problem, we can either let \( u = x \) and \( {dv} = \cos \left( x\right) {dx} \), or let \( u = \cos \left( x\right) \) and \( {dv} = {xdx} ... | Yes |
Find the volume of the solid of revolution generated when the region \( R \) bounded by \( y = 4 - {x}^{2} \) and the \( x \) -axis is revolved about the \( x \) -axis. | Solution. First, we observe that \( y = 4 - {x}^{2} \) intersects the \( x \) -axis at the points \( \left( {-2,0}\right) \) and \( \left( {2,0}\right) \). When we take the region \( R \) that lies between the curve and the \( x \) -axis on this interval and revolve it about the \( x \) -axis, we get the three-dimensio... | Yes |
Find the volume of the solid of revolution generated when the finite region \( R \) that lies between \( y = 4 - {x}^{2} \) and \( y = x + 2 \) is revolved about the \( x \) -axis. | Solution. First, we must determine where the curves \( y = 4 - {x}^{2} \) and \( y = x + 2 \) intersect. Substituting the expression for \( y \) from the second equation into the first equation, we find that \( x + 2 = 4 - {x}^{2} \) . Rearranging, it follows that\n\n\[ \n{x}^{2} + x - 2 = 0 \n\] \n\nand the solutions ... | No |
Find the volume of the solid of revolution generated when the finite region \( R \) that lies between \( y = \sqrt{x} \) and \( y = {x}^{4} \) is revolved about the \( y \) -axis. | Solution. We observe that these two curves intersect when \( x = 1 \), hence at the point \( \left( {1,1}\right) \) . When we take the region \( R \) that lies between the curves and revolve it about the \( y \) -axis, we get the three-dimensional solid pictured at left in Figure 6.2.8.\n\n\n\nFigure 6.2.10: The solid of revolution described in Example 6.2.9.\n\ncurves in the first quadrant between their points of intersection \( \\left( {\\left( {... | Yes |
Consider a trapezoid-shaped dam that is 60 feet wide at its base and 90 feet wide at its top, and assume the dam is 25 feet tall with water that rises to within 5 feet of the top of its face. Water weighs 62.4 pounds per cubic foot. How much force does the water exert against the dam? | Solution. First, we sketch a picture of the dam, as shown in Figure 6.4.6. Note that, as in problems involving the work to pump out a tank, we let the positive \( x \) -axis point down.\n\n\n\nFigure 6.4.6: A trapezo... | Yes |
Find all functions \( y \) that are solutions to the differential equation\n\n\[ \frac{dy}{dt} = \frac{t}{{y}^{2}} \] | Solution. We begin by separating the variables and writing\n\n\[ {y}^{2}\frac{dy}{dt} = t \]\n\nIntegrating both sides of the equation with respect to the independent variable \( t \) shows that\n\n\[ \int {y}^{2}\frac{dy}{dt}{dt} = \int {tdt} \]\n\nNext, we notice that the left-hand side allows us to change the variab... | Yes |
Example 7.4.2. Solve the differential equation\n\n\\[ \n\\frac{dy}{dt} = {3y} \n\\] | Solution. Following the same strategy as in Example 7.4.1, we have\n\n\\[ \n\\frac{1}{y}\\frac{dy}{dt} = 3 \n\\]\n\nIntegrating both sides with respect to \\( t \\) ,\n\n\\[ \n\\int \\frac{1}{y}\\frac{dy}{dt}{dt} = \\int {3dt} \n\\]\n\nand thus\n\\[ \n\\int \\frac{1}{y}{dy} = \\int {3dt} \n\\]\n\nAntidifferentiating an... | Yes |
Let’s determine how well the 100th partial sum \( {S}_{100} \) of\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{\left( -1\right) }^{k + 1}}{k} \]\n\napproximates the sum of the series. | Solution. If we let \( S \) be the sum of the series \( \mathop{\sum }\limits_{{k = 1}}^{\infty }\frac{{\left( -1\right) }^{k + 1}}{k} \), then we know that\n\n\[ \left| {{S}_{100} - S}\right| < {a}_{101} \]\n\nNow\n\n\[ {a}_{101} = \frac{1}{101} \approx {0.0099} \]\n\nso the 100th partial sum is within 0.0099 of the s... | Yes |
Determine the third order Taylor polynomial for \( f\left( x\right) = {e}^{x} \), as well as the general \( n \) th order Taylor polynomial for \( f \) centered at \( x = 0 \) . | Solution. We know that \( {f}^{\prime }\left( x\right) = {e}^{x} \) and so \( {f}^{\prime \prime }\left( x\right) = {e}^{x} \) and \( {f}^{\prime \prime \prime }\left( x\right) = {e}^{x} \) . Thus,\n\n\[ f\left( 0\right) = {f}^{\prime }\left( 0\right) = {f}^{\prime \prime }\left( 0\right) = {f}^{\prime \prime \prime }\... | Yes |
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