Calculus III MATH 313 Lectures PDF

Lectures of Calculus III MATH 313 Course Description : Part (1): Chapter 10 : Parametric Equations and Polar curves Chapter 11 : Three - Dimensional Space ; Vectors Part (2): Chapter 16 : Partial Differentiation Chapter 17 : Multiple Integrals Chapter 18 : Vecto...

Lectures of Calculus III MATH 313 Course Description : Part (1): Chapter 10 : Parametric Equations and Polar curves Chapter 11 : Three - Dimensional Space ; Vectors Part (2): Chapter 16 : Partial Differentiation Chapter 17 : Multiple Integrals Chapter 18 : Vector Calculus Text Book of Part (1) : Howard Anton , Irl Bivens , Stephen Davis , '' Calculus '' Early Transcendentals , John Wiley & Sons , Inc., 10 th Edition , 2012. Text Book of Part (2) : Earl W. Sowokowski , '' Calculus '' , Thomson Advantage Books, 5th Edition , 1991. Grade Distribution : Home Works 5% Participation 5% Quizzes 10 % First Periodic Exam 20 % Second Periodic Exam 20 % Final Exam 40 % Prepared by : Dr. Fayad Galal 1436 /1437 H 1 2  CHAPTER (10) PARAMETRIC AND POLAR CURVES ; CONIC SECTIONS 10.1 PARAMETRIC EQUATIONS ; TANGENT LINES AND ARC LENGTH FOR PARAMERTIC CURVES : Page (692) Parametric Equations : * Suppose that a particle moves along a curve C in the xy-plane in such a way that its x- and y-coordinates; as functions of time, are x  f t  , y  g t  We call these the parametric equations of motion for the particle and refer to C as the trajectory of the particle or the graph of the equations (Figure 10.1.1). The variable t is called the parameter for the equations. Figure 10.1.1 Example 1 : Page (692) Sketch the trajectory over the time interval of the particle whose parametric equations of motion are 3  x  t  3 sin t , y  4  3 cos t 1  Solution * One way to sketch the trajectory is to choose a representative succession of times, plot the (x,y) coordinates of points on the trajectory at those times, and connect the points with a smooth curve. The trajectory in Figure 10.1.2 was obtained in this way from the data in Table 10.1.1 in which the approximate coordinates of the particle are given at time increments of 1 unit. Observe that there is no t-axis in the particle; the values of t appear only as labels on the plotted points, and even these are usually omitted unless it is important to emphasize the locations of the particle at specific times. Table 10.1.1 Figure 10.1.2 * Although parametric equations commonly arise in problems of motion with time as the parameter, they arise in other contexts as well. Thus, unless the problem dictates that the parameter t in the equations x  f t  , y  g t  represents time, it should be viewed simply as an independent variable that varies over some interval of real numbers. (In 4  fact, there is no need to use the letter t for the parameter; any letter not reserved for purpose can be used). If no restrictions on the parameter are stated explicitly or implied by the equations, then it is understood that it varies from   to  . To indicate that a parameter t is restricted to an interval  a ,b , we will write x  f  t  , y  g  t   a  t  b Example 2 : Page (693) Find the graph of the parametric equations x  cos t , y  sin t  0  t  2   2 Solution * One way to find the graph is to eliminate the parameter t by noting that x 2  y 2  sin2 t  cos 2 t  1 Remember that : * sin2 t  cos 2 t  1 Thus, the graph is contained in the unit circle x 2  y 2  1. Geometrically, the parameter t can be interpreted as the angle swept out by the radial line from the origin to the point  x, y    cos t , sin t  on the unit circle (Figure 10.1.3). As t increases from 0 to 2 , the point traces the circle counterclockwise, starting at  1,0  when t  0 and completing one full revolution when t  2. One can obtain different portions of the circle by varying the interval over which the parameter varies. For example, x  cos t , y  sin t 0  t     3 represents just the upper semicircle in Figure 10.1.3. 5  Figure 10.1.3 Orientation : Page (694) * The direction in which the graph of a pair of parametric equations is traced as the parameter increases is called the direction of increasing parameter or sometimes the orientation imposed on the curve by the equations. Thus, we make a distinction between a curve, which is a set of points, and a parametric curve, which is a curve with an orientation imposed on it by a set of parametric equations. For example, we saw in Example 2 that the circle represented parametrically by i traced counterclockwise as t increases and hence has counterclockwise orientation. As shown in Figure 10.1.2 and 10.1.3, the orientation of a parametric curve can be indicated by arrowheads. * To obtain parametric equations for the unit circle with clockwise orientation, we can replace t by  t in  2  and use the identities cos   t   cos t and sin   t    sin t. This yields x  cos t , y   sin t  0  t  2  6  Here, the circle is traced clockwise by a point that stats at when and completes one full revolution when (Figure 10.1.4). Figure 10.1.4 Example 3 : Page (694) Graph the parametric curve x  2t  3 , y  6 t  7 by eliminating the parameter, and indicate the orientation on the graph. Solution * To eliminate the parameter we will solve the first equation for t as a function of x , and then substitute this expression for t into the second equation:  1 t     x  3 2  1 y  6    x  3  3 x  2  2 * Thus, the graph is a line of slope 3 and y-intercept 2. 7  Remember that : y  mx  c * is a line of slope m and y  int ercept c * To find the orientation we must look to the original equations; the direction of increasing t can be deduced by observing that x increases as t increases or by observing that y increases as t increases. Either piece of information tells us that the line is traced left to right as shown in Figure 10.1.5. Figure 10.1.5 Expressing Ordinary Functions Parametrically : Page (694) * An equation y  f  x  can be expressed in parametric form by introducing the parameter x  t ; this yields the parametric equations x  t , y  f t For example, the portion of the curve y  cos x over the interval   2 , 2  can be expressed parametrically as x  t , y  cos t   2  t  2  8  (Figure 10.1.7). Figure 10.1.7 * If a function f is one-to-one, then it has an inverse function f  1. In this case the equation y  f  1  x  is equivalent to x  f  y . We can express the graph of f  1 in parametric form by introducing the parameter y  t ; this yields the parametric equations x  f t  , y  t For example, Figure 10.1.8 shows the graph of f  x   x 5  x  1 and its inverse. Figure 10.1.8 The graph of can be represented parametrically as 9  x  t , y  t5  t  1 and the graph of f  1 can be represented parametrically as x  t5  t  1 , y  t Tangent ines to Parametric Curves : Page (695) * We will be concerned with curves that are given by parametric equations x  f t  , y  g t  in which f  t  and g  t  have continuous first derivatives with respect to t. It can be proved that if dy / dt  0 , then y is a differentiable function of x, in which case the chain rule implies that dy dy / dt  4 dx dx / dt This formula makes it possible to find dy / dx directly from the parametric equations without eliminating the parameter. Example 4 : Page (695) Find the sloe of the tangent line to the unit circle x  cos t , y  sin t  0  t  2  at the point where t   / 6 (Figure 10.1.9). 10  Figure 10.1.9 Solution * From  4  , the slope at a general point o the circle is dy dy / dt cos t     cot t 5 dx dx / dt  sin t Thus, the slope at t   / 6 is dy    cot   3 dx t  / 6 6 Remember that :  180  *   30 , cot  3 6 6 6 Example 6 : Page (697) * The curve represented by the parametric equations x  t2 , y  t3     t    is called a semicubical parabola. The parameter t can be eliminated by cubing x and squaring y, from which it follows that y 2  x 3. * The graph of this equation, shown in Figure 10.1.13, consists of two branches: an upper branch obtained by graphing and a lower branch obtained by graphing y   x 3 / 2. 11  Figure 10.1.13 * The two branches meet at the origin, which corresponds to t  0 in the parametric equations. This is a singular point because the derivatives dx / dt  2t and dy / dt  3t 2 are both zero here. Example 7 : Page (697) Without eliminating the parameter, find dy / dx and d 2 y / dx 2 at  1,1  and  1,  1 on the semicubical parabola given by the parametric equations in Example 6. Solution * From  4  we have dy dy / dt 3t 2 3    t t  0 7  dx dx / dt 2 t 2 and from  4  applied to y'  dy / dx we have 12  d 2 y dy' dy'/ dt 3 / 2 3 2     8 dx dx dx / dt 2t 4t * Since the point  1,1 on the curve corresponds to t  1 in the parametric equation, it follows from 7  and  8  that dy 3 d2 y 3  and  dx t 1 2 dx 2 t 1 4 * Similarly, the point  1,  1 corresponds to t   1 in the parametric equations, so applying 7  and  8  again yields dy 3 d2 y 3   and   dx t1 2 dx 2 t1 4 * Note that that values we obtained for the first and second derivatives are consistent with the graph in Figure 10.1.13, since at  1,1  on the upper branch the tangent line has positive slope and the curve is concave up, and at  1,  1 on the lower branch the tangent line has negative slope and the curve is concave down. * Finally, observe that we were able to apply Formulas 7  and  8  for both t  1 and t   1 , even though the points  1,1 and  1,  1 lie on different branches. In contrast, had we chosen to perform the same computations by eliminating that parameter, we would have had to obtain separate derivative formulas for y  x 3 / 2 and y   x 3 / 2. 13 Arc Lengh of Parametric Curves : Page (697) * The following result provides a formula for finding the arc length of a curve from parametric equations for the curve. Its derivation is similar to that of Formula  3  in Section 6.4 and will be omitted. 10.1.1 Arc Length for Parametric Curves : Page (698) If no segment of the curve represented by the parametric equations x  x t  , y  yt   a  t  b Is traced more than once as t increases from a to b, and if dx / dt and dy / dt are continuous functions for a  t  b , then the arc length L of the curve is given by b 2 2  dx   dy  L      dt 9 a  dt   dt  Example 8 : Page (698) Use to find the circumference of a circle of radius a from the parametric equations x  a cos t , y  a sin t  0  t  2  Solution x  a cos t , y  a sin t dx dy   a sin t ,  a cos t dt dt Remember that : d d *  cos x    sin x ,  sin x   cos x dx dx 14 * The circumference of a circle of radius a is b 2 2  dx   dy  L      dt a  dt   dt  2     a sin t  2   a cos t  2 dt 0 2 a  sin 2 t  cos 2 t dt 0 Remember that : * sin2 t  cos 2 t  1 2 2 a  dt  a  t  0  a  2  0   2 a 0 Remember that : x r 1 *  x dx  r C r  1  r 1 b *  f  x  dx  F  x   a  F b   F  a  b a 15 EXERCISE SET 10.1: (Home Work) Page (701) 3,4,5 Sketch the curve by eliminating the parameter, and indicate the direction of increasing t. 3. x  3t  4 , y  6 t  2 4. x  t  3 , y  3t  7 0  t  3 5. x  2 cos t , y  5 sin t  0  t  2  13,17 Find parametric equations for the curve, and check your work by generating the curve with a graphing utility. 13. A circle of radius 5, centered at the origin, oriented clockwise. 17. The portion of the parabola x  y 2 joining  1,  1 and  1,1 , oriented down to up. 45,49 Find dy / dx and d 2 y / dx 2 at the given point without eliminating the parameter. 45. x  t , y  2t  4 ; t  1 49. x    cos  , y  1  sin ; t  / 6 53. Find all values of t at which the parametric curve x  2 sin t , y  4 cos t  0  t  2  has (a) a horizontal tangent line and (b) a vertical tangent line. 65,67 Find the exact arc length of the curve over the stated interval. 1 3 65. x  t 2 , y t  0  t  1 3 67. x  cos 3t , y  sin 3t 0  t    16 10.2 POLAR COORDINATES : Page (702) Polar Coordinate Systems : * A polar coordinate system in a plane consists of a fixed point O , called the pole (or origin), and a ray emanating from the pole, called the polar axis. In such a coordinate system we can associate with each point P in the plane a pair of polar coordinates  r ,  , where r is the distance from P to the pole and  is an angle from the polar axis to the ray OP (Figure 10.2.1). Figure 10.2.1 * The number r is called the radial coordinate of P and the number  the angular coordinate (or polar angle) of P. In Figure 10.2.2, the points  6 , / 4  ,  5,2 / 3  ,  3,5 / 4  , and  4,11 / 6  are plotted in polar coordinate systems. If P is the pole, then r  0 , but there is no clearly defined polar angle. We will agree that an arbitrary angle can be used in this case; that is,  0 ,  are polar coordinates of the pole for all choices of . Figure 10.2.2 17 * The polar coordinates of a point are not unique. For example, the polar coordinates  1,7  / 4  ,  1,   / 4  , and  1,15 / 4  all represent the same point (Figure 10.2.3). Figure 10.2.3 * In general, if a point P has polar coordinates , then  r ,  2n  and  r ,  2n  are also polar coordinates of P for any nonnegative integer n. Thus, every point has infinitely many pairs of polar coordinates. * As defined above, the radial coordinate r of a point P is nonnegative, since it represents the distance from P to the pole. However, it will be convenient to allow for negative values of r as well. To motivate an appropriate definition, consider the point P with polar coordinates  3,5 / 4 . As shown in Figure 10.2.4, we can reach this point by rotating the polar axis through an angle of 5 / 4 and then moving 3 units from the pole along the terminal side of the angle, or we can reach the point P by rotating the polar axis through an angle of / 4 and then moving 3 units from the pole along the extension of the terminal side. 18  Figure 10.2.4 This suggests that the point  3,5 / 4  might also be denoted by   3, / 4  , with the minus sign serving to indicate that the point is on the extension of the angle's terminal side rather than on the terminal side itself. * In general, the terminal side of the angle    is the extension of the terminal side of  , so we define negative radial coordinates by agreeing that   r ,  and  r ,    are polar coordinates of the same point. Relatinship Between Polar and Rectangular Coordinates : Page (706) * Frequently, it will be useful to superimpose a rectangular xy- coordinate system on top of a polar coordinate system, making the positive x-axis coincide with the polar axis. If this is done, then every point P will have both rectangular coordinates  x , y  and polar coordinates  r , . As suggested by Figure 10.2.5, these coordinates are related by the equations x  r cos  , y  r sin  1 19  Figure 10.2.5 * These equations are well suited for finding x and y when r and  are known. However, to find r and  when x and y are known, it is preferable to use the identities sin2   cos 2   1 and tan  sin / cos  to rewrite  1  as y r 2  x2  y2 , tan   2 x Example 1 : Page (707) Find the rectangular coordinates of the point P whose polar coordinates are  r ,    6 , 2 / 3  (Figure 10.2.6). Figure 10.2.6 Solution * Substituting the polar coordinates r  6 and   2 / 3 in  1  yield 20   1 x  r cos   6      3  2 2  3 y  r sin  r sin 6   3 3 3  2  Remember that : * x  r cos  , y  r sin * Thus, the rectangular coordinates of P are   x , y    3, 3 3.  Example 2 : Page (707) Find the polar coordinates of the point P whose rectangular   coordinates are  2,  2 3 (Figure 10.2.7). Figure 10.2.7 Solution * We will find the polar coordinates  r ,  of P that satisfy the conditions r  0 and 0    2. From the first equation in  2 ,   2 r  x  y    2   2 3 2 2 2 2  4  12  16 21  Remember that : * r 2  x2  y2 so r  4. From the second equation in  2 . y 2 3 tan     3 x 2 Remember that : y * tan  x * From this and the fact that    2,  2 3 lies in the third quadrant, it follows that the angle satisfying the requirement 0    2 is   4 / 3. * Thus,  r ,    4 , 4 / 3  are polar coordinates of P. All other coordinates of P are expressible in the form  4      4 ,  2n  or   4 ,  2n   3   3  where n is an integer. Graphs in Polar Coordinates : Page (707) * We will now consider the problem of graphing equations in r and  , where  is assumed to be measured in radians. some examples of such equations are r  1 ,    / 4 , r   , r  sin , r  cos 2 * In a rectangular coordinate system the graph of an equation in x and y consists of all points whose coordinates  x , y  satisfy the equation. However, in a polar coordinates system, points have infinitely many different pairs of polar coordinates, so that a given point may have some polar coordinate that satisfy 22  an equation and others that do not. Given an equation in r and  , we define its graph in polar coordinates to consist of all points with at least one pair of coordinates  r ,  that satisfy the equation. Example 3 : Page (708) Sketch the graphs of  (a) r  1 (b)   4 Solution (a) For all values of  , the point  1,  is 1 unit away from the pole. Since  is arbitrary, the graph is the circle of radius 1 centered at the pole (Figure 10.2.8a). Figure 10.2.8a (b) For all values of r, the point  r , / 4  lies on a line that makes an angle of  / 4 with the polar axis (Figure 10.2.8b). Positive values of r correspond to points on the line in the first quadrant and negative values of r to points on the line in the third quadrant. Thus, in absence of any restriction on r, the graph is the entire line. Observe, however, that had we imposed the restriction r  0 , the graph would have been just the ray in the first quadrant. 23  Figure 10.2.8b * Equations r  f   that express r as a function of  are especially important. One way to graph such an equation is to choose some typical values of  , calculate the corresponding values of r, and then plot the resulting pairs  r ,  in a polar coordinate system. The next two examples illustrate this process. Example 4 : Page (708) Sketch the graph of r     0  in polar coordinates by plotting points. Solution * Observe that as  increases, so does r; thus, the graph is a curve that spirals out from the pole as  increases. A reasonably accurate sketch of the spiral can be obtained by plotting the points that correspond to values of  that are integer multiples of  / 2 , keeping in mind that the value of r is always equal to the value of  (Figure 10.2.9). 24  Figure 10.2.9 Example 5 : Page (708) Sketch the graph of the equation r  sin in polar coordinates by plotting points. Solution * Table 10.2.1 shows the coordinates of points on the graph at increments of  / 6. Table 10.2.1 * These points are plotted in Figure 10.2.10. Figure 10.2.10 25 * Note, however, that there are 13 points listed in the table but only 6 distinct plotted points. This is because the pairs from r   on yield duplicates of the preceding points. For example,   1 / 2 ,7  / 6  and  1 / 2, / 6  represent the same point. * Observe that the points in Figure 10.2.10 appear to lie on a circle. We confirm that this is so by expressing the polar equation r  sin in terms of x and y. To do this, we multiply the equation through by r to obtain r 2  r sin which now allows us to apply Formula  1  and  2  to rewrite the equation as x2  y2  y Rewriting this equation as x 2  y 2  y  0 and then completing the square yields 2  1  1 x2   y     2 4 1  1 which is a circle of radius centered at the point  0 ,  in the 2  2 xy-plane. * It is often useful to view the equation r  f   as an equation in rectangular coordinates (rather than polar coordinates) and graphed in a rectangular  r -coordinate system. For example, Figure 10.2.11 shows the graph of r  sin displayed using rectangular  r -coordinate. 26  Figure 10.2.11 This graph can actually help to visualize how the polar graph in is generated:  At   0 we have r  0 , which corresponds to the pole on the polar graph.  As  varies from 0 to  / 2 , the value of r increases from 0 to 1, so the point  r ,  moves along the circle from the pole to the high point at  1, / 2 .  As  varies from  / 2 to  , the value of r increases from 1 back to 0, so the point  r ,  moves along the circle from the pole to the high point back to the pole.  As  varies from  to 3 / 2 , the values of r negative varying from 0 to  1. Thus, the point  r ,  moves along the circle from the pole to the high point at  1, / 2  , which is the same as the point   1,3 / 2 . This duplicates the motion that occurred for 0     / 2.  As  varies from 3 / 2 to 2 , the value of r varies from  1 to 0, so the point  r ,  moves along the circle from the high point back to the pole, duplicating the motion that occurred for  / 2    . 27 Example 6 : Page (709) Sketch the graph of r  cos 2 in polar coordinates. Solution * Instead of plotting points, we will use the graph of r  cos 2 in rectangular coordinates (Figure 10.2.12) to visualize how the pole graph of this equation is generated. Figure 10.2.12 * The analysis and the resulting polar graph are shown in Figure 10.2.13. This curves is called a four-petal rose. 28  Figure 10.2.13 Symmetry Tests : Page (710) * Observe that the polar graph of r  cos 2 in Figure 10.2.13 is symmetric about the x-axis and the y-axis. This symmetry could have been predicted from the following theorem, which is suggested by Figure 10.2.14 (we omit the proof). Figure 10.2.14 10.2.1 Theorem (Symmetry Tests) : Page (710) (a) A curve in polar coordinates is symmetric about the x-axis if replacing  by   in its equation produces an equivalent equation (Figure 10.2.14a). (b) A curve in polar coordinates is symmetric about the y-axis if replacing  by    in its equation produces an equivalent equation (Figure 10.2.14b). (c) A curve in polar coordinates is symmetric about the origin if replacing  by    , or replacing r by  r in its equation produces an equivalent equation (Figure 10.2.14c). 29 Example 7 : Page (710) Use Theorem 10.2.1 to confirm that the graph of r  cos 2 in Figure 10.2.13 is symmetric about the x-axis and y-axis. Solution * To test for symmetry about the x-axis, we replace  by  . This yield r  cos   2   cos 2 Thus, replacing  by   does not alter the equation. * To test for symmetry about the y-axis, we replace  by   . This yield r  cos 2       cos  2  2   cos  2   cos 2 Thus, replacing  by    does not alter the equation. Example 8 : Page (711) Sketch the graph of r  a  1  cos   in polar coordinates, assuming a to be a positive constant. Solution * Observe first that replacing  by   does not alter the equation, so we know in advance that the graph is symmetric about the polar axis. Thus, if we graph the upper half of the curve, then we can obtain the lower half by reflection about the polar axis. * As in our previous examples, we will first graph the equation in rectangular  r -coordinates. This graph, which is shown in Figure 10.2.15a, can be obtained by rewriting the given equation as r  a  a cos  , from which we see that the graph in rectangular  r -coordinates can be obtained by first reflecting the graph of r  a cos  about the x-axis to obtain the graph of r   a cos  , and then translating that graph up 30  a units to obtain the graph of r  a  a cos . Now we can see the following:  As  varies from 0 to  / 3 , r increases from 0 to a / 2.  As  varies from  / 3 to  / 2 , r increases from a / 2 to a.  As  varies from  / 2 to 2 / 3 , r increases from a to 3a / 2.  As  varies from 2 / 3 to  , r increases from 3a / 2 to 2a. This produces the polar curve shown in Figure 10.2.15b. * The rest of the curve can be obtained by continuing the preceding analysis from  to 2 or, as noted above, by reflecting the portion already graphed about the x-axis (Figure 10.2.15c). This heart-shaped curve is called a cardioid (from the Greek word kardia meaning "heart"). Figure 10.2.15 Example 9 : Page (711) Sketch the graph of r 2  4 cos 2 in polar coordinates. Solution * This equation does not express r as a function of  , since for r in terms of  yields two functions: 31  r  2 cos 2 and r   2 cos 2 Thus, to graph the equation r 2  4 cos 2 we will have to graph the two functions separately and then combine those graphs. * We will start with the graph of r  2 cos 2. Observe first that this equation is not changed if we replace  by   or if we replace  by   . Thus, the graph is symmetric about the x-axis and the y-axis. This means that the entire graph can be obtained by graphing the portion in the first quadrant, reflecting that portion about the y-axis to obtain the portion in the second quadrant, and then reflecting those two portions about the x-axis to obtain the portions in the third and fourth quadrants. * To begin the analysis, we will graph the equation r  2 cos 2 in rectangular  r -coordinates (see Figure 10.2.16a). Note that there are gaps in that graph over the intervals  / 4    3 / 4 and 5 / 4    7  / 4 because cos 2 is negative for those values of . From this graph we can see the following:  As  varies from 0 to  / 4 , r increases from 2 to 0.  As  varies from  / 4 to  / 2 , no points are generated on the polar graph. * This produces the portion of the graph shown in Figure 10.2.16b. As noted above, we can complete the graph by a reflection about the y-axis followed by a reflection about the x-axis (Figure 10.2.16c). The resulting propeller-shaped graph is called a lemniscate (from the Greek word lemniscos for a looped ribbon resembling the number 8). We leave it for you to verify that the equation r  2 cos 2 has the same 32  graph as r   2 cos 2 , but traced in a diagonally opposite manner. Thus, the graph of the equation r 2  4 cos 2 consists of two identical superimposed lemniscates. Figure 10.2.16 Families of Lines and Rays Through the Pole : Page (712) * If  0 is a fixed angle, then for all values of r the point  r ,0  lies on the line that makes an angle of   0 with the polar axis; and, conversely, every point on this line has a pair of polar coordinates of the form  r ,0 . Thus, the equation   0 represents the line that passes through the pole and makes an angle of  0 with the polar axis (Figure 10.2.17a). If r is restricted to be nonnegative, then the graph of the equation   0 is the ray that emanates from the pole and makes an angle of  0 with the polar axis (Figure 10.2.17b). Thus, as  0 varies, the equation   0 produces either a family either a family of lines through the pole or a family of rays through the pole, depending on the restriction on r. 33  Figure 10.2.17 Families of Circles : Page (712) * We will consider three families of circles in which a is assumed to be a positive constant: r  a , r  2a cos  , r  2a sin 3  5 The equation r  a represents a circle of radius a centered at the pole (Figure 10.2.18a). Thus, as a varies, this equation produces a family of circles centered at the pole. For families  4  and  5  , recall from plane geometry that a triangle that is inscribed in a circle with a diameter of the circle for a side must be a right triangle. Thus, as indicated in Figure 10.2.18b and 10.2.18c, the equation r  2a cos  represents a circle of radius a , centered on the x-axis and tangent to the y-axis at the origin; similarly, the equation r  2a sin represents a circle of radius a , centered on the y-axis and tangent to the x-axis at the origin. Thus, as a varies, Equations  4  and  5  produce the families illustrated in Figure 10.2.18d and 10.2.18e. 34  Figure 10.2.18 Families of Rose Curves : Page (713) * In polar coordinates, equations of the form r  a sin n , r  a cos n 6  7  In which a  0 and n is a positive integer represent families of flower-shaped curves called roses (Figure 10.2.19). The rose consists of n equally spaced petals or radius a if n is odd and 2n equally spaced petals of radius a if n is even. It can be shown that a rose with an even number of petals is traced out exactly once as  varies over the interval 0    2 and a rose with an odd number of petals is traced out once as  35  varies over the interval 0    . A four-petal rose of radius 1 was graphed in Example 6. Figure 10.2.19 Families of Cardioids and Limacons : Page (713) * Equations with any of the four forms r  a  b sin , r  a  b cos  8  9 In which a  0 and b  0 represent polar curves called limacons (from the latin word limax for a snail-like creature that is commonly called a "slug"). There are four possible shapes for a limacon that are determined by the ratio a / b (Figure 10.2.20). If a  b (the case a / b  1 ), then the limacon i called a cardioid because of its heart-shaped apprearance, as noted in Example 8. 36  Figure 10.2.20 Example 10 : Page (714) * Figure 10.2.21 shows the family of limacons r  a  cos  with the constant a varies from 0.25 to 2.25 in steps of 0.25. In keeping with Figure 10.2.20, the limacons evolve the loop type to the convex type. As a increases from the starting value of 0.25, the loops get smaller and smaller until the cardioid is reached at a  1. As a increases further, the limacons evolve through the dimpled type into the convex type. Figure 10.2.21 37 Families of Spirals : Page (714) * A spiral is a curve that coils around a central point.Spirals generally have "left-hand" and "right-hand" versions that coil in opposite direction, depending on the restrictions on the polar angle and the signs of constants that appear in their equations. Some of the more common types of spirals are shown in Figure 10.2.22 for nonnegative values of  , a , and b. Figure 10.2.22 38 EXERCISE SET 10.2: (Home Work) Page (716) 3. Find the rectangular coordinates of the points whose polar coordinates are given. (a)  6 , / 6  (b) 7 ,2 / 3  (c)   6 ,  5 / 6  5. In each part, a point is given in rectangular coordinates. Find two pairs of polar coordinates for the point, one pair satisfying r  0 and 0    2 , and the second pair satisfying r  0 and  2    0. (a)   5 , 0   (b) 2 3 ,  2  (c)  0 ,  2  9. Identify the curve by transforming the given polar euation to rectangular coordinates. (a) r  2 (b) r sin  4 6 (c) r  3 cos  (d) r  3 cos   2 sin  11. Express the given equations in polar coordinates. (a) x  3 (b) x 2  y 2  7 (c) x 2  y 2  6 y  0 (d) 9 x y  4 17,19 Find an equation for the given polar graph. [Note: Numerical labels on these graphs represent distances to the origin.] 17. 39 19. 21,23,24,25,27, Sketch the curve in polar coordinates 29,38,39,44,46  21.   23. r  3 3 24. r  4 cos  25. r  6 sin 27. r  3  1  sin  29. r  4  4 cos  38. r 2  cos 2 39. r 2  16 sin 2 44. r  3 sin 2 46. r  2 cos 3 40 10.3 TANGENT LINES, ARC LENGTH, AND AREA FOR POLAR CURVESOLAR COORDINATES : Page (719) Tangent Lines to Polar Curves : * Our first objective in this section is to find a method for obtaining slopes of tangent lines to polar curves of the form r  f   in which r is a differentiable function of . We showed in the last section that a curve of this form can be expressed parametrically in terms of the parameter  by substituting f   for r in the equations x  r cos  and y  r sin. This yields x  f   cos  , y  f   sin from which we obtain dx dr   f   sin   f '   cos    r sin  cos  d d  1 dy dr  f   cos   f '   sin   r cos   sin d d Remember that : d d d *  f  x .g  x    f  x .  g  x    g  x .  f  x   dx  dx dx * Thus, if dx / d and dy / d are continuous and if dx / d  0 , then y is a differentiable function of x, and Formula  4  in section 10.1 with  in place of t yields dr r cos   sin  dy dy / d d    2 dx dx / d dr  r sin   cos  d 41 Example 1 : Page (720) Find the slope of the tangent line to the circle r  4 cos  at the point where    / 4. Solution * From  2  with r  4 cos  , so that dr / d   4 sin , we obtain dy dy / d 4 cos 2   4 sin 2  cos 2   sin2     dx dx / d  8 sin cos  2 sin cos  * Using the double-angle formula for sine and cosine, Remember that : * sin 2 x  2 sin x cos x , cos 2 x  cos 2 x  sin 2 x dy cos 2    cot 2 dx sin 2 Remember that : cos x *  cot x sin x * Thus, at the point where    / 4 the slope of the tangent line is dy  m   cot  0 dx    / 4 2 Remember that :  * cot  cot 90  0 2 which implies that the circle has a horizontal tangent line at the point where    / 4 (Figure 10.3.1). 42  Figure 10.3.1 Example 2 : Page (720) Find the points on the cardioid r  1  cos  at which there is a horizontal tangent line, a vertical tangent line, or a singular point. Solution * A horizontal tangent line will occur where dy / d  0 and dx / d  0 , a vertical tangent line where dy / d  0 and dx / d  0 , and a singular point where dy / d  0 and dx / d  0. * We could find these derivatives from the formulas in  1 . However, an alternative approach is to go back to basic principles and express the cardioid parametrically by substituting r  1  cos  in the conversion formulas x  r cos  and y  r sin. This yields x   1  cos   cos  , y   1  cos   sin  0    2  * Differentiating these equations with respect to  and then simplifying yields dx   1  cos    sin    sin  cos    2 cos   1 sin d dy   1  cos   cos    sin  sin   1  cos   1  2 cos   d 43  1 * Thus, dx / d  0 if sin  0 or cos   , and dy / d  0 if 2 1 cos   1 or cos   . 2 * The solutions of dx / d  0 on the interval 0    2 are dx  5 0:  0 , , , , 2 d 3 3 Remember that :  5 1 * sin0  sin   sin 2  0 , cos  cos  3 3 2 y  sin x y  cos x * The solutions of dy / d  0 on the interval 0    2 are dy 2 4 0:  0 , , , 2 d 3 3 44  Remember that : 2 4 1 * cos 0  cos 2  1 , cos  cos  3 3 2 * Thus, horizontal tangent lines occur at   2 / 3 and   4 / 3 ; vertical tangent lines occur at    / 3 ,  , and   5 / 3 ; and singular points occur at   0 and   2 (Figure 10.3.2). Figure 10.3.2 * Note, however, that r  0 at both singular points, so there is really one singular point on the cardioid-the pole. Tangent Lnes to Polar Curves at the Origin : Page (721) * Formula  2  reveals some useful information about the behavior of a polar curve r  f   that passes through the origin. If we assume that r  0 and dr / d  0 when   0 , then it follows from Formula  2  that the slope of the tangent line to the curve at   0 is 45  dr 0  sin  0 dy  d  sin  0  tan  dx dr cos  0 0 0  cos  0 d (Figure 10.3.3). Figure 10.3.3 However, tan0 is also the slope of the line   0 , so we can conclude that this line is tangent to the curve at the origin. Thus, we have established the following result. 10.3.1 Theorem : Page (721) If the polar curve r  f   passes through the origin at   0 , and if dr / d  0 at   0 , then the line   0 is tangent to the curve at the origin. * This theorem tells us that equations of the tangent lines at the origin to the curve r  f   can be obtained by solving the equation f    0. It is important to keep in mind, however, that r  f   may be zero for more than one value of  , so there may be more than one tangent line at the origin. This is illustrated in the next example. 46 Example 3 : Page (721) * The three-petal rose r  sin 3 in Figure 10.3.4 has three tangent lines at the origin, which can be found by solving the equation sin 3  0 Figure 10.3.4 * We note that the complete rose is traced once as  varies over the interval 0     , so we need only look for solutions in this interval. We leave it for you confirm that these solutions are  2   0 ,   , and   3 3 * Since dr / d  3 cos 3  0 for these values of  , these three lines are tangent to the rose at the origin, which is consistent with the figure. 47 Arc Length of a Polar Curve : Page (721) * A formula for the arc length of a polar curve r  f   can be derived by expressing the curve in parametric form and applying Formula  2  of Section 10.1 for the arc length of a parametric curve. We leave it as an exercise to show the following. 10.3.2 Arc Length Formula for Polar Curves : Page (722) If no segment of the polar curve r  f   is traced more than once as  increases from  to  , and if dr / d is continuous for      , then the arc length L from    to    is   2  dr   f      f '     3 2 2 L  d   r2    d    d  Example 4 : Page (722) Find the arc length of the spiral r  e in Figure 10.3.5 between   0 and   . Figure 10.3.5 Solution * The arc length is  2  dr  L  r2    d   d  48      e   e   2  2 d  2  e d 0 0 Remember that : 1 *  e ax  b dx  e ax  b  C a       2  e   2 e  e 0  2 e  1  31.3 0 Remember that : * e0  1 Example 5 : Page (722) Find the total arc length of the cardioid r  1  cos . Solution * The cardioid is traced out once as  varies from   0 to   . Thus,  2  dr  L  r   d 2   d  2    1  cos   2    si n   2 d 0 2    1  2 cos   cos     sin   d 2 2 0 Remember that : * cos 2 x  sin2 x  1 2  2   1  cos   d 0 49  Remember that : 1 * cos 2 x   1  cos 2 x  2 2 2 1 1  2  cos  d  2  cos  d 2 0 2 0 2 1 * Since cos  changes sign at  , we must split the last 2 integral into the sum of two integrals: the integral from 0 to  plus the integral from  to 2. However, the integral from  to 2 is equal to the integral from 0 to  , since the cardioid is symmetric about the polar axis (Figure 10.3.6). Figure 10.3.6 * Thus, 2  1 1 L2  cos  d  4  cos  d 0 2 0 2 Remember that : 1 *  cos  ax  b  dx  sin  ax  b   C a   1      8  cos    8  sin  sin0   8  1  0   8  2 0  2  50  Remember that :  * sin0  0 , sin  1 2 Area in Polar Coordinates : Page (722) * We begin our investigation of area in polar coordinates with a simple case. 10.3.3 Area Problem in Polar Coordinates : Page (722) Suppose that  and  are angles that satisfy the condition       2 and suppose that f   is continuous and nonnegative for     . Find the area of the region R enclosed by the polar curve r  f   and the rays    and    (Figure 10.3.7). Figure 10.3.7 * In rectangular coordinates we obtained areas under curves by dividing the region into an increasing number of vertical strips, approximating the strips by rectangles, and taking a limit. In polar coordinates rectangles are clumsy to work with, and it is better to partition the region into wedges by using rays   1 ,    2 ,... ,    n 1 such that 51     1   2 ...   n 1   (Figure 10.3.8). Figure 10.3.8 * As shown in that figure, the rays divide the region R into n wedges with area A1 , A2 ,... , An and central angles 1 ,  2 ,... ,  n. The area of the entire region can be written as n A  A1  A2 ...  An   Ak 4 k 1 * If  k is small, then we can approximate the area Ak of the kth wedge by the area of a sector with central angle  k and   radius f  k* , where    k* is any ray that lies in the kth wedge (Figure 10.3.9). Figure 10.3.9 * Thus, from  4  and Formula  5  of Appendix B for the area of a sector, we obtain 52    n n 1 2 A   Ak    f k *    k 5 k 1 k 1 2 * If we now increase n in such a way that max  k 0 , then the sectors will become better and better approximations of the wedges and it is reasonable to expect that  5  will approach the exact value of the area A (Figure 10.3.10); that is,    n 1 2 1  *    2 A lim f      2 f   d  max  k 0 k  1 2  k  k  Figure 10.3.10 * Note that the discussion above can easily be adapted to the case where f   is nonpositive for     . We summarize this result below. 10.3.3 Area in Polar Coordinates : Page (723) If  and  are angles that satisfy the condition       2 and if f   is continuous and either nonnegative or nonpositive for      , then the area A of the region R enclosed by the polar curve r  f   (     ) and the lines    and    is   1 1 2 A    f    d   r d 6  2  2  2 53 * The hardest part of applying  6  is determining the limits of integration. This can be done as follows: Area in Polar Coordinates: Limits of Integration : Page (723) Step 1. Sketch the region R whose area is to be determined. Step 2. Draw an arbitrary "radial line" from the pole to the boundary curve r  f  . Step 3. Ask, "Over what interval of values must  vary in order for the radial line to sweep out the region R ?" Step 4. Your answer in Step 3 will determine the lower and upper limits of integration. Example 6 : Page (724) Find the area of the region in the first quadrant that is within the cardioid r  1  cos . Solution * The region and a typical radial line are shown in Figure 10.3.11. Figure 10.3.11 54 * For the radial line to sweep out the region,  must vary from 0 to  / 2. Thus, from  6  with   0 and    / 2 , we obtain   /2 1 1 A   r 2 d    1  cos   2 d  2 2 0 Remember that : *  a  b  2  a 2  2ab  b 2 /2  1 2    1  2 cos   cos 2  d 0 Remember that : 1 * cos 2 x   1  cos 2 x  2 /2 1  1     1  2 cos    1  co s 2   d 2 0 2  1  / 2 3 1     d 2 0  2    2 cos  cos 2 2  Remember that : 1 *  cos  ax  b  dx  sin  ax  b   C a /2 1 3 1    2   2 sin   sin 2  2 4 0 1  3     1  3 1     2  2  2   2 sin  sin    2   0  2 sin0  s in0  2 4   4  55  Remember that :  * sin0  0 , sin  1 , sin   0 2 1  3      3       2      1  0  0   1 2  2  2    8 Example 7 : Page (724) Find the entire area within the cardioid of Example 6. Solution * For the radial line to sweep out the entire cardioid,  must vary from 0 to 2. Thus, from  6  with   0 and   2 ,  2 1 1 A   r 2 d    1  cos   2 d  2 2 0 * If we proceed as in Example 6, this reduces to 2 1 3 1  A    2 cos   cos 2  d 2 0  2 2  2 1 3 1      2 sin  sin 2  2 2 4 0 1  3 1  3 1     2   2 sin 2  sin 4     0   2 sin0  sin0   2  2 4  2 4  Remember that : * sin n  0 ,where n is int eger 1  3 1   3   2  2   2  0    0     0     2  4   2 56 Alternative Solution. * Sine the cardioid is symmetric about the x-axis, we can calculate the portion of the area above the x-axis and double the result. In the portion of the cardioid above the x-axis, ranges from 0 to  , so that   1 2 3 A  2  r d    1  cos   d   2 0 2 0 2 Using Symmetry : Page (724) * Although Formula  6  is applicable if r  f   is negative, area computations can sometimes be simplified by symmetry to restrict the limits of integration to intervals where r  0. This is illustrated in the next example. Example 8 : Page (724) Find the area of the region enclosed by the rose curve r  cos 2. Solution * Referring to Figure 10.2.13 and using symmetry, Figure 10.2.13 the area in the first quadrant that is swept out for is one - eighth of the total area inside the rose. Thus, from Formula 6  57    /4 1 2 1 A   r d  8  cos 2 2 d  2 0 2 Remember that : 1 * cos 2 x   1  cos 2 x  2  /4 1  4   1  cos 4  d 0 2 Remember that : 1 *  cos  ax  b  dx  sin  ax  b   C a  /4  1   2   sin 4   4 0   1   1   2   sin     0  sin0    4 4   4  Remember that : * sin0  0 , sin   0   1   1    2    0     0   0      4 4   4  2 * Sometimes the most natural way to satisfy the restriction       2 required by Formula  6  is to use a negative value for . For example, suppose that we are interested in finding the area of the shaded region in Figure 10.3.12a. The first step would be to determine the intersections of the cardioid r  4  4 cos  and the circle r  6 , since this information is needed for the limits of integration. To find the points of intersection, we can equate the two expressions for r. This yields 58  1 4  4 cos   6 or cos   2 which is satisfied by the positive angles  5  and   3 3 * However, there is a problem here because the radial lines to the circle and cardioid do not sweep through the shaded region in Figure 10.3.12b as  varies over the interval  / 3    5 / 3. There are two ways to circumvent this problem-one is to take advantage of the symmetry by integrating over the interval 0     / 3 and doubling the result, and the second is to use a negative lower limit of integration and integrate over the interval   / 3     / 3 (Figure 10.3.12c). The two methods are illustrated in the next example. Figure 10.3.12 59 Example 9 : Page (725) Find the area of the region that is inside of the cardioid r  4  4 cos  and outside of the circle r  6. Solution Solution Using a Negative Angle. * The area of the region can be obtained by subtracting the areas in Figures 10.3.12d and Figures 10.3.12e: Figures 10.3.12  /3  /3 1 1 A   4  4 cos   d    6  2 d 2  / 3 2  / 3 2 1 /3  2      d 2  / 3  2  4  4 cos  6  Remember that : *  a  b  2  a 2  2ab  b 2 1 /3    2  / 3  16   32 cos   16 cos 2     36   d /3    16 cos   8 cos 2    10 d  / 3 60  Remember that : 1 * cos 2 x   1  cos 2 x  2 /3    16 cos   4  1  cos 2   10  d  / 3 /3    16 cos   4 cos 2  6  d  / 3 Remember that : 1 *  cos  ax  b  dx  sin  ax  b   C a x r 1 *  x dx  r C r  1  r 1 /3   16 sin  2 sin 2  6    / 3           16 sin     2 sin 2    6      3  3  3            16 sin      2 sin 2     6       3  3  3  Remember that : * sin   x    sin x          2  16 sin    2 sin 2    6      3  3  3  Remember that :  2 * sin    x   sin x , sin    sin    3  3  61     * 3  180 3  3    60 , sin    sin 60  2 3 y  sin x  3 3   2  16 2  2   18 3  4  2 2  Solution Using Symmetry. * Using symmetry, we can calculate the area above the polar axis and double it. This yields  / 3 1 /3 1  A  2    4  4 cos   d    6  d  2 2  0 2 0 2  /3    4  4 cos   2   6  2  d   0 /3    32 cos   8 cos 2  12  d 0 /3   32 sin  4 sin 2  12  0 62         32 sin    4 sin 2    12    3  3  3 3 3  32 4  4  18 3  4 2 2 which agree with the preceding result. Intersections of Polar Graphs : Page (726) * In the last Example we found the intersections of the cardioid and circle by equating their expressions for r and solving for . However, because a point can be represented in different ways in polar coordinates, this procedure will not always produce all of the intersections. For example, the cardioids r  1  cos  and r  1  cos  7  intersect at three points: the pole, the point  1, / 2  and the point  1,3 / 2  (Figures 10.3.13). Figures 10.3.13 * equating the right-hand sides of the equations in 7  yields 1  cos   1  cos  or cos   0 , so     k , k  0 ,  1 ,  2 ,... 2 63 Substituting any of these values in 7  yields r  1, so that we have found only two distinct points of intersection,  1, / 2  and  1,3 / 2  : the pole has been missed. This problem occurs because the two cardioids pass through the pole at different values of  -the cardioid r  1  cos  passes through the pole at   0 , and the cardioid r  1  cos  passes through the pole at   . 64 EXERCISE SET 10.3: (Home Work) Page (726) 1,2 Find the slope of the tangent line to the polar curve for the given value of . 1. r  2 sin ;    / 6 2. r  1  cos  ;    / 2 20,22 Use Formula  3  to calculate the arc length of the polar curve. 20. The entire circle r  2a cos  22. r  e 3 from   0 to   2 25 Write down, but do not evaluate, an integral for the area of each shaded region. 36 Find the area of the shaded region. 39,40,41,43 Find the area of the region described. 39. The region inside the circle r  3 sin and outside the cardioid r  1  sin. 65 40. The region outside the cardioid r  2  2 cos  and inside the circle r  4. 41. The region inside the cardioid r  2  2 cos  and outside the circle r  3. 1 43. The region between the loops of the limacon r   cos . 2 66  CHAPTER (11) THREE-DIMENSIONAL SPACE ; VECTORS 11.1 RECTANGULAR COORDINATES IN 3-SPACE ; SHERES ; CYLINDRICAL SURFACES : Page (767) Rectangular Coordinate Systems : * In the remainder of this text we will call three-dimensional space 3-space, two dimensional (a plane) 2-space, and one- dimensional space (a line) 1-space. Just as points in 2-space can be placed in one-to-one correspondence with pairs of real numbers using two perpendicular coordinate lines, so points in 3-space can be placed in one-to-one correspondence with triples of numbers by using three mutually perpendicular coordinate lines, called the x-axis, the y-axis, and the z-axis, positioned so that their origins coincide (Figure 11.1.1). The three coordinate axes from a three-dimensional rectangular coordinate system (or Cartesian coordinate system). The point of intersection of the coordinate axes is called the origin of the coordinate system. Figure 11.1.1 * Rectangular coordinate systems in 3-space fall into two categories: left-handed and right-handed. A right-handed 67  system has the property that when the fingers of the right hand are cupped so that they curve from the positive x-axis toward the positive y-axis, the thumb points (roughly) in the direction of the positive z-axis (Figure 11.1.2). A similar property holds for a left-handed coordinate system (Figure 11.1.2). We will use only right-handed coordinate system in this text. Figure 11.1.2 * The coordinate axes, taken in pairs, determine three coordinate planes: the xy-plane, the xz-plane, and the yz- plane (Figure 11.1.3). Figure 11.1.3 To each point P in 3-space we can assign a triple of real numbers by passing three planes through P parallel to the coordinate planes and letting a, b, and c be the coordinates of the intersections of those planes with the x-axis, y-axis, and z- axis, respectively (Figure 11.1.4). 68  Figure 11.1.4 We call a, b, and c the x-coordinate, y-coordinate, and z- coordinate of P, respectively, and we denote the point P by  a,b,c  or by P  a,b,c . Figure 11.1.5 shows the points  4,5,6  and   3,2,  4 . Figure 11.1.5 * Just as the coordinate axes in a two-dimensional coordinate system divide 2-space into four quadrants, so the coordinate planes of a three-dimensional coordinate system divide 3-space into eight parts, called octants. The set of points with three positive coordinates forms the first octant; the remaining octants have no standard numbering. * You should be able to visualize the following facts about three- dimensional rectangular coordinate systems: 69  Region Description xy-plane Consists of all points of the form  x, y,0  xz-plane Consists of all points of the form  x,0,z  yz-plane Consists of all points of the form  0, y,z  x-axis Consists of all points of the form  x,0,0  y-axis Consists of all points of the form  0, y,0  z-axis Consists of all points of the form  0, y,z  Distance in 3-Space; Spheres : Page (768) * Recall that in 2-space the distance d between the points P1  x1 , y1  and P2  x2 , y2  is d   x2  x1    y2  y1   1 2 2 The distance formula in 3-space has the same form, but it has a third to account for the added dimension. (We will see that this is a common occurrence in extending formulas from 2- space to 3-space). The distance between the points P1  x1 , y1 , z1  and P2  x2 , y2 , z2  is d  x2  x1  2   y2  y1  2   z2  z1  2  2 Example 1 : Page (769) Find the distance d between the point  2,3,  1 and  4,  1, 3 . Solution * From Formula  2  d  x2  x1  2   y2  y1  2   z2  z1  2 70    4  2  2    1  3  2   3  1 2  36  6 * Recall that the standard equation of the circle in 2-space that has center  x0 , y0  and radius r is  x  x0    y  y0   r 2 2 2  3 This follows from distance formula  1  and the fact that the circle consists of all points in 2-space whose distance from  x0 , y0  is r. Analogously, the standard equation of the sphere in 3-space that has center  x0 , y0 , z0  and radius r is  x  x0  2   y  y0  2   z  z0  2  r 2 4 * This follows from distance formula  2  and the fact that the sphere consists of all points in 3-space whose distance from  x0 , y0 , z0  is r. Note that  4  has the same form as the standard equation for the circle in 2-space, but with an additional term to account for the third coordinate. Some examples of the standard equation of the sphere are given in the following table: Equation Graph  x  3  2   y  2  2   z  1 2  9 Sphere with center  3,2,1 and radius 3  x  32  y 2   z  4 2  5 Sphere with center   1,0,  4  and radius 5 x2  y2  z2  1 Sphere with center  0,0,0  and radius 1 * If the terms in are expanded and like terms are collected, then the resulting has the form 71  x 2  y 2  z 2  Gx  Hy  Iz  J  0 5 The following example shows how the center and radius of a sphere that is expressed in this form can be obtained by completing the squares. Example 2 : Page (769) Find the center and radius of the sphere x 2  y 2  z 2  2 x  4 y  8 z  17  0 Solution * We can put the equation in the form of  4  by completing the squares: x 2  y 2  z 2  2 x  4 y  8 z  17  0 x 2       2 x  1  y 2  4 y  4  z 2  8 z  16  17  21  x  1 2   y  2  2   z  4  2  4 which is the equation of the sphere with center  1,2,  4  and radius 2. * In general, completing the squares in  5  produces an equation of the form  x  x 0 2   y  y0 2   0  k z  z 2 If k  0 , then the graph of this equation is a sphere with center  x0 , y0 , z0  and radius k. If k  0 , then the sphere has radius zero, so the graph is the single point  x0 , y0 , z0 . If k  0 , the equation is not satisfied by any values of x, y, and z, so it has no graph. 72 11.1.1 Theorem : Page (770) An equation of the form x 2  y 2  z 2  Gx  Hy  Iz  J  0 Represents a sphere, a point, or has no graph. Cylindrical Surfaces : Page (770) * Although it is natural to graph equations in two variables in 2-space and equations in three variables in 3-space, it is also possible to graph equations in two variables in 3-space. For example, the graph of the equation y  x 2 in an xy-coordinate system is a parabola; however, there is nothing to prevent us from inquiring about its graph in an xyz-coordinate system. To obtain this graph we need only observe that the equation does not impose any restrictions on z. Thus, if we find values of x and y that satisfy this equation, then the coordinates of the point  x, y, z  will also satisfy the equation for arbitrary values of z. Geometrically, the point  x, y, z  lies on the vertical line through the point  x, y,0  in the xy-plane, which means that we can obtain the graph of y  x 2 in an xyz-coordinate system by first graphing the equation in the xy-plane and then translating that graph parallel to the z-axis to generate the entire graph (Figure 11.1.6). 73  Figure 11.1.6 * The process of generating a surface by translating a plane curve parallel to some line is called extrusion, and surfaces that are generated by extrusion are called cylindrical surfaces. A familiar example is the surface of a right circular cylinder, which can be generated by translating a circle parallel to the axis of cylinder. The following theorem provides basic information about graphing equations in two variables in 3- space: 11.1.2 Theorem : Page (770) An equation that contains only two of the variables x, y, and z represents a cylindrical surface in an xyz-coordinate system. The surface can be obtained by graphing the equation in the coordinate plane of the two variables that appear in the equation and then translating that graph parallel to the axis of the missing variable. 74 Example 3 : Page (770) Sketch the graph x 2  z 2  1 in 3-space. Solution * Since y does not appear in this equation, the graph is a cylindrical surface generated by extrusion parallel to the y-axis. In the xz-plane the graph of the equation x 2  z 2  1 is a circle. * Thus, in 3-space the graph is a right circular cylinder along the y-axis (Figure 11.1.7). Figure 11.1.7 Example 4 : Page (770) Sketch the graph z  sin y in 3-space. Solution * (See Figure 11.1.8). Figure 11.1.8 75 EXERCISE SET 11.1: (Home Work) Page (771) 3,4,5 Sketch the curve by eliminating the parameter, and indicate the direction of increasing t. 3. x  3t  4 , y  6 t  2 4. x  t  3 , y  3t  7 0  t  3 5. x  2 cos t , y  5 sin t  0  t  2  13,17 Find parametric equations for the curve, and check your work by generating the curve with a graphing utility. 13. A circle of radius 5, centered at the origin, oriented clockwise. 17. The portion of the parabola x  y 2 joining  1,  1 and  1,1 , oriented down to up. 45,49 Find dy / dx and d 2 y / dx 2 at the given point without eliminating the parameter. 45. x  t , y  2t  4 ; t  1 49. x    cos  , y  1  sin ; t  / 6 53. Find all values of t at which the parametric curve x  2 sin t , y  4 cos t  0  t  2  has (a) a horizontal tangent line and (b) a vertical tangent line. 65,67 Find the exact arc length of the curve over the stated interval. 1 3 65. x  t 2 , y t  0  t  1 3 67. x  cos 3t , y  sin 3t 0  t    76 11.2 VECTORS : Page (773) Vectors in Physics and Engineering : * A particle that moves along a line can move in only two directions, so its direction of motion can be described by taking one direction to be positive and the other negative. Thus, the displacement or change in position of the point can be described by a signed real number. For example, a displacement of 3    3  describes a position change of 3 units in the positive direction, and a displacement of describes a position change of 3 units in the negative direction. However, for a particle that moves in two dimensions or three dimensions, a plus or minus sign is no longer sufficient to specify the direction of motion-other methods are required. one method is to use an arrow, called a vector, that points in the direction of motion and whose length represents the distance from the starting point to the ending point; this is called the displacement vector for the motion. For example, Figure 11.2.1a shows the displacement vector of a particle that moves from point A to point B along a circuitous path. Figure 11.2.1 Note that the length of the arrow describes the distance between the starting and ending points and the actual distance traveled by the particle. 77 * Arrows are not limited to describing displacements-they can be used to describe any physical quantity that involves both a magnitude and a direction. Two important examples are forces and velocities. For example, the arrow in Figure 11.2.1b represents a force vector of 10 Ib acting in a specific direction on a block, and the arrows in Figure 11.2.1c show the velocity vector of a boat whose motor propels it parallel to the shore at 2 mi/h and the velocity vector of a 3 mi/h wind acting at an angle of 45 with the shoreline. Intuition suggests that the two velocity vectors will combine to produce some net velocity for the boat at an angle to the shoreline. Thus, our first objective in this section is to define mathematical operations on vectors that can be used to determine the combined effect of vectors. Vectors Viewed Geometrically : Page (774) * Vectors can be represented by arrows in 2-space or 3-space; the direction of the arrow specifies the direction of the vector, and the length of the arrow describes its magnitude. The tail of the arrow is called the initial point of the vector, and the tip of the arrow the terminal point. We will denote vectors with lowercase boldface type such as a,k ,v ,w,and x. When discussing vectors, we will refer to real numbers a scalars. Scalars will be denoted by lowercase italic such as a, k, v, w, and x. Two vectors, v and w , are considered to be equal (also called equivalent) if they have the same length and same direction, in which case we write v  w. Geometrically, two vectors are equal if they are translations of one another; thus, the three vectors in Figure 11.2.2a are equal, even though they are in different positions. 78  Figure 11.2.2 * Because vectors are not affected by translation, the initial point of a vector v can be moved to any convenient point A by making an appropriate translation. If the initial point of v is A and the terminal point is B, then we write v  AB when we want to emphasize the initial and terminal points (Figure 11.2.2b). If the initial and terminal points of a vector coincide, then the vector has length zero: we call this the zero vector and denote it by 0. The zero vector does not have a specific direction, so we will agree that it can be assigned any convenient direction in a specific problem. * There are various algebraic operations that are performed on vectors, all of whose definitions originated in physics. We begin with vector addition. 11.1.2 Definition : Page (774) If v and w are vectors, then the sum v  w is the vector from the initial point of v to the terminal point of w when the vectors are positioned so the initial point of w is at the terminal point of v (Figure 11.2.3a). * In Figure 11.2.3b we have constructed two sums, v  w (from purple arrows) and w  v (from green arrows). It is evident that 79  vw  wv and that the sum (gray arrow) coincides with the diagonal of the parallelogram determined by v and w when these vectors are positioned so they have the same initial point. Figure 11.2.3 Since the initial and terminal points of 0 coincide, it follows that 0v  v0  v 11.1.3 Definition : Page (774) If v is a nonzero vector and k is a nonzero real number (a scalar), then the scalar multiple kv is defined to be the vector whose length is k times the length of v and whose direction is the same as that of v if k  0 and opposite to that of v if k  0. We define kv  0 if k  0 or v  0. * Figure 11.2.4 shows the geometric relationship between a vector v and various scalar multiples of it. Observe that if k and v are nonzero, then the vectors v and kv lie on the same line if their initial points coincide and lie on parallel or coincident lines if they do not. 80  Figure 11.2.4 * Thus, we say that v and kv are parallel vectors. Observe also that the vector   1  v has the same length as v but oppositely directed. We call   1  v the negative of v and denote it by v (Figure 11.2.5). In particular,  0    1 0  0. Figure 11.2.5 * Vector subtraction is defined in terms of addition and scalar multiplication by vw  v w   The difference v  w can be obtained geometrically by first constructing the vector  w and then adding v and  w , say the parallelogram method (Figure 11.2.6a). However, if v and w are positioned so their initial points coincide, then v  w can be formed more directly, as shown in Figure 81  11.2.6b, by drawing the vector from the terminal point of w (the second term) to the terminal point of v (the first term). In the special case where v  w the terminal points of the vectors coincide, so their difference is 0 ; that is,   v  v  v v  0 Figure 11.2.6 Vectors in Coordinate Systems : Page (775) * Problems involving vectors are often best solved by introducing a rectangular coordinate system. If a vector v is positioned with its initial point at the origin of a rectangular coordinate system, then its terminal point will have coordinates of the form  v1 ,v 2  or  v1 ,v2 ,v3  , depending on whether the vector is in 2-space or 3-space (Figure 11.2.7). Figure 11.2.7 * We call these coordinates the components of v , and we write v in component form using the bracket notation 82  v  v1 ,v 2 or v  v1 ,v 2 ,v 3 2  space 3  space In particular, the zero vectors in 2-space and 3-sp