Lecture Notes. Math 263 Calculus III. (Part II)

Transcription

1 Math 63 Calculus III Lecture Notes Math 63 Calculus III (Part II)

2 Math 63 Calculus III i Table Of Contents Lecture Lecture Lecture Lecture Lecture Lecture

3 Math 63 Calculus III Lecture 10 Sections 16.1 and 16. Integrals In single variable calculus we presented the integral of a continuous function as the opposite of differentiation and showed how it was related to the area under the curve defined b the graph of the function. The link between area and the integral was illustrated b partitioning or dividing some interval over which the function is defined into n parts and then using the maimum or minimum value of the function on each partition to define the height of a rectangle. The area under the curve defined b the function on the given interval is then approimated b summing the areas of these rectangles. If the maimum value of the function is used to define the height of each rectangle, then the approimated area is greater than the actual area under the curve. If the minimum values are used, the then approimated area is less than the actual area. B choosing finer and finer partitions i.e. larger values of n, the accurac of this approimation increases. In fact, the actual area under the curve is bounded b the two areas found b using the minimum and maimum values of the function on each partition. If A min (n) represents the area found b partitioning the interval into n parts and using the minimum values of the function on each partition, and A ma (n) represents the area found b partitioning the interval into n parts and using the maimum values of the function on each partition, and A represents the actual area the we have: A min (n) A A ma (n) The first known use of this technique of partitioning an interval and using rectangles to approimate the area under a curve was b the Greek Mathematical Archimedes (87BC - 1BC). In fact there is currentl a discussion among scholars that Archimedes anticipated both Newton and Leibniz in the discover/invention of the calculus. Archimedes (87BC - 1BC)

4 Math 63 Calculus III 3 Approimatel 000 ears ago Archimedes used a similar technique to estimate the area of the circle. Instead of using n-rectangles Archimedes used n-sided polgons however the idea is essentiall the same. Figure 1 shows the modern version of approimating the area of one quarter of a circle using rectangular approimations. Figure 1. Approimating the area of a circle with rectangles In the case of Figure 1, the maimum value of the circle was used on each interval and onl n = 6 partitions of the interval were used. Figure shows the same partitioning with the minimum value of the circle on each partition being used. In both cases, the approimation is clearl not ver accurate, however b doubling the number of partitions i.e. letting n = 1, it is clear that the approimation is better. The idea of partitioning an interval and taking the minimum and maimum values of the function on the each partition and then summing the product of the partition width and the maimum or minimum value was first presented b Riemann ( ) and is known ad the Riemann sum. Georg Friedrich Bernhard Riemann ( )

5 Math 63 Calculus III 4 Figure. Area using minimum values. Figure 3. Partition with n = 1

6 Math 63 Calculus III 5 More formall for a general function, f, we write the Riemann sums as: n A n f min ma i1 n A n f i1 i i i i Where i is the place where f takes on the minimum value of the i th partition, i is where f takes on the maimum value on the i th partition, and i is the size or width of the i th partition. Note: although we often construct the Riemann sums using the same size partitions, this is not required and is reflected b associating the subscript i with the partition size. We define the definite integral over the interval [a, b] as the limit of either the upper or lower Riemann sum as the partition size tends towards zero or the number of partitions tends towards infinit: b a n i i i i n 0 i1 i lim lim f d f f For functions of two variables we etend this idea of partitioning and summing to define a double Riemann sum and the integral over a region. To see this, we consider the graph of the surface defined b the function f(, ). In this case, we can partition the area in the -pane under the surface into squares (or rectangles) and form boes whose heights are given b either the minimum or maimum value of the function over the rectangular partitions. Figure 4 illustrates the partitioning for a quarter of a sphere. Like with the circle, we can choose the height of the boes to like entirel under the surface of the sphere b choosing the minimum value of the sphere over each square in the -plane or we can choose the height of the boes to lie entirel above the surface b choosing the maimum value of sphere over each square in the -plane. In both cases, we can compute the volume of each bo and then sum these volumes to approimate the volume, V, under the surface.

7 Math 63 Calculus III 6 z Figure 4. Approimating Sphere with boes If we partition the -ais into m-parts and the -ais into n-parts, and if we let V min (m, n) be the volume obtained b summing the volume of the boes formed b taking the minimum value of the sphere and we let V ma (m, n) be the volume obtained b taking the maimum value of the sphere on each partition, then we have: V min (m, n) V V ma (m, n)

8 Math 63 Calculus III 7 And it is clear that the volume of the sphere is bounded b V min and V ma over the specified region of the -plane. More formall for a general function, f, we write these Riemann sums as: V m n f min ma m n,, i1 j1 m n,, i1 j1 i j i j V m n f i j i j Analogous to what we did for the single variable case, we define the definite integral over the region, R, as the limit of either the upper or lower Riemann sums as the partition sizes tends towards zero or the number of partitions tends towards infinit: R m n fda lim f i, ji j lim f i, ji j mn,, 0 i1 j1 i, j The general notation R fda implies integration over of an arbitrar shaped region (area) e.g. triangle, circle, etc. In our presentation above we chose a rectangular shaped region for convenience since it makes the summation limits and partitioning of the -pane into squares or rectangles simpler. However, this need not be the case. We can instead picture covering an arbitrar shaped region in the -plane with rectangles so that either all the rectangles lie just inside the region or the rectangles etend just outside the region (so that the region is contained inside our rectangles). We can then compute either the minimum or maimum value of the function on each rectangle and compute the volume of the boes, and sum. We further note that no particular ordering is implied in the summation. We can picture, taking all the boes and laing them end to end in a line in arbitrar order and summing their volumes. In short, the order of summation is not important since the same volumes of the same boes are ultimatel summed together. This will have an important consequence with regards to the order of integration which we will discuss shortl. Another important point to consider is that if we choose a function f(, ) = 1 everwhere then our integral becomes:

9 Math 63 Calculus III 8 R fda da A R for f(, ) =1 That is, when f(, ) = 1, the integral gives us the area of the region we are integrating over. Finall we note that, using the definition of continuit and the -technique for limits one can formall prove that mn, ma lim V m, n V m, n 0 min The formal proof is beond the scope of the course, but we provide the following brief outline: Since f(, ) is continuous on the closed region R, then one can show that f(, ) is uniforml continuous on R. That is for an given an > 0 we can find > 0 such that for an two points ( 1, 1 ) and (, ) in R, if, then,, 1 1 f f. We note that if we 1 1 choose our i j to lie completel inside R then m n i j A R where A R is i1 j1 the area of R. Since f(, ) is uniforml continuous then if we choose 0 then we can find > 0 such that f, f 1, 1 whenever A R AR 1 1. In particular, we choose i and j so that on each square or rectangle of our partition we have f i, j f i, j. Therefore: A R V mn V m n f f ma m n m n, min,,, i j i j i j i j i1 j1 i1 j1 m n f f i1 j1 R i1 j1,, m n ij A i j i j i j

10 Math 63 Calculus III 9 ====================================================================== Integrated Integrals As we mentioned above, the value of the Riemann sum in independent of the order of the summation. This is a direct result of the commutative, associative, and distributive properties of real numbers. As such we are free to sum the terms in which ever wa we choose. We can use this freedom to show that the integral over a region is equivalent to a double integral over two intervals. To see this we go back to our definition of the integral over a region as the limit of a double Riemann sum: R m n m n fda lim f i, ji j lim f i, j j i mn, mn, i1 j1 i1 j1 m n lim i lim f i, j j m n i1 j1 m lim f(, ) d m i 1 i c d i Note: we have assumed that i is constant (in the sense we did when taking partial derivative) in going from the second to the last step in the equation above. That is we assumed that f was a function of the form f( i, ) and could be treated as if it were a single variable function, We then applied the definition of the integral for a single variable function as being a Riemann sum over an interval, in this case [c, d]. We now let d (, ) F i process: c f d i and, substituting into the epression above, repeat this m b b d fda lim F F d f (, ) dd i i m a a c R i 1 Thus we have epressed the integral over a region as the double integral over two intervals. Please note, that we can repeat this argument, reversing the order of the summation so that we sum over i first and j second (i.e. over first and second) so the result has the order of integration reversed. That is we can show that the order of integration in the resulting

11 Math 63 Calculus III 10 double integral can also be reversed. Hence we have the following general result: If f(, ) is a continuous function on the region R defined b the intervals [a, b] and [c, d] (written R = [a, b] [c, d]) then: Eample 1. Let [0, 16]. R (, ) (, ) b d d b fda f d d f d d a c c a f, 1. Find the integral over the region R = [0, 8] 4 8 Using the formula above for the integral: R fda 1 d d 1 d d As noted above, the region of integration need not be rectangular but ma be of arbitrar shape. In the formula above where we epressed the integral over the region R as the double integral over two intervals, the formula (does not assume) but is implicitl stated as the integral over a rectangular region. For non-rectangular regions, the formula still applies, but must be stated with more care. In particular, the limits of the inner integral become functions of the variable of the outer integral: If f(, ) is a continuous function on the region R then: R 1 1 b g d h fda f (, ) d d f (, ) d d a g c h A couple of points are note worth here. First, the order of integration can still be reversed, however their ma be a preferred order in the sense that

12 Math 63 Calculus III 11 the boundar functions (g 1 (), g (), h 1 (), and h ()) ma be significantl more straight forward to find or compute for one order as opposed to the other. That is, doing the integral over first sa, might be easier in the sense that h 1 (), and h () are either given, are obvious from the problem, or are relativel simple functions whereas g 1 () and g () might be difficult to determine or be ver comple functions. Second, the double integral itself ma be significantl simpler to evaluate using one order rather than the other. We shall give eamples of these points below. Finall, the limits on the outer integral must alwas be constants. Thus the chosen order of integration will depend on the problem and is usuall determined b the function being integrated and the shape of the region R. The order of integration which results in the simplest evaluation of the integrals is the one that is preferred. Eample. Let f(, ) =. Integrate f(, ) for the triangular region bounded b the -ais, the -ais, and the line =. Making a graph of R often helps to visualize the region of integration. Figure 5 show the region of integration for this eample. = R 1 Figure 5: Triangular region of integration

13 Math 63 Calculus III 1 In this case our function is relativel simple and smmetric in the variables, so integrating in either order is equivalent. However, since one of the boundar functions is given in terms of, i.e. g () = =, integrating over first saves us the effort of inverting this function. Graphicall or geometricall integrating over first is equivalent to moving along the ais from 0 to 1 and integrating from = 0 to = -. That is, summing up the vertical strips as shown in Figure 6. Integrate over these vertical strips = 1 Figure 6: Integrate over first corresponding to the vertical strips Hence, the limits of integration are a = 0, b = 1, g 1 () = 0, and g () =. We therefore have: 1 1 d d 0 d d d d

14 Math 63 Calculus III 13 If we choose to do the integral in the opposite order, then we need to invert the = i.e. epress as function of. In this case this is relativel straightforward and we get: = 1 - ½ = h (). Integrating in this order corresponds to integrating from = 0 to = along horizontal strips ranging from = 0 to = 1 - ½, as shown in Figure 7. Integrate over these horizontal strips = 1 ½ 1 Figure 7: Integrate over first corresponding to the horizontal strips The limits of integration are c = = 0, d = =, h 1 () = = 0, and h () = = 1 - ½. Therefore: dd d d 1 d d

15 Math 63 Calculus III 14 We note that in this eample, the order of integration is equall simple in both directions. However, this is not alwas the case, as will be demonstrated below. Eample 3. Integrate and the parabola =. f, 1 over the region bounded b the line = Figure 8 shows the region of integration. In this case, it s fairl straightforward to identif the limits of integration as follows: c = 0, d = 1, g 1 () =, g () =. = 1 R = Figure 8: Region of Integration defined b = and = We therefore have: 1 1 d d d d

16 Math 63 Calculus III 15 Eample 4. Evaluate d d 0 3. In this case we note that the order of integration is specified b the wa the integral is written. However, evaluating the integral in this order is nontrivial since the integral of 3 1 has no elementar anti-derivative. We would therefore like to reverse the order of integration in the hopes of simplifing the problem. However, we cannot just reverse the order since the lower limit of the inner integral is a function of. We must therefore determine the region of integration first, so we can epress the limits of integration in terms of functions of for the inner integral. The limits of integration are = 0, = 6, = /3, and =. Figure 9 shows the region bounded b these lines. = = 6 R = /3 = 0 6 Figure 9: Region of Integration defined b =, = /3, = 0, = 6 From the Figure 9 we see the region of integration is the triangular shaped region bounded b the -ais, and the lines = and = /3. To reverse the order of integration we want to integrate from = 0 to = while integrating along horizontal strips from = 0 to = 3. Thus we have:

17 Math 63 Calculus III d d 0 1d d d 1 d Thus reversing the order of integration allowed us to evaluate the double integral in this case. It is important to note that this required us to determine the region of integration from the integration limits of the original integral.

18 Math 63 Calculus III 17 Lecture 11 Sections 16.3 and 16.4 Double & Triple Integrals & Polar Coordinates Triple Integrals In the previous lecture we showed how a function of two variables can be integrated over a region in -space i.e. an area and how integration over a region is equivalent to an iterated or double integral over two intervals. This concept and be etended to integration over a solid region or volume of 3- space using triple integrals. In this case, we have a continuous function, f(,, z) of three variables that is integrated over a solid region W in 3-space. We define our integral in terms of a triple Riemann sum over a solid region that has been partitioned into boes having volume Vijk i jzk. If W is a bo-shaped region then b a d c f e we can let i, j, and zk z. That is the l m n solid region W = [a, b] [c, d] [e, f]. If we let ( i, j, z k ) be a point in ijk th bo, then we sa: If f is a continuous function of three variables, then we define the integral over the solid region W as: l m n fdv lim f,, z z W lmn,, i 1 j 1 k 1 i j k i i k We can show that the integral over the solid region W is equivalent to a triple integral over three intervals. The argument is similar to the argument we used for the double integral and is therefore not repeated here. We state formall that:

19 Math 63 Calculus III 18 If f(,, z) is a continuous function on the solid region W defined b the intervals [a, b], [c, d], [e, f] (written W = [a, b] [c, d] [e, f]) then: W f d b fdv f (,, z) d d dz e c a Eample 1. Integrate f(,, z) = 1 + z over the cube of length 4. From above: W fdv z d d dz z d dz z d dz z dz zdz 16z 3z We note that if f(,, z) = 1, the triple integral is just the volume of the solid region W. Eample. Find the volume under the plane z = 1 - ¼ - over the region [0, 8] [0, 16]. We wish to epress the double integral of the function z = 1 - ¼ - over the region [0, 8] [0, 16] as a triple integral of the function f(,, z) = 1 over a solid region W. We therefore must find W. Figure 10 shows the volume bounded b the region [0, 8] [0, 16] and the plane z = 1 - ¼ -. From Figure 10 we see that that as goes from 0 to 8 and from 0 to 16, z ranges from 0 to the plane z = 1 - ¼ -. Thus the limits of integration are a = 0, b = 8, c = 0, d = 16, e = 0, and f = 1 - ¼ -.

20 Math 63 Calculus III 19 z z = 1 - ¼ - 1 R 16 8 Figure 10. Volume of integration defined b the region R and a plane We therefore have: V dz d d z d d d d 1 d d Note: In this case we reversed the order of integration so that we integrated over z first and last. In addition, the limits of integration over z were functions of and. This is similar to the results we found for double integrals.

21 Math 63 Calculus III 0 Eample 3. Compute the integral for f(,, z) = z over the solid region bounded b the cone z and the plane z = 3. First we need to determine the solid region of integration based on the information above. Figure 11 shows the cone and the plane specified b the problem. z Level Surface i.e. circle of radius z z = 3 3 z Circle of radius z Figure 11. Solid region of integration defined b cone and plane z= 3 The intersection of the cone and the plane z = 3 corresponds to the level surface for the cone at z = 3. Thus we have: 3 or 9 For values of z between 0 and 3 we will get similar level surfaces. We can view the solid region of integration as just built from the areas of the stacked circles of radii between 0 and 3. The top circle will have radius of 3 and therefore the limits of integration for and are between -3 and 3 corresponding to the circular region in the -plane as shown in Figure 11. However for an particular circle, its radius will be z corresponding to the equation of the level surface radius z we let var from -z to z with given b z. To get the area of the circle with z (the plus sign corresponding to the area of the upper half of the circle while the minus

22 Math 63 Calculus III 1 sign corresponds to the area of the lower half of the circle). Thus as z varies from 0 to 3, varies from z to z and from Therefore the limits of integration are: z to z. z between 0 and 3 between z and z between z and z. Hence the triple integral is: W 3 z z fdv z d d dz 0 z z We can evaluate the integral b integrating over these limits: W z 3 z z 3 z fdv z d d dz z d dz 0 z z 0 z z z 3 z zz d dz z z z s dz 0 z in 0 z z 81 zdz z We can also compute this integral in reverse order i.e. with the inner integral over z and the outer integral over. To do this we note, that for a particular and, z can var between z and z = 3. Furthermore since the allowed values of an must lie in the circular region the -plane, for a particular, can var between Finall can var between -3 and 3. Thus we get: 9 and 9. 9 in W fdv 3 9 zdzdd

23 Math 63 Calculus III We can integrate to get the value of the integral: W fdv zdzdd z dd 9 dd d d sin Finall we summarize our findings from these eamples: The limits for the outer integral are constants. The limits for the middle integral can involve onl one variable (that in the outer integral) The limits for the inner integral can involve two variables (those on the two outer integrals). In essence, all the results for the double integral appl to the triple integral with the appropriate generalizations resulting from having three variables as opposed to two variables. ====================================================================== Polar Coordinates Polar coordinates provide and alternate (from Cartesian coordinates) reference-coordinate sstem for epressing the location of objects and points in the plane. Cartesian coordinates give the location of a point p relative to a defined origin on an equall spaced rectangular grid. We thus write p where and are the rectangular (Cartesian) coordinates of, the point. We can also epress this same location as the distance r from the

24 Math 63 Calculus III 3 origin and a direction as given b an angle measured relative to the -ais, as shown in Figure 1. (, ) = (r, ) r Figure 1. Cartesian and Polar Coordinates The two coordinates, (, ) and (r, ) are equivalent and given one pair, the other pair can be computed. Hence, we note from Figure 1 that: rcos rsin Using some basic trigonometr, we have: r tan 1 The choice of coordinate sstem is a matter of convenience since both sstems are equivalent. However, it is often the case, that (depending on the particulars of a problem or situation), one sstem is preferred over the other due to the resulting simplifications in the problem or because the problem is given or stated using a particular sstem.

25 Math 63 Calculus III 4 In the case where the preferred sstem is the polar sstem, we would like to be able to work entirel in that sstem. In particular, if we need to perform an integration we would like to be able to integrate in polar coordinates as opposed to converting to Cartesian coordinates, integrating, and then converting back to polar coordinates. Now in Cartesian coordinates, the elemental area A is given b i.e. a small rectangle of length and height. In polar coordinates, the elemental area must be epressed in terms of r and. However the shape of this area is not rectangular, but annular, as shown in Figure 13. r r Figure 13. Elemental area in polar coordinates If we consider a small annular area in polar coordinates of etent r and, then if r and are small enough the annulus is approimatel rectangular. The width of this rectangle is r and its height is r. Therefore the area is: A = rr Thus in polar coordinates the integral over the region R is: R b fda f r, rdrd a

26 Math 63 Calculus III 5 Eample 4. Compute the area of a circle of radius 1 using Cartesian and polar coordinates. In Cartesian coordinates the region bounding the circle is given b the equation 1 with both and varing (not independentl) between -1 and 1. If we allow to var between -1 and 1 then can var between 1. Thus we have: A dd d d sin In polar coordinates r varies between 0 and 1 and between 0 and. Therefore: 1 1 r 1 1 rdrd d d

27 Math 63 Calculus III 6 Lecture 1 Sections 16.5 and 16.6 Integrals in Spherical & Clindrical Coordinates; Probabilit Clindrical and Spherical Coordinates As with polar coordinates (see previous lecture) in -space, clindrical and spherical coordinates provide two alternatives to Cartesian coordinates in 3- space. These coordinate sstems are equivalent to Cartesian coordinates but can provide significant simplifications in some problems especiall where distance and angle provide the primar or preferred method for describing the location of objects and points in 3-space. Perhaps the best eample of this the latitude and longitude coordinate sstem used for locating aircraft, ships, satellites, and other objects on or above the surface of the earth. Latitude and longitude are essentiall part of a spherical coordinate sstem used for identifing the locations of objects above or on the surface of the earth. Just as with polar coordinates, it is desirable to perform integrations directl in these coordinates as opposed to converting to Cartesian coordinates, integrating, and converting back to either clindrical or spherical coordinates. In order to do this we must epress the elemental volume V in clindrical or spherical coordinates so that we can integrate directl in these coordinate sstems. Before we do this, we describe both these coordinate sstems and their relationship to Cartesian coordinates. Figure 14 shows the location of a point p in both clindrical and Cartesian coordinates. In Cartesian coordinates the location of p is given b rectangular coordinates (,, z) which specif with distances from the origin along each of the three coordinate aes. In clindrical coordinates, the location is specified b given the distance in the -plane to the place directl below the point. The direction or angle measured from the -ais to this place is used to identif the location below the point in the -plane. Finall the Cartesian z- coordinate is used to identif the height, h, above the -plane where the point is located.

28 Math 63 Calculus III 7 z z (,, z) = (,, h) h Figure 14: Clindrical Coordinates From Figure 14 we see that the relationship between the clindrical and Cartesian coordinates is as follows: cos sin z h Using some basic trigonometr, we have: tan 1 h z

29 Math 63 Calculus III 8 In spherical coordinates the distance r from the origin to the point is used along with the azimuth angle and the elevation angle. The azimuth angle,, is measured from the -ais to the line connecting the origin to the place in the -plane directl below where the point is located. The elevation angle,, is measured from the z-ais along the line connecting the origin to the point. Figure 15 shows the location of a point in both spherical and Cartesian coordinates. z z r (,, z) = (r,, ) Figure 15: Spherical Coordinates From Figure 15 we see that the relationship between the spherical and Cartesian coordinates is as follows:

30 Math 63 Calculus III 9 rsin rsin z rcos cos sin Using some basic trigonometr, we have: r z tan 1 tan 1 z With a bit more trigonometr we can show that clindrical and spherical coordinates are related as follows: r sin h rcos Or: r h h 1 tan The elemental volume in clindrical coordinates is similar to the elemental area in polar coordinates. It resembles a pie-shaped wedge with the point removed. Alternativel we can view it as the annulus in polar coordinates with height added to make it three-dimensional. Figure 16 shows the elemental clindrical volume.

31 Math 63 Calculus III 30 z V = ( ) h h Figure 16: Clindrical Elemental Volume If, h, and are sufficientl small we can view the clindrical elemental volume as a bo of height h, length and width. The elemental volume is therefore V = ( ) h. The integral over the solid region W in clindrical coordinates is: W d b fdv f,, h d d dh c a The elemental volume in spherical coordinates is shown in Figure 17. In this case, two angular components define this small solid region. As in the previous cases if r,, and are small enough, then we can regard this volume as approimatel bo-shaped. The height of the bo is r, the width rsin(), and the length r. The elemental volume is therefore: V = r sin() r

32 Math 63 Calculus III 31 The integral over the solid region W in spherical coordinates is: W b a,, sin fdv f r r dr dd z r r V = (r sin()) r (r) r sin() r rsin() Figure 17: Spherical Elemental Volume Eample 1. Find the volume of a hemisphere of radius a. Use clindrical coordinates to perform the integration. Figure 18 shows the solid region of integration. From Figure 18 we see that the polar radius,, is a function of h, the distance from the -plane. The polar angle must sweep out a circle to cover the hemisphere.

33 Math 63 Calculus III 3 z h Figure 18: Hemisphere Volume For Integration Thus the limits of integration are: h 0 to a 0 to 0 to a h The volume is therefore: W fdv a a h a a h a dhd a h dhd 0 0 a 3 3 h a 1 ah d d 3 3 a a ddhd 1

34 Math 63 Calculus III 33 Eample. Find the volume of a hemisphere of radius a. Use spherical coordinates to perform the integration. In this case, spherical coordinates are the natural sstem to use. The limits of integration follow directl from the coordinate sstem: r 0 to a 0 to / to We therefore have: W a fdv sin 0 r drdd 0 a 3 3 sin d d sin a a cosd d 3 3 a a a dd ====================================================================== Probabilit Probabilit is the likelihood or chance that a particular event will occur. The word event is used in a broad and general sense to include just about an situation, eperience, or incident of interest. The onl restriction placed on an event is that it be random. That is, an specific single occurrence of the event is unpredictable. The probabilit, P, that an event, E, will occur is defined as: P(E) = N(E)/T where:

35 Math 63 Calculus III 34 N(E) - is the number of was event E can occur; T - is the total number of possible outcomes (of all possible events of a certain tpe or class and which includes the event E). Classical eamples of probabilities involve games of chance: coin-flipping, dice, cards, etc. Consider the following eamples: 1. Coin-flipping - the probabilit of a flipped coin coming up heads. Since there is onl one wa for this event to occur, N(E) = 1. There are eactl two possible outcomes: heads or tails. Thus P(E) = ½.. Dice the probabilit of rolling a one or a two. In this case, there are eactl two possible was for this event to occur i.e. rolling a one or rolling a two. There are si possible outcomes (rolling a one, two, three, four, five, or si). Thus P(E) = /6 =. 3. Cards the probabilit of drawing a face card (Jack, Queen, or King). Since there are four suits (heart, diamonds, spades, clubs) and 3 face cards in each suit, there are 1 was of drawing a face card. The total number of possible outcomes is 5 (the number of cards in a standard deck), therefore P(E) = 1/5 = 3/13. Although there is great interest in determining the likel outcomes for games of chance (think Las Vegas), probabilit is used widel across man fields: phsics, medicine, insurance, and finance, to name just a few. The medical and insurance industries for eample, collect data on disease (cardiovascular, cancer, etc) and mortalit (due to accidents, suicide, murder, etc) in the general population and use this information to predict the likel outcomes of these events e.g. having a heart attack, being in an automobile accident, etc. The medical industr ma use this information to concentrate its efforts on development of new drugs, education, or treatment of disease in the general population, while the insurance industr ma use this information to set polic rates. Often data is not just collected for the population in general, but is categorized b one or more parameters that characterize the population. For eample, the insurance industr ma collect data on the number of automobile accidents. These ma be categorized b driver age, gender,

36 Math 63 Calculus III 35 income, and the seriousness of the accident i.e. fatal, serious injuries, minor injuries, no injuries. B doing this, the insurance industr can estimate the likelihood that sa, a 16 old male driver will be involved in a fatal accident versus the likelihood that a 35 ear old woman will be in such an accident. It can then set rates accordingl. The table below gives accident data for 00 categorized b driver age and Blood Alcohol Content (BAC). Alcohol and Accidents* Driver Age BAC < 0.08 g/dl BAC > 0.08 g/dl %.4% % 3.7% % 5.7% % 5.0% % 4.1% 64 > 10.5% 0.6% *Data taken from Traffic Safet Facts 00 Alcohol, US Department of Transportation National Highwa Safet Administration, Figure 19 show a histogram plot of the data in the table. It should be noted that the sum of the data in the table is 100% and it therefore includes all fatal accidents for the ear 00.

37 Math 63 Calculus III 36 Figure 19. Histogram plot of Accident data If we view the boes of the histogram in Figure 19 as the boes in a Riemann sum, with = age difference, = BAC difference, and p(, ) = percent accidents for age and BAC, then the histogram is an approimation to the volume of accidents in 00. In fact, we could think of collecting this data with at finer granularit, sa partitioning the ages into one ear intervals and BAC into units of 0.01 g/dl. If we did this, then the boes in Figure 19 would have smaller footprints in the -plane i.e. A = would be smaller and there would be man more boes in the graph. We could theoreticall keep decreasing the granularit b partitioning the ages b months, then das and the BAC at thousandths and then ten-thousandths g/dl. The result would be a closer and closer approimation to a smooth and continuous function p(, ) giving the percent of accidents for each age and BAC. In the limit where and become continuous, p(, ) would be a continuous and smooth function giving the percent of accidents of a particular age and BAC. However since the total number of accidents must remain at 100%, the integral (instead of the sum, now) of p(, ) over all and must still be 100%. We also note that p(, ) 0 for all values of and. That is, p(, ) cannot be negative since it makes no sense to have, sa, -1% accidents at a particular age and BAC. Functions that are non-negative everwhere and whose integrals over their domains equal one (or 100%) are called probabilit densit functions. That is, such functions can be used to represent or model the likelihood that an event (defined b one or more parameters i.e. variables) will occur. If probabilit densit is a function of two or more variables, this it is called a joint densit function. Formall we make the following definition: A function p(, ) is called a joint densit function of and if: p, dd 1 and p(, ) 0 for all and We note that the ± limits on the integral represent (i.e. are placeholders for) the valid ranges for the variables and. These ma (or ma not) be infinit.

38 Math 63 Calculus III 37 The product p(,) is the probabilit that and will occur. For eample in the table above, the probabilit that an accident will involve a driver between the ages of 45 to 64 with a BAC greater than 0.08, is 4.1%. In this case, = 45, = 0.08, = = 19, = 0.08, and p(, ) = 1/( ) = 16 %/(ear-g/dl). In terms of the integral, the limits determine the probabilit for a range of events. Thus for a generic densit function, we write: The probabilit that is between a and b and is between c and d is: d b P a b, c d p, dd c a In the accident data, we would write that the probabilit that an accident involved a driver between the ages of 45 and 64 with a BAC between 0.08 and 1.0 is: P 45 64, p, dd Assuming that p(, ) could be deduced from the data and represented as a continuous, smooth function we could compute the integral and obtain this probabilit. In practice, a continuous and smooth function is usuall fitted to the data since it is not practical to measure or obtain data at etremel fine granularit. If the data is of sufficient qualit, it will tpicall be well represented b a continuous and smooth function. Eample 3. Let p(, ) = + on the region 0 1, 0 1. Verif that p is a probabilit densit function. To show that p is a probabilit densit function we must show that p, dd 1 and p(, ) 0 for all and. Since 0 and 0, then + 0 and hence p(, ) 0.

39 Math 63 Calculus III p, dd dd d d Thus p is a probabilit densit function. Eample 4. Find the probabilit that ½ and the probabilit that ½ and ½, for the densit function of Eample P dd d d d P, dd d Eample 5. A machine factor produces components 10 cm long and 5 cm in diameter. To be useable i.e. pass qualit control, the length and diameter must be within a 0.1 cm tolerance of these values. The probabilit densit function p, e e gives the likelihood that a component of length and diameter will result as a variation in the production process. What is the probabilit that a component is useable? To be useable, the component must have length between 9.9 and 10.1 cm and diameter between 4.9 and 5.1 cm. Therefore: P , e e d d

40 Math 63 Calculus III 39 Note: the integral cannot be solved in closed form. The result was found b using tabulated numerical results. The probabilit that the component will be useable is approimatel 57.5%.

41 Math 63 Calculus III 40 Lecture 13 Sections 16.7 and 17.1 Change of Variables & Parameterized Curves Change of Variable In single variable calculus, it is often the case that changing the variable of integration will simplif the evaluation of an integral. In fact, we have alread looked at changing variables during integration when we discussed polar, clindrical, and spherical coordinates. Instead of viewing integration in, sa spherical coordinates, as a change in variable we looked at it as a change in coordinates. We can also view this change in coordinates as a change in the variables of the integration. The two views are equivalent, since changing the variable of integration is equivalent to a change in coordinate sstem and vise versa. Now in single variable calculus, changing the variable of integration involved three steps: Substituting the new variable into the function being integrated; Computing the new differential and making the appropriate substitution into the integral; Changing the limits of integration to reflect the equivalent limits in terms of the new variable. This procedure is essentiall identical for integrals involving multivariable functions, ecept that the second step, computing the new differential, is a bit more involved since there are multiple differentials and associated variables in the integral. Let s look at a single variable eample to illustrate this procedure. Suppose we wish evaluate the following integral: cos d 0 This integral can be readil evaluated b changing variables. Let t = +. The first step is to substitute the new variable into the integral in the function cos( + ). Thus we have:

42 Math 63 Calculus III 41 cos d cos t d 0 0 The net step is to compute the differential dt in terms of d and substitute. Since t = +, we have dt = d. Making the substitution we have: cos d cos t d cos t dt Finall we need to change the limits of integration to reflect the new variable. When = 0, t = and when = (/) ½, t = 3/. Thus we have: cos d cos t dt 0 0 The integral can now be evaluated and we obtain: cos d cos t dt sin t We can generalize this change of variable for an function as follows. Let f() be function we wish to integrate on the interval [a, b] and let = (t) be a change in variable we wish to make for. Then the differential is d d dt and we have: dt b a tb d f d f t dt ta dt Note that the change of variable introduces the derivative of with respect to t into the integral as part of the substitution for the differential, d. In the case of the eample above, the function t t, takes the interval [0, (/) ½ ] and maps it into another interval [0, 3/]. However the mapped interval is now curved instead of linear. In essence, the change of variable has deformed the original interval into a new interval and is characterized b the mapping defined b the change of variable, that is in the case of our

43 Math 63 Calculus III 4 eample, the mapping t t. We can also view a change of variable as a change in coordinates. If we recall the transformation from Cartesian coordinates to polar coordinates, we encountered eactl this tpe of deformation as our elemental rectangular area A = was deformed into the annulus rr. Let s take a closer look at this process. Consider the function f(, ) = + over the region, R, bounded b the and aes and the circle + = 16. We are in essence integrating over the quarter circle in the first quadrant of the Cartesian coordinate sstem, as shown in Figure = 16 R 4 Figure 0: Region of integration in Cartesian plane Our integral in Cartesian coordinates is: R 4 16 da dd 0 0 If we decide to use polar coordinates instead, then we get: R cos sin da dd r r rdrd In the r-plane the region of integration is rectangular, as shown in Figure 1.

44 Math 63 Calculus III 43 / / Figure 1: Region of integration in Polar plane 4 r Thus b changing coordinates we have deformed the rectangular region of integration into a quarter circular region (see Figure ). The change of coordinates from Cartesian to polar defines a mapping of a region R in the Cartesian plane to a region T in the polar plane and vise versa. We therefore suspect that we can also view the change in coordinates from Cartesian to polar as a change in variable. (r, ) =rcos() (r, ) =rsin() / / 4 + = 16 R 4 r 4 Figure : Mapping from Cartesian to Polar Coordinates

45 Math 63 Calculus III 44 To confirm our suspicions let s eamine this transformation as a change in variable. We have: = (r, ) = rcos() and = (r, ) = rsin() For a small region in the r-plane, sa a rectangle of width r and height, this region will map into the Cartesian plane according the transformations (r, ) and (r, ). Let us consider the mapping of the four corner of this region: (r, ) ( rcos(), rsin() ) (r+ r, ) ( (r + r) cos(), (r + r )sin() ) (r+ r, + ) ( (r + r) cos(+), (r + r ) sin(+) ) (r, + ) ( rcos(+), rsin(+) ) The vector along the bottom edge of this region is: (r, 0) (r+ r, ) - (r, ) = ( (r + r) cos(), (r + r )sin() ) - ( rcos(), rsin() ) = ( r cos(), r sin() ) And the vector along the left edge of this region is: (0, ) (r, + ) - (r, ) = ( rcos(+), rsin(+) ) - ( rcos(), rsin() ) = ( r (cos(+) - cos()), r(sin(+) - sin()) ) = ( r (cos(+) - cos())/, r (sin(+) - sin())/ ) = ( -r sin(), r cos() ) The area of this region in the -plane is: (r cos(), r sin()) (-r sin(), rcos()) = rr =

46 Math 63 Calculus III 45 Where we have approimated the region in the -plane as a parallelogram with sides defined b the vectors in the cross product above. Thus we have that: f(,) f( (r, ),(r, ) ) rr = f(rcos(), rsin() ) ) rr and hence our integral becomes: R,, f dd f r rdrd T Thus we can regard the change from Cartesian to polar coordinates as a change in variable from and to r and. In general, we would like to be able to change the variables and to some arbitrar variables, sa u and v: = (u, v) and = (u, v) and do so using a procedure similar to that for the single variable case. We will do this b repeating the above argument for the general case. Consider a rectangular region in the uv-plane corresponding to some, in general, nonrectangular region in the -plane, similar to what we found for polar coordinates. If we consider a small region of the uv-plane, sa a rectangle of width u and height v, then the corners of these regions will be as shown in Figure 3, where the region in the -plane is approimated b a parallelogram. The parallelogram will be a good approimation if u and u are small enough and our functions are continuous and smooth. (u, v) (u,v) v v T u v u ( (u,v+v), (u,v+v) ) b ( (u,v), (u,v) ) a R ( (u+u,v+v), (u+u,v+v) ) ( (u+u,v), (u+u,v) ) u Figure 3: Mapping resulting from general variable change

47 Math 63 Calculus III 46 Now in the uv-pane the area of the elemental region is just A = uv. In the -plane, the area of the elemental region is the area of the parallelogram. This area (from previous lectures) is just a b where a and b are the displaced vectors defining the sides of the elemental parallelogram region. Now: a uu v u v i u u v u v ˆj,, ˆ,, uu, v u, v u u, v u, v uiˆ u ˆ ui uj ˆ u u u b uvv uv i uvv uv ˆj,, ˆ,, u, v v u, v uv, v uv, viˆ v ˆ vi vj ˆ v v v uj ˆ vj ˆ Thus: A ab u v u v u v u v uv u v u v u v, uv uv uv, u v

48 Math 63 Calculus III 47 u v We have introduced some matri notation here, with being the u v determinant of matri formed b the partials. In addition, the short-hand, notation for this determinant is and is called the Jacobian. uv, Thus under a change of variable the elemental area A is transformed a follows: u v, A uv uv uv, u v If we make these substitutions into our integral then: Carl Gustav Jacob Jacobi ( ) R, f (, ) dd f u, v, u, v dudv T uv, To convert an integral from, to u, v coordinates/variables we make the following three changes: Substitute for and in the integral in terms of u and v; Compute the Jacobian and use its absolute value to make the appropriate change in the differential area i.e. dd, uv, dudv Change the region of integration to reflect the equivalent limits for the new variables We note that this procedure is essentiall identical to that of the single variable case, ecept the change in the differential now involves the Jacobian instead of just the derivative of old variable with respect to the new.

49 Math 63 Calculus III 48 Finall we note that this eact same procedure applies to integrals involving three variables ecept that the Jacobian is given b: u v w z,, uvw,, u v w z z z u v w Eample 1. Compute the Jacobian for polar coordinates., r cos r sin rcos rsin r, sin r cos r r Eample. Compute the Jacobian for spherical coordinates. r cos sin sin sin cos cos z,, sin sin cos sin sin cos r,, r z z z r r r r r cos 0 r sin r r r r r r in cos 0 r sin cos 0 cos sin sin sin cos cos cos sin sin sin sin sin cos sin sin cos sin sin cos s r cos sin r cos sin sin r sin sin r cos sin cos r r 3 3 r cos sin sin sin sin Eample 3. Find the volume of the ellipsoid z 1. a b c Let = au, = bv, and z = cw. Then the Jacobian becomes:

50 Math 63 Calculus III 49 u v w z,, uvw,, u v w z z z u v w a b 0 abc 0 0 c Thus: dddz abc du dv dw W S Where W is the solid region defined b the ellipse and S is the solid region defined b the sphere since in uvw-coordinates we have u + v + w = 1. Thus 4 we have dddz W abc du dv dw abc du dv dw abc S S 3. ====================================================================== Parameterized Curves Consider a smooth and continuous function of a single variable, f(). If we consider the graph of this function = f(), then we will obtain a smooth and continuous curve in the -plane, as shown in Figure 4. (, ) Figure 4. Point moving on a curve in the -plane If we consider an arbitrar point (, ) on the curve, then we can view the curve (graph of f(, )) as the path that this point will take if it were to

51 Math 63 Calculus III 50 move subject to the constraint that it sta on the curve. In essence, the curve is the path this point will follow or trace out if = f(). An eample of this is car on a road. The car is our point and road is the curve or path that the car takes as it moves. Now with the car, we are not particularl interested in knowing what -coordinate the car is at for a given - coordinate, rather we could like to know where the car is i.e. both and coordinates at a particular time, t. Mathematicall, both and are functions of the time t, that is = (t) and = (t). As the car moves, it traces out a path defined b the road, and at a particular time, t, it is located at ((t), (t)). To make our eample a little more concrete, sa ou leave school at noon and start driving home. Five minutes later ou stop in front of dr cleaners to pick up some clothes. If we consider t = 0 to correspond to noon, when ou left school, then at t = 5 ((5), (5)) will be the location on the road where the dr cleaners is located. As ou continue on our wa, we could pole our location ever five minutes to see where ou are: passing the grocer store, the park, the public librar, etc. Each of these places lies on the curve that defines that path ou taking, as shown in Figure 5. ( (5), (5) ) ( (0), (0) ) ( (5), (5) ) Store Dr Cleaners ( (15), (15) ) Park ( (10), (10) ) Home Librar ( (0), (0) ) School Figure 5. Path from school to Home Note that the points all still lie on the curve defined b our function = f() i.e. the road. However, what we are interested in is not so much the relationship between and, but rather the path ((t), (t)) that a point on this curve is following. In the case of the road eample, the time is the independent variable and and are the dependant variables. In effect, the time is the parameter we wish to var or input, and the location ( and ) is the desired output e.g. where ou are (or where ou will be) at time t.

52 Math 63 Calculus III 51 A function viewed in this manner as a path dependent on a parameter (or parameters) is called parameterized curve. Of course, we do not need to restrict ourselves to functions of a single variable and paths or curves need not be confined to -space. In fact, in our road eample, unless the ground under the road is perfectl flat, the road will rise and fall with the terrain, in effect defining a path in 3-space. Think about a road winding though the mountains. In fact, it is often the case, that paths or curves in 3-pace are completel defined b their parameterization. That is, the function relating,, and z together is not particularl interesting. Rather the parameterized curve that defines the path in 3-space (or even -space) is what is primaril of interest. B wa of eample, consider the parameterized curve defined b (t) = cos(t), (t) = sin(t), and z(t) = t. This curve is known as a heli and is show in Figure 6. A point moving on this curve follows a rising circular path around the z-ais. Figure 6: Heli Perhaps the simplest curve in both two and three space is a line. What is the parameterized equation for a line in 3-space? Well consider at point

53 Math 63 Calculus III 5 p (,, z) passing through or starting at a point p,, z o o o o and traveling in a straight line. To travel in a straight line, the point must move in some fied direction through and wa from p o. Now a fied direction is given b a vector, v which defines an orientation relative to a coordinate sstem. A point moving in this direction will move a distance equal to v in some unit of time. If it starts at p o then at time t, the point will be located at: p p v t at time t later. Thus we have the following result: o The parametric equation of a line through the point po o, o, zo parallel to the vector (or in the direction of the vector) v abc,, p p vt at, bt, z ct o o o o and is: Or (t) = o + at, (t) = o + bt, z(t) = z o + ct Eample 4. Find the parameterized equation of the line through the points (1,, -1) and (3, 3, 4). We use the points to compute the displaced vector giving the direction between the points: v 3,3, 4 1,, 1,1,5 The parameterized equation of the line is: (t) = t + 1, (t) = t +, z(t) = 5t 1 Now let s consider the general parameterized curve. An point in the Cartesian plane is defined b its position vector: r iˆ j ˆzkˆ If,, and z are functions of a parameter t, then we have:

54 Math 63 Calculus III 53 = f(t), = g(t), and z = h(t) and we can write these as a single parametric equation in vector form called a parameterization: As the parameter t varies, the vector r t f t i g t j h t k ˆ ˆ ˆ rt traces out a path or curve in 3- r t p vt. Thus: space. For eample the parametric equation of a line is o The line through the point po o, o, zo v abc,, is: r t p vt o in the direction of the vector We can use the parametric equations for a curve to find where a curve intersects a surface or another curve. Consider the following eamples. Eample 5. Find the places where the line = t, = t, z = t + 1, intersects the surface of the sphere + + z = 100. We are looking for the values of t where t + 4t + (t+1) = 100. Or: t + 4t + t + t + 1 = t + t 99 = t Thus the points of intersection are t -4.3 and t Eample 6. Do the lines = t - 1, = t + 1, z = 5 t and = t +, = t + 4, z = t + 3 intersect? For the lines to intersect we must find t 1 and t such that:

55 Math 63 Calculus III 54 t 1-1 = t + t = t t 1 = t + 3 Multipling the first equation b and subtracting from the second we get: 3 = -3 t or t = -1. If we multipl the second equation b and subtract from the first we get -3t 1-3 = -6 or t 1 = 1. For these lines to intersect, these values must also satisf the third equation: 5 t 1 = t Thus the lines do not intersect.

56 Math 63 Calculus III 55 Lecture 14 Sections 17. and 17.3 Motion, Velocit, Acceleration & Vecto r Fields Motion, Velocit, and Acceleration In our first lecture we talked about velocit and acceleration and showed that the were vectors. That is, the both had a magnitude and a direction associated with them. In particular the velocit of a moving object e.g. a particle has the following properties: The velocit of a moving object is a vector v such that: The magnitude of v is the speed of the object; The direction of v is the direction of motion; The speed of the object is v, and the velocit is tangent to the object s path Phsicall, the velocit is defined as the derivative of the position with respect to time. Since position (as we also noted in the first lecture) is a vector, then the velocit will also be a vector. Recalling from the previous lecture that the position vector r iˆ j ˆzkˆ we have: The velocit of an object is the derivative of the position vector with respect to time: dr d ˆ d ˆ dz v i j kˆ v ˆ ˆ ˆ i v j vzk dt dt dt dt Where v, v, and v z are the components of the velocit vector and are equal to time derivative of the components of the position vector.

57 Math 63 Calculus III 56 Eample 1. The position of a particle moving in a circle of radius R is given b r t R cos t iˆ Rsin t ˆj where is the angular velocit (how fast the particle is rotating in radians per second). That is, the period (how long it takes the particle to make one revolution is. Find the velocit of the particle, its speed, and show that it is perpendicular to the position. The velocit is : dr d d v Rcos t i R dt dt dt Rsin t i R t ˆj ˆ sin ˆ cos t ˆj. The speed of the particle is: v Rsin t i R cos t j R sin t R cos t R ˆ ˆ We note that: sin ˆ cos ˆ cos ˆ sin R sin tcost R sin tcost 0 vr R t i R t j R t i R t ˆj Thus the velocit and the position are perpendicular. As we noted above, the velocit is tangent to the position vector. If the position vector is given or known as a parameterized path or curve, then we can compute the tangent to the path at an point using the velocit. Eample. Let r t t iˆt ˆj tk. Find the tangent line at (1, 1, ). 3 ˆ First we note that the point (1, 1, ) lies on the path defined b particular r 1 (1, 1, ). rt. In To find the tangent line we first compute the velocit:

58 Math 63 Calculus III 57 d v t r t ti t j k dt ˆ3 ˆ ˆ At t = 1 we have: v 1 iˆ3 ˆj kˆ. Now the tangent line is of the form: ˆ ˆ ˆ ˆ ˆ ˆ o o o 3 i j z k i j k t and at t = 1 we want: Therefore: ˆ ˆ o o o iˆ ˆj z k iˆ 3ˆj k iˆ ˆj kˆ iˆ ˆj z kˆ iˆ ˆj kˆ iˆ 3 ˆj kˆ iˆ ˆj o o o Therefore the equation of the tangent line is: i ˆ 3 ˆ jk ˆ t i ˆ ˆ j The acceleration is defined as the derivative of the velocit with respect to time and is equal to the second derivative of the position with respect to time. Since velocit (as we also noted above) is a vector, then the acceleration is a vector. Thus we have: The acceleration, respect to time: a, of an object is the derivative of the velocit with dv dv dv ˆ ˆ dvz a i j kˆ a ˆ ˆ ˆ i a j azk dt dt dt dt Where a, a, and a z are the components of the acceleration vector and are equal to time derivative of the components of the velocit vector. Since the velocit is the derivative of the position we also have: a iˆ ˆj kˆ iˆ ˆj dt dt dt dt dt dt dt dt dv d d d dz d d d z k ˆ

59 Math 63 Calculus III 58 Eample 3. Show that the acceleration is parallel to the position and perpendicular to the velocit for a particle moving in a circle. Compute the magnitude of the acceleration. From Eample 1 we have: r t R t i R t ˆj cos ˆ sin v Rsin t i R t ˆ cos ˆj The acceleration is: dv d a R t i R t j R t i R dt dt Rcostiˆ R sin t ˆj r sin ˆ cos ˆ cos ˆ sin t Since the acceleration is a multiple of the position, these vectors are parallel. Note the minus sign in the above epression means that the acceleration points in the opposite direction of r, that is towards the center of the circle. Now: av rv 0 (using the results from Eample 1). Thus the acceleration and the velocit are perpendicular. The magnitude of the acceleration is: ˆj ˆ ˆ cos sin cos sin t a R t i R t j R t R R v R R R

60 Math 63 Calculus III 59 Combining the results of Eamples 1 and 3 we have: A particle moving continuall in a circle is said to be in uniform circular motion which is describe b: r t R t i R t ˆj cos ˆ sin The radius of the circle is R and the period of the motion is ; The velocit is tangent to the circle and the speed v is constant and equal to R ; v The acceleration points towards the center of the circle with a ; R The velocit is perpendicular to both the position and the acceleration. In uniform circular motion the magnitude of the velocit is constant but its direction is alwas changing during the motion. For uniform linear motion (i.e. motion in a straight line), the magnitude of the velocit is changing but its direction is constant. The acceleration is parallel to the velocit vector (versus perpendicular in uniform circular motion) and points in the same direction if the particle is speeding up and in the opposite direction if the particle is slowing down ˆ Eample 4. Let 1ˆ 3 ˆ acceleration. r t t t i t t j t t k. Compute velocit, d 43 1ˆ 33 1 ˆ 3 1 ˆ 3 14ˆ 3 ˆ kˆ v t r t t i t j t k t i j dt d a t v t 6t 4i ˆ 3 ˆj kˆ dt We note that the acceleration and velocit vectors are parallel since the are both multiples of the same vector: 4iˆ 3ˆj kˆ. For t < 0, the acceleration is negative, so the object is slowing down and for t > 0 the acceleration is positive and the object is speeding up.

61 Math 63 Calculus III 60 We also note that: ˆ 3 ˆ r t t t i t t j t t kˆ 3 ˆ t t 4iˆ 3ˆj k iˆ 6 ˆj So that rt is also a multiple of 4iˆ 3ˆj kˆ. Thus the motion is linear. A particle/object whose motion is described b: r t r f t v o o is in uniform linear motion i.e. moving in a straight line. This motion has the following characteristics: The motion is along a straight line through the point r o and parallel to v ; The velocit and acceleration are parallel to each other and the vector v From the definition of velocit we have: dr v d iˆ d ˆj dz kˆ dt dt dt dt And the speed is therefore: v d d dz dt dt dt The distance, d, of the particle along this curve is the integral of the speed: d t t1 v t dt Since the distance along he curve is equal to the length of that segment or part of the curve, we use this formula to give us the length of an part of the curve or even the entire curve.

62 Math 63 Calculus III 61 If the curve C is given parametricall for a t b b smooth functions and if the velocit v is not zero for a t b, then the length L of C is: L b a v t dt Eample 5. Find the circumference of the ellipse (t) = cos(t), (t) = sin(t), 0 t. We compute the velocit: v (t) = -sin(t) v (t) = cost(t) o L t t dt 4sin cos 9.96 Note: the integral needs to be evaluated numericall. ====================================================================== Vector Fields Up to this point we have discussed vectors and functions of one, two, and three variables. Although the functions we have presented thus far have contained two or three variables, the have been scalar functions. That is, the have taken a point in -space (, ) or 3-space (,, z) and converted or mapped these three values to a single (or scalar) value. Mathematicall we write: f : or f : or f : 3 where stands for the collection or set of all real numbers and is the 3 collection of all ordered pairs of real numbers (Cartesian plane) and is the collection of all ordered triples of real numbers (Cartesian 3-space). The notation f : means that f is a mapping or function that takes a real number and maps it to another real number i.e. f is a function of a single

63 Math 63 Calculus III 6 variable. f : i.e. f is a function of two variables; and means that f maps a pair of numbers to a single number f : 3 means f maps a triple of numbers to a single number i.e. f is a function of three variables. We need not however confine our discussion to scalar functions; instead we can look at functions of the form: 3 f : 3 or f : or f : 3 Such functions are known as vector-valued functions. In the case where f assigns a vector to each point in 3-space we call f a vector field. In fact, we have alread encountered eamples of vector-valued functions and vector fields, although we have not called them as such. The gradient f of a scalar function is an eample of a vector field. If f is a scalar function of two variables then: f : 3 That is f maps a pair of values ( and ) into a vector. Another eample of a vector-valued function or vector field is a parameterized curve in 3-space. In this case we have: C : 3 That is C(t) takes a single value, t, and maps it to a vector. An eample of this was the heli: C t t i t j tk cos ˆsin ˆ ˆ An eample of a phsical vector field is the ocean current known as the Gulf Stream (See Figure 7). The Gulf Stream is an eample of a velocit vector field. That is, at each point in the ocean, there is a vector giving the velocit of the ocean current at that point. Vector fields are drawn b placing an arrow at each point on grid or map showing the direction and strength of the field at that point. In the case of Figure 7, the strength of the current i.e.

64 Math 63 Calculus III 63 its speed is indicated b the length of the arrow and its direction b the orientation of the arrow. Figure 7: Gulf Stream. Beginning in the Caribbean and ending in the northern North Atlantic, the Gulf Stream Sstem is one of the world's most intensel studied current sstems. This current plas an important role in the poleward transfer of heat and salt and serves to warm the European subcontinent. Image and tet taken from: Another eample of a vector field is a Force Field. A force field is a vector field giving the force (magnitude and direction) at ever point in space. An eample of a force field is the earth s gravitational field. That is, the force eerted on another bod at each point in the space surrounding the earth. It is the gravitational field of the earth that keeps the moon and man-made satellites in orbit about the earth. Figure 8 depicts the earth s gravitational field which shows that the force lines are directed towards the center of the earth. The length of the arrows get shorter as the distance from the center of the earth increases indicating that the force field gets weaker.

65 Math 63 Calculus III 64 Figure 8. Earth s Gravitational (Force) Field. Another eample of a force field is the earth s magnetic field, shown in Figure 9. In this case, the vectors point around the earth starting at the south-pole towards the north-pole. Figure 9. Earth s magnetic field. Note: images taken from With these eamples we formall define a vector field as follows:

66 Math 63 Calculus III 65 A vector field in -space is a function F, whose value at a given point (, ) is a 3-dimensional vector. Similarl, a vector field in 3-space is a function F,, z whose values are 3-dimesional vectors. The arrow over the function F indicates that its value is a vector, not a scalar. We can also represent the point (, ) or (,, z) as a vector r and write F r. Vector fields can be visualized b sampling the vector function at various points and drawing arrows in the appropriate directions. The length of the arrows is proportional to the magnitude of the vector function at the point. F, i j. Eample 6. Sketch the vector field ˆ ˆ In this case the magnitude of the vector field is F, and we note that F, is perpendicular to the position vector r iˆ j ˆ i.e. F, r iˆ j ˆ iˆ j ˆ 0. The vector field will therefore look like a spiral of arrows centered around the origin with the length of the arrows increasing as the distance from the origin increases, as shown in Figure 30. F, i j Figure 30: Vector field ˆ ˆ

67 Math 63 Calculus III 66 F, j Eample 7. Sketch the vector fields ˆ G, i. and ˆ The vector ˆ j is parallel to the -direction, pointing up when is positive and down when is negative. The larger the is the longer the vector. The vectors in the field are constant along vertical lines since the vector field does not depend on. See Figure 31. F, j Figure 31: Vector field ˆ The vector ˆ i is parallel to the -direction, pointing right when is positive and left when is negative. The larger the is the longer the vector. The magnitudes of vectors in the field are constant along horizontal lines since the vector field does not depend on. See Figure 3. G, i Figure 3: Vector field ˆ

68 Math 63 Calculus III 67 Newton s Law of Gravitation states that the magnitude of the gravitational force eerted b an object of mass M on an object of mass m is proportional to M and m and inversel proportional to the square of the distance between them. The direction of the force is from m to M along the line connecting them: GMm r GMm F z F r r rr r r,, ˆ where r rr r force is attractive. and rˆ r. Note that the minus sign means that the r

69 Math 63 Calculus III 68 Lecture 15 Sections 17.4 and 17.5 Flow and Parameterized Surfaces Flow of a vector Field Let s re-eamine to notion of a vector field as a velocit field. If we take a look at a velocit field, sa the Gulf Stream (as shown in Figure 33), then each point in the field gives the speed and direction of the flow of water in the Atlantic Ocean. Figure 33: Gulf Stream. Beginning in the Caribbean and ending in the northern North Atlantic, the Gulf Stream Sstem is one of the world's most intensel studied current sstems. This current plas an important role in the pole-ward transfer of heat and salt and serves to warm the European subcontinent. Image and tet taken from: If we look at the white area of Figure 33, it shows the flow off the east cost of the United States, etending from Florida up through Maine and out into the North Atlantic. We note that the arrows in this zone of the map are relativel long, indicating a strong or fast current and pointing in generall

70 Math 63 Calculus III 69 the same direction: north-east. Consider an object placed in the water off the coast of Florida sa an empt bottle or a raft. What can we sa about the path of this object? Well from the magnitude and direction of the velocit field our bottle or raft will be swept up the east coast of the United States and out into the north Atlantic. We epect the speed of the bottle or raft to be approimatel the speed of the current which is given b the velocit field. If G, is the Gulf Stream velocit field, and v is the velocit of our bottle or raft, then: v G, Where and are the position of the bottle or raft in the ocean. Since the bottle or raft will move with the flow of the ocean water,,, and v are functions of time. That is we could think of setting our bottle adrift in the water on Monda morning and then asking where is it on Tuesda morning, Wednesda morning, etc. Because the bottle will be carried along b the currents of the Gulf Stream, it position will change over time. In the case where the bottle offers no resistance to the water, its speed will match that of the water eactl and we will have: v t G t t, But the velocit is just time derivative of the position, r, with respect to time, so we have: dr t G t, t G r t dt Hence the velocit of the bottle is determined b its position in the velocit field and it will follow a path determined b the field. That is, it will follow a flow line or streamline. The white area in Figure 33 is an area of the Gulf Stream with a clear streamline or flow line where water is moving from the south near Florida to the north-east. We generalize this concept with the following definition: A flow line of a vector field vt G r t equals vt. Thus: is a path rt whose velocit

71 Math 63 Calculus III 70 dr t dt G r t The flow of a vector field is the famil of all its flow lines. Flow lines are also called streamlines or integral curves. The equation dr t Gr t is known as a differential equation (Math 75) since it dt relates the derivates of its variables to a function of those variables. That is it is an equation that contains both variables and their derivatives. For G, G i, ˆ G, ˆ j, we have: vector fields in -space, d t dt d t dt G t, t G t, t In general, differential equations are difficult to solve and numerical, G, are not too methods must be used. However if G and complicated, closed form solutions can be found. We consider some eamples. Eample 1. Let if (, ) = (1, ) at t= 0. In this case we have: G, 3i 4 ˆj, that is a constant vector field. Find and ˆ d t dt d t dt 3 4 So that:

72 Math 63 Calculus III 71 (t) = 3t + o and (t) = 4t + o. Since (0) = o = 1 and (0) = o =, we have: r t t i t ˆj (t) = 3t + 1 and (t) = 4t +. or 3 1 ˆ4 Eample. Let G, i j ˆ and let (0) = -, and (0) =. Find r t. ˆ In this case we have: d t dt d t dt 1 () t From the first equation we have: (t) = t + o. And since (0) = o = -, we have: (t) = t -. Putting this into the second equation we get: d t dt t Or (t) = ½t t + o. Since (0) = o =, we get: (t) = ½t t +. Thus: 1 r t t iˆ t t ˆj G, i j ˆ. Find Eample 3. Let ˆ rt.

73 Math 63 Calculus III 7 We have: d t dt d t dt () t t In this case we note that if we let (t) = acos(t) and (t) = asin(t), then these equations will be satisfied. Hence rt a costiˆa sin t ˆj. As we noted above, in general, the differential equation for the flow line will not have a closed form solution and will require numerical methods to determine a solution. In such cases, Euler s method is often used in a first attempt at finding a solution. Euler s method makes used of a local linearization to approimation to rt using its first derivative. That is dr t rtt r t t. In this case we note that since dt dr t Gr t we have: r tt r t tg r t. Starting dt at t = 0 we can construct a series of values for using this approimation. Hence we have: rt Leonhard Euler ro r0 r1 ro tg r r r tg r o 1 1 And so on. In general we can write the n th term as r n r n1tg r n1 which will give us the position at time nt. B graphing the points r, r, r, r we can plot the flow line. As usual, as we make t smaller and smaller, our approimation well be better and better. Although this method will work in principle, if the vector field G is a complicated function, more sophisticated o 1 n

74 Math 63 Calculus III 73 numerical techniques (beond the scope of this course) are required. For eamples and a more comprehensive discussion see, for eample: Numerical Recipes in C++: The Art of Scientific Computing b William H. Press, Saul A. Teukolsk, William T. Vetterling, Brian P. Flanner, Cambridge Universit Press, 00, ISBN: ====================================================================== Parameterized Surfaces Consider the heli again, as shown in Figure 34. We recall that the heli is a parameterized curve given b vector rt costiˆ sin t ˆj tkˆ. Although the heli eists in 3-space, it is a curve and is therefore one-dimensional. That is, we an think of unwinding or un-twisting the heli and straightening it out so that it is just a straight line, which is of course a one-dimensional object. Figure 34: The Heli

75 Math 63 Calculus III 74 Surfaces on the other hand are two-dimensional objects. Just like the heli, the ma eist in three dimensions but we can think of pounding them flat or stretching them out again so that the lie flat or in a plane, which is twodimensional. Unlike curves however, surfaces depend on two variables or parameters. Thus consider the curve: (t) = cos(t) (t) = sin(t) z(t) = 0 This is just a circle in the -plane. If we set z to some other constant value, sa 10, then we still have a circle ecept that this circle will be in a plane parallel to the -plane located 10 units above it. However if we were to allow z to var freel i.e. make it a parameter independent of t, sa s, then we will no longer have curve but rather a surface and this surface will be a clinder. Thus: (s, t) = cos(t) (s, t) = sin(t) z(s, t) = s are the parametric equations that define a clinder. The parameters s and t are independent of each other and as such can var independentl over their allowed ranges. In the case of the clinder we have: 0 t and - s. In general we can epress the points on a surface in terms of two parameters, s and t as follows: (s, t) = f 1 (s, t) (s, t) = f (s, t) z(s, t) = f 3 (s, t) As the values of s and t var over their allowed ranges, the point (,, z) will sweep out the surface. The functions f 1 (s, t), f (s, t), and f 3 (s, t) are called the parameterization of the surface. As with the curve, we can use the position vector r to define the location of all the points on the surface: r s t f st i f st j f st kˆ,, ˆ, ˆ, 1 3 We note that the graphs of functions of two variables are surfaces in 3- space. That is we let z = f(, ). In this case we can parameterize these

76 Math 63 Calculus III 75 surfaces b letting (s, t) = f 1 (s, t) = t, (s, t) = f (s, t) = s, and z(s, t) = f 3 (s, t) = f(s, t). Eample 4. Give the parameterization of the lower hemisphere + + z = 1. In this case we have where s t 1. z 1. We let = t, = s and z 1 t s Consider a plane containing two non-parallel vectors v 1 and v and a point p o. We can get to an point on the plane b starting at p o and moving parallel to v 1 and then to v. Since sv 1 is parallel to v 1 and tv is parallel to v we have: The parametric equations of a plane through the point non-parallel vectors v 1 and v is: r s t p sv tv, o 1 p o and containing the If p iˆ ˆj z kˆ, v ai ˆ a ˆ j a k, and o o o o ˆ v bi b j b k, then ˆ ˆ 1 ˆ 3 (s, t) = o + sa 1 + tb 1 (s, t) = o + sa + tb z(s, t) = z o + sa 3 + tb 3 Note that,, z are linear functions of s and t. Eample 5. Write the parametric equations for the plane through the point (, -1, 3) and containing the vectors v ˆ 1 iˆ 3ˆ j k and v ˆ iˆ4 ˆj 5k. Using the results above we have: (s, t) = + s + t (s, t) = s - 4t z(s, t) = 3 s + 5t

77 Math 63 Calculus III 76 ******************************************************************* In our discussion of spherical coordinates we showed that the distance r from the origin to a point along with the azimuth angle and the elevation angle were an alternative to Cartesian coordinates for locating points in 3- space. Figure 35 shows these angles and their relationship to an arbitrar point (,, z) in Cartesian 3-space. z z r (,, z) = (r,, ) Figure 35: Spherical Coordinates Figure 35 shows that the relationship between the spherical and Cartesian coordinates is as follows: rsin rsin z rcos cos sin We can use the angles and as alternative parameters to the Cartesianlike parameters s and t presented above when developing parametric equations for surfaces. The angle is the azimuth angle and can be

78 Math 63 Calculus III 77 identified with the longitude angle used for locating points on or near the surface of the earth. This angle ranges either between 0 and or equivalentl between and (the later is used for longitude). The elevation angle can be identified with the latitude angle. This angle ranges between 0 and. We thus have: 0 and 0. Points in the Northern hemisphere range between 0 and / i.e. 0 /. Points in the Southern hemisphere range between / and i.e. /. If a point is located at ( o, o ) then the point located on the other side of the sphere on the line through this point and the center is called the antipodal point. In azimuth this point is located at o +. In elevation, the antipodal point is located at - o. Eample 6. Find the parametric equations for the sphere with center at the point (, -1, 3) and having radius. The sphere with radius has the following parametric equations: = cos()sin() = sin()sin() z = cos() The center of this sphere is at (0, 0, 0). To shift it to (, -1, 3) we add this point to the parameterization above: = cos()sin() + = sin()sin() - 1 z = cos() + 3 ******************************************************************* The man cases, surfaces have an ais of rotational smmetr with circular cross sections perpendicular to that ais. These surfaces are referred to as surfaces of revolution. Consider the following case: what is the parameterization of the cone whose base is the circle of radius a in the plane and whose verte is height h above the -plane? In this case we note that the radius of the cone is a function of the height, s, above the -plane, as shown in Figure 36.

79 Math 63 Calculus III 78 h s r a Figure 36: Cross section for the cone In the -plane the radius of the cone is a and at height h above the plane, the radius is zero since we are at the verte of the cone. The triangle in Figure 36 is a cross section of half the cone. If we were to rotate this triangle about its vertical leg, we would sweep out the cone. The vertical or z-ais is therefore the rotational ais of smmetr. If we find the radius r as a function of the height parameter s and then rotate this radius as it moves between the -plane and the verte we will sweep out the cone. Using similar triangles we have: So that 1 s a r h h s r s a. Rotating about this z-ais is parametricall h equivalent to computing the and components of the parameterization b taking r(s) and multipling b the cosine and sine of the rotation angle, t. Thus we have: (s, t) = r(s)cos(t) (s, t) = r(s)sin(t) Finall the z-component of the parameterization is just s, the height above the -plane. Thus we get:

80 Math 63 Calculus III 79 (s, t) = a(1 s/h) cos(t) (s, t) = a(1 s/h) sin(t) z(s, t) = s where 0 s h and 0 t. This is the parameterization of the cone. Note the parameterization essentiall uses clindrical coordinates. ******************************************************************* Finall we talk about parameter curves. These are essentiall parameterized curves on a surface. That is, unlike a heli which is a parameterized curve in 3-space, a parameter curve eists in 3-space like the heli, but it is contained entirel on surface. On a parameterized surface, such a curve is obtained b setting one of the parameters equal to a constant and letting the other var. Thus if the surface is parameterized b: r s t f st i f st j f st kˆ,, ˆ, ˆ, 1 3 then there are two families of parameter curves on the surface, one famil with t constant and the other with s constant. Eample 7. For the vertical clinder (s, t) = cos(t) (s, t) = sin(t) z(s, t) = s describe the parameter curves with t constant and s constant. If t is constant, then (s, t) = o = constant and (s, t) = o = constant. Thus and are fied. We note that the constants o and o are such that 1 and therefore this point lies on the surface of the clinder. Since o o z(s, t) = s, then z will var up and down above the point (o, o ) on the surface of the clinder. This curve is therefore a vertical line on the surface, as are all the curves in this famil, as shown in Figure 37. If s is constant then we have: (s, t) = cos(t) (s, t) = sin(t) z(s, t) = constant. In this case, (s, t) and (s, t) define a circle on the surface of the clinder that is in a plane parallel the -plane and located at a height equal to the

81 Math 63 Calculus III 80 constant value of s above or below the -pane. This famil of curves in shown in Figure 38. z t = constant famil Figure 37: The constant t parameter curve famil on the clinder z s = constant famil Figure 38: The constant s parameter curve famil on the clinder

82 Math 63 Calculus III 81 Eample 8. Describe the families of parameter curves for the sphere with parameterized surface given b: (, ) = cos()sin() (, ) = sin()sin() z(, ) = cos(). In this case our parameters are and. If we let = constant then we have: (, ) = sin() (, ) = sin() z(, ) = cos() where and are constants such that = cos() and = sin(). We note that sin so that and z define a circle running from pole to pole in the plane that contains the z-ais and is defined b the constant angle. Figure 39 shows this famil of curves. On the surface of the earth, these would be lines of longitude. z = constant famil Figure 39: The constant parameter curve famil on the sphere If we now let = constant we get: (, ) = acos() (, ) = asin() z(, ) = b where a and be are constants. In this case an define a circle of radius a = sin() in a plane parallel to the -plane and at height b above or below the -plane. These are lines of constant latitude on the surface of the earth and are shown in Figure 40.