Multivariable Calculus MAS Dr Stephen Britton September 3
Introduction The aim of this course to introduce you to the basics of multivariable calculus. The early part is dedicated to discussing curves and conic sections in two dimensions, and then moves to three dimensions, covering lines, planes and quadric surfaces. After this, we will move on to discussing vector-valued functions and the curves they represent. You will learn how to use differentiation to find tangent lines, normal vectors, slopes and more. The final part of the course deals with multiple integrals, specifically double and triple integrals to find areas and volumes. These notes are based on the textbook Calculus, Late Transcendentals, th edition, by Howard Anton, Irl Bivens and Stephen Davis, and published by Wiley.
Curves and Conic Sections in Two Dimensions To begin with in this course, we will only consider two directions. Until section 3, we can assume that everything is described in the xy-plane.. Parametric Curves If we consider a generic curve C in the xy-plane, is there a way to parameterise the motion using a single variable? We define the parametric equations of the motion of a particle as x = f(t), y = g(t), () where t is the parameter which describes where we are on the curve. The curve itself is called the graph of the parametric equations, or alternatively, the trajectory of the particle. An example is given in Figure. Generally y 3 x,y 4 6 8 x C Figure : Trajectory C. speaking, t represents time in this course, but it could be any parameter that makes sense in a given problem. Often, t will be restricted to an interval [a, b], but if no interval is given, assume that the range is (, ). This will be written x = f(t), y = g(t), (a t b). () Problems will often ask you to sketch the graph of parametric equations. To do this, eliminate t. The resulting equation involving x and y should be something you can recognize.
Example: Sketch the graph of the equations x = cos t, y = sin t, ( t π). Solution: The easiest way to understand what this looks like is to notice that we can eliminate t by the doing the following x + y = ( cos t) + ( sin t) = 4(cos t + sin t) = 4. (3) This is clearly the equation of a circle of radius. To see what portion of the circle is graphed, lets look at a few values: t (x, y) t = : (, ) t = π : (, ) t = π : (, ) t = 3π : (, ) t = π : (, ) In other words, we have an entire circle, as is sketched in Figure. Example: Sketch the graph of the equations x = t, y = t + 3, ( t π). Solution: To eliminate t, solve both equations for t, i.e. t = x +, t = (y 3). We then equate these and simplify to get one equation. x + = (y 3) x + = y 3 y = x + 5. 3
x,y y t x Figure : Circle of radius. 5 5 4 4 x Figure 3: Plot of y = x + 5. We now can easily see that this is a line with slope which intersects the y-axis at y = 5. It is plotted in Figure 3. Next, we define the orientation of a graph to be the direction of increasing parameter. In the example of the line above, the orientation is up the slope as that is the direction of increasing t. A curve with orientation is called a parametric curve. For the example of the circle, we had an anti-clockwise orientation. If we replaced t with t, then x( t) = cos( t) = cos t, y( t) = sin( t) = sin t, which has clockwise orientation. 4
.. Expressing equations parametrically We can always describe the equation y = f(x) parametrically. The simplest way to do this is to take x = t and then y = f(t). Example: y = x +. Solution: x = t y = t +. Of course, for something like a circle you can of course do the same. However, there are often better parameterisations to chose. In the case of a circle of radius a, the parameterisation x = a cos t, y = a sin t is often convenient... Tangent lines for parametric curves Recall that the tangent line to a curve y = f(x) at a point P is the line that passes through P that has slope m = df(x). For a parametric curve, we dt can find the slope using the parameter t via the chain rule: There are three special types of tangent lines: dy dx = dy/dt dx/dt. (4) If dy/dt = and dx/dt, the slope is and we have a horizontal tangent line. If dx/dt = and dy/dt, the slope is infinite and we have a vertical tangent line. If dx/dt = and dy/dt =, the slope is indeterminate and we have a singular point. These require special treatment. Let s look at an example. Example: Find the tangent line of x = t, y = t for t =. Solution: Of course, we can see that this is equivalent to y = x, but let s do this using the new method. First, the point we are interested in is P = (, 4). We know that a line has an equation y = y + m(x x ), and we can find the slope of the tangent lines since dy dx = dy/dt dx/dt = t 5 = t. (5)
Therefore, at P m P = () = 4. (6) The equation of the tangent line is then This is sketched in Figure 4. y = 4 4(x ) = 4 4x. (7) y 4 4 x 5, 4 5 Figure 4: Tangent line to the function y = x...3 Arc Length of a Parametric Curve For a parametric curve the arc length is given by x = x(t), y = y(t), (a t b), (8) L = b a (dx ) + dt ( ) dy dt. (9) dt Example: Find the circumference of the circle x = a cos t, y = a sin t, ( t π). 6
Solution: We find L = π π ( a sin t) + (a cos t) dt = at dt = πa. If the curve doubles over, this will not work: in other words no point on the curve should be graphed by two values of t. In our example of the circle, if we integrated from to 3π we would find 3πa, which is obviously not the circumference, because the values between π and 3π repeat those between and π.. Polar Coordinates Up until now we have been dealing with the coordinates x and y, which are known as Cartesian coordinates or rectangular coordinates. We can equally describe a plane using any pair of coordinates so that each position on the plane is uniquely described by the pair. In particular a useful set of coordinates are polar coordinates... Defining Polar Coordinates To define polar coordinates, we begin by fixing a point O, called the origin or pole. We then define a half-line (or ray) which begins at and continues to infinity in a given direction, called the polar axis. This is shown in Θ r Origin O Θ Figure 5: Polar coordinates Figure 5. The distance from the origin to the point P is called the radial coordinate, r, and the angle that the line OP makes with the polar axis is called the angular coordinate, θ, more commonly known as the polar 7
angle. It is important to notice that the polar angle returns to its original position when the angle is π. Therefore, any angle greater than or equal to π is equivalent to an angle in the range ( θ π). In fact, more generally, the angles θ π n, θ and θ + π n are equivalent if n is an integer... Relationship to Cartesian Coordinates It is quite easy to make a link between polar coordinates and Cartesian coordinates. The trick is to make the polar axis coincide with the x-axis. Then, we can see from Figure 6 that the following relationship holds y P r,θ r y r sin Θ Θ O Figure 6: Polar coordinates in terms of Cartesian coordinates which can also be written in the form x = r cos θ, y = r sin θ, () r = x + y, tan θ = y x. () Change the Cartesian coordinates (4, 4 3) into polar coordi- Example: nates. Solution: r = 4 + (4 3) = 64 r = 8, 8
and tan θ = 4 3 = 3 θ = π 4 3. The polar coordinates are therefore (r, θ) = (8, π/3). Change the polar coordinates (3, 3π/4) into Cartesian coordi- Example: nates. Solution: x = 3 cos 3π 4 = 3, and y = 3 sin 3π 4 = 3, The Cartesian coordinates are therefore (x, y) = ( 3/, 3/ ). If you are asked to graph an equation in polar coordinates, the easiest thing Π..5 Π 4 Π..5.5..5 3Π. 4 3 r 3 3 4 Figure 7: Unit circle and the half-line θ = π 4. to do is to simply plot some points and see what happens. Later on, you may begin to recognise some of the graphs and be able to plot them from memory. For now, let s look at some simple examples. If we fix r, say to r =, we get a circle, see Figure 7. Note also that once we hit θ = π, the graph repeats, and so we do not need to continue. Also in the same figure we see the plot of θ = π/4. We can either imagine that we can take negative values of r, 9
or as we would normally expect we could impose r and plot a half-line, as is done here. Finally, let s look at an example where both coordinates vary, say r = sin θ. Then we get the plot in Figure 8, where we see that the plot repeats values after θ = π, because the Cartesian coordinates are x = sin t cos θ, y = sin θ sin θ = sin θ.. Π 3Π 4.5 Π 4 Θ or Θ Π..5.5. Figure 8: Plot of r = sin θ...3 Symmetry Tests Symmetry tests can be performed in polar coordinates by appropriate choice of shifts. A curve in polar coordinates is symmetric about the x-axis if replacing θ with θ leaves the equations unchanged. A curve in polar coordinates is symmetric about the y-axis if replacing θ with π θ leaves the equations unchanged. A curve in polar coordinates is symmetric about the origin if replacing r with r or equivalently θ with θ + π leaves the equations unchanged. Let s look at an example. Example: Test the symmetry properties of r = sin θ. Solution: If we replace θ with θ, we get sin( θ) = sin θ, so it is not symmetric about the x-axis. If we replace θ with π θ, we get sin(π θ) = sin θ, so it is symmetric about the y-axis. Finally, if we replace θ with θ + π we
. r, Θ r, Π Θ. r, Θ.5.5.5.5. r, Θ. r, Θ..5 r, Θ Π or r, Θ.5. Figure 9: Symmetry test get sin(θ + π) = sin θ, or if we replace r with r we get the same, so it is not symmetric about the origin...4 Families of curves Circles: The following describe circles of radius a, r = a, r = a cos θ, r = a sin θ. () The first of these is a circle centered on the origin, the second centred on (a, ), and the last centred on (, a). Figure shows these. Rose curves: The following describe rose curves for a > and n integer, r = a sin nθ, r = a cos nθ. (3) If n is odd, the pattern is drawn when the values ( θ < π) are taken, and for n even, the pattern is drawn when the values ( θ < π) are taken. See what the look like: Section. of the text book has images.
.5.5...5.5.5..5.5..5.5..5..5.5. r a. r a Cos Θ.5.5.5..5..5.5..5.5..5 r a Sin Θ Figure : Families of circles Limaçons: The following describe limaçons for a > and b >, r = a ± b sin nθ, r = a ± b cos nθ. (4) If a = b, these becomes cardioids. Again, see textbook for images. Spirals: Although there are many types of spiral, the only one we are interested in is the Archimedean spiral, which is the one you most likely imagine when someone says the word spiral. It has the equation r = a θ. (5)
..5 Tangent Lines in Polar Coordinates If we take a general curve of the form r = f(θ) and write it in terms of its x and y coordinates, we would have x = f(θ) cos θ, y = f(θ) sin θ. (6) This is now a parametric curve in parameter θ, and so we can find the slope of the tangent line provided f(θ) is differentiable. We have and the slope is dx dθ = f(θ) sin θ + f (θ) cos θ = r sin θ + dr dθ cos θ, dy dθ = f(θ) cos θ + f (θ) sin θ = r cos θ + dr (7) dθ sin θ, dy dy dx = dθ = dx dθ dr r cos θ + sin θ dθ. (8) r sin θ + dr cos θ dθ Example: Find the slope of the tangent line to the circle r = cos θ at θ = π/3. Solution: We get dr dθ dy dx = = sin θ, and so cos θ sin θ sin θ cos θ sin θ cos θ Therefore, the slope at θ = π/3 is = cos θ sin θ m = cot π 3 = 3. = cot θ...6 Tangent Lines to Polar Curves at the Origin At the origin, we have r = with angle θ, so assuming that dr dθ origin, the equation becomes at the dy dx = + dr sin θ dθ + dr cos θ = tan θ. (9) dθ The line with this slope is the line θ = θ, and so this line is tangent to the curve at the origin. To recall what this line looks like, return to section... 3
Example: Find the tangent lines at the origin of the four-petal rose r = sin θ. Solution: We must find all values of θ less than π for which r =. This gives us θ =, θ = π/, θ = π, θ = 3π/. These are the tangent lines and are shown in Figure. Note that the lines θ = and θ = π coincide, as do θ = π/ and θ = 3π/.. Θ Π, 3 Π.5 Θ, Π..5.5..5. Figure : r = sin θ 4
..7 Arc Length in Polar Coordinates To find the arc length in terms of polar coordinates, we recall formula (9) and substitute the results of (7), giving ( ) dx + dθ = ( ) dy dθ ( r sin θ + dr dθ cos θ ) + ( r cos θ + dr dθ sin θ = r (sin θ + cos θ) + r dr cos θ sin θ dθ r dr ( ) dr cos θ sin θ + (sin θ + cos θ) dθ dθ ( ) dr = r +, dθ ) () which results in Let s look at an example. L = β α r + ( ) dr dθ () dθ Example: Find the arc length of the logarithmic spiral r = ae bθ between θ = and θ = π. Solution: L = = π π (ae bθ ) + (abe bθ ) dθ a + b e bθ dθ = a b..8 Area in Polar Coordinates + b e bθ π = a b + b (e bπ ). () In order to calculate area in polar coordinates, we need to find a way to parameterise the area. The first thing to remember is that any polar curve can be written in the form r = f(θ), and therefore we can parameterise in terms of θ. Then, if we choose two rays θ = α and θ = β such that the angles satisfy α < β α + π (in other words, the angles are different and are less than π apart, so that the area is only covered once), provided f(θ) 5
is continuous in this region, R, we can find the area. This concept is shown in Figure. We can obtain the area under the curve r = f(θ) by diving 6 5 4 r f Θ Θ Β 3 R Θ Α 4 4 Figure : Area in polar coordinates this region into wedges using a set of rays, θ = θ, θ = θ, θ = θ n, which divide the region into small areas A, A, A n with angles θ, θ, θ n. This is shown in Figure 3. The total area is then the sum of these small 6 5 4 r f Θ 3 A A Θ Β A n Θ n Θ Θ Θ Α 4 4 areas, Figure 3: Area in polar coordinates divided into wedges A = A + A + + A n = n A k. (3) Each of these is the area of a sector, which has area given by r θ. Using r = f(θ), and θ = θ k for any ray in the wedge of area A k, we arrive at the 6 k=
discrete version of the area formula A = n A k k= n k= [f(θ k)] θ k. (4) We can take the limit where these angles θ k vanish to give an integral form A = lim k n k= β [f(θ k)] θ k = α [f(θ)] dθ. (5) Thus, the area of a region R under a continuous curve r = f(θ) and between θ = α and θ = β is A = β α β [f(θ)] dθ = α r dθ. (6) The most difficult parts of this is generally finding the limits of integration. Sometimes you will be told explicitly. Important note: Always draw the region to help you identify the limits. Let s look at a simple example. Example: Find the area of the circle r = a sin θ. Solution: This looks like the third types of circle Figure. Since we want the area of the whole circle, the limits are θ π since the entire circle is above the x-axis. Note that for a circle centered on the origin r = a, the limits would be θ π. The area is A = π = a π (a sin θ) dθ = a π ( cos θ) dθ = a sin θ dθ ( θ π ) = πa. We can often make use of the symmetry of the region to simplify a problem. For example, the upper and lower semi-circles of a full circle are mirror images and obviously have the same area. Hence π π a dθ = a = πa, (7) 7
is clearly true. Let s look at a more complicated example to help understand this. Example: Find the area of the cardioid r = 4 + 4 cos θ outside the circle r = 6. Solution: The region is the coloured region in Figure 4. The rays θ = θ 5 Θ Θ 5 5 5 Θ Θ Figure 4: Area of the cardiod r = 4 + 4 cos θ outside the circle r = 6 and θ = θ represent the limits of integration. We can find these limits by finding the intersections of the circle with the cardiod, 4 + 4 cos θ = 6 cos θ = θ = ±π 3. We can now see that there are two equally valid approaches: integrate from π to π, or use symmetry to take twice the area from to π. Obviously, 3 3 3 we must subtract the area of the circle in this region from the area of the cardiod in this region. Let s look at both methods. 8
Integration over whole region: A = = = = π/3 π/3 π/3 π/3 π/3 π/3 π/3 π/3 (4 + 4 cos θ) dθ π/3 π/3 (6) dθ (6 + 3 cos θ + 6 cos θ 36) dθ (6 cos θ + 8 cos θ ) dθ (6 cos θ + 4( + cos θ) ) dθ = (6 sin θ + 4(θ + sin θ) θ) π/3 π/3 ( ( 3 = 6 + 4 π 3 + ) 3 π ) ( ( 3 6 3 + 4 π 3 ) 3 + π ) 3 = 8 3 4π. Notice that the first and second terms in the penultimate line are in fact the same. This is due to the symmetry about the x-axis. Using symmetry: π/3 A = (6 cos θ + 8 cos θ ) dθ ( ( 3 = 6 + 4 π 3 + ) 3 π ) = 8 3 4π. 3 The procedure for equating the equations of the curves can fail to find some intersections. This is why it is always important to draw diagrams to ensure that you know how many intersection points to expect..3 Conic Sections A conic section (or conic) is any curve that can be obtained as the set of points of intersection between a plane and a double-napped circular cone. These are the ellipse, parabola and hyperbola (note that a circle is a special case of an ellipse). The degenerate conic sections arise when the plane 9
passes through the vertex of the cone, and are the point, line and intersecting lines. The non-degenerate conic sections are shown in Figures 5 and the degenerate conic sections are shown in Figure 6. Figure 5: Conic sections
Figure 6: Degenerate conic sections
.3. Types of Conic Section A parabola is the set of points in the plane equidistant from a fixed line called a directrix and a fixed point called the focus. A parabola is symmetric about the axis perpendicular to the directrix which runs through the focus, called the axis of symmetry. The parabola intersects this axis at its vertex. See Figure 7. 6 5 4 Focus3 Axis x Figure 7: A parabola An ellipse is the set of all points in the plane that such that the sum of the distance from a point on the ellipse to two fixed points called foci is a constant. This constant must be greater than the distance between the foci. The midpoint of the line line joining the foci is called the centre. Note that a circle is a special case of an ellipse when the distance between the foci is, i.e. the foci coincide with the centre. The line segment across the ellipse through the foci is called the major axis, and the line segment across the ellipse perpendicular to this through the centre is called the minor axis. Finally, the endpoints of the major axis are called vertices. See Figure 8...5..5 Major axis Focus Centre.5 Focus.5 Minor axis. Figure 8: An ellipse
A hyperbola is the set of all points in the plane that such that the difference of the distance from a point on the hyperbola to two fixed points called foci is a positive constant. This constant must be less than the distance between the foci. This difference must be positive, and as a result there are two branches which curve around the different foci. The midpoint of the line segment between the foci is called the centre, and the line through the foci is called the focal axis. The line perpendicular to the focal axis and through the centre is called the conjugate axis. The intersections of the hyperbola with the focal axis are called vertices. A hyperbola also has a pair of associated lines called asymptotes, which are line such that as the hyperbola move towards ± away from the centre, the distance between the asymptotes and the hyperbola approaches zero. See Figure 9. 4 4 5 5 5 5 4 4 Figure 9: A hyperbola. The points marked are the centre (middle) and foci (left and right). The horizontal red line is the focal axis and the vertical red line is the conjugate axis..3. Equations of parabolas We denote the distance between the vertex and focus of a parameter as p. Then, the distance from the vertex to the directrix is also p. It is sometimes convenient to denote the focus as F (x, y). The standard positions of a parabola are those when the vertex coincides with the origin. Figure sketches the standard parabolas. You should therefore recognise any equation of the form y = ±4px, or x = ±4py, (8) as the standard equations of a parabola. To understand these equations, we return to the definition of a parabola. The distance from the directrix to 3
y y F p,.5..5. x F p,..5..5 y 4px y 4 y y 4px x 3 F,p x x 4py 3 4 5 F, p Figure : Standard equations of a parabola a point must be the same as the distance to the focus. Look at Figure. The distance to the directrix to Q is always simply y + p. The distance from the focus, F, to Q is (y p) + x. Therefore (y p) + x = (y + p) y py + p + x = y + py + p x = 4py, (9) as you would expect from the diagram. To sketch a parabola, determine which of the four types in equation (8) it is, and then use Figure to determine which way it curves. Determine p from the equation and plot a couple of convenient points to ensure you are close. Useful points are the intersections of the parabola with a line parallel to the directrix running through the focus, see Figure. Please note that this is basically the same method given in the textbook, but I think the explanation is less confusing. Also, notice that we are always assuming that the vertex of the parabola is at the origin. If this wasn t the case, we would have to find it. Example: Find the focus of the parabola y = 6x and sketch the 4
y 4 3 F,p Q x y p Figure : Equations a parabola y 4 3 p,p,p p,p x y p Figure : Sketching a parabola parabola, focus and directrix. Solution: We see that we have a parabola of type y = 4px and so p = 4. Since we are looking at the second type in Figure, the focus is at ( 4, ), and the parabola should be to the left of the y-axis. It is shown in Figure 3. More generally, we can have a parabola which has its vertex at any point. They have the following equations: Vertex (h, k), parallel to x-axis (y k) = 4p(x h), opens to the right. (y k) = 4p(x h), opens to the left. (3) 5
4,8 5 y 4, x 4 8 6 4 4 x 5 5 4, 8 5 Figure 3: Sketching the parabola y = 6x Vertex (h, k), parallel to y-axis (x h) = 4p(y k), opens above. (x h) = 4p(y k), opens below. (3) You should be able to recognise these equations as parabolas and reduce appropriate quadratic equations to these forms. It is often necessary to complete the squares to find the correct form of the equation. Example: Determine if y 6y x + = represents a parabola and if so find p and the vertex position. Solution: We complete the square to get y 6y x + = (y 3) 9 x + = (y 3) x 8 = (y 3) = (x + 4). Hence, p = and the vertex is at ( 4, 3)..3.3 Equations of ellipses We denote the length of the major axis of an ellipse by a and the length of the minor axis by b, with the distance between the foci denoted by c. In this context, it is sometimes convenient to denote the foci as F (x, y ) and F (x, y ). a is referred to as the semi-major axis and b as the semi-minor 6
axis. This is just terminology: obviously a number is not the same as a line! These numbers uniquely define the ellipse. Their ratio give the shape, and the magnitude gives the size of the ellipse. This is shown in the first diagram in Figure 4. If we consider the red triangles in the second diagram, we see c a c b 3 3 a b c Q b P 3 3 c Figure 4: The semi-major axes and semi-minor axes for ellipses that the distance from the two foci to Q is simply b + c. Since the sum of the distances from the foci to any point is the same, this should be the same as the sum of the distances from the foci to P, which is (a+c)+(a c) = a. We therefore find the relations a = b + c, or c = a b. (3) As for a parabola, the easiest types of ellipse to imagine are those centred on the origin. These are the standard positions of the ellipse, which give rise to the standard equations. These standard ellipses are shown in Figure 5. You should recognise any equation of the forms x a + y b =, or x b + y a =, (33) as a standard ellipse. To see where these come from, consider Figure 6. The 7
y 3 a F,c a F c, y b b F c, 3 3 x x a y a b b F, c 3 a x b y b a Figure 5: Standard ellipses sum of the distances from the foci to Q should be a. We therefore have (x + c) + y + (x c) + y = a (x + c) + y = a (x c) + y (x + c) + y = (x c) + y 4a (x c) + y + 4a 4cx = 4a (x c) + y + 4a (squaring) a cx a = (x c) + y ( a cx ) = (x c) + y (squaring) a a cx + c a x = x cx + c + y a c = a x ( a c ) + y (34) = x a + y a c = x a + y b, 8
. Q.5 c,.5 c,. Figure 6: Equations of ellipses due to the relations (3). This is clearly the correct equation for the ellipse in Figure 6. To sketch an ellipse, you will want to determine whether the major axis is the x or y axis, i.e. which type of parabola in Figure 5 we are dealing with. Determine a and b, and plot the points where the ellipse intersects the x- and y-axes. More generally, we want to consider ellipses with centres that are not the origin. In this case, the equations look like (assuming b < a) Ellipse with centre (h, k), and major axis parallel to x-axis (x h) (y k) + =, a b (35) Ellipse with centre (h, k), and major axis parallel to y-axis (x h) (y k) + =, b a (36) and to sketch the ellipse we follow more or less the same procedure with the new centre and using the points where the ellipse meets its major and minor axes for reference. Example: Determine if 9x + 4y 8x + 4y + 9 = represents an ellipse. Sketch the resulting curve. 9
Solution: Completing the squares, we get 9x + 4y 8x + 4y + 9 = (3x 3) 9 + (y + 6) 36 + 9 = 9(x ) + 4(y + 3) = 36 (x ) (y + 3) + 4 9 =, (37) which is an ellipse centred on (, 3), with the major axis parallel to the y-axis, and a = 3, b =. Also, c = 3 = 5. It is sketched in Figure 7. y 3 x F 3 Centre 4 5 F 6 Figure 7: The ellipse (x ) + 4 + (y+3) 9 = 3
.3.4 Equations of hyperbolas When defining hyperbolas, we denote the distance between the vertices by a, the distance between the foci by c, and in analogy to the ellipse, we define a constant b via the formulae b = c a, or c = a + b. (38) These quantities are drawn in Figure 8. We see that b is defined in such a c 5 5 a 4 c a 4 b 5 5 V V a c 4 4 Figure 8: Defining a hyperbola way that the asymptotes intersect the circle with diameter c. We sometimes find it convenient to denote the foci as F (x, y ) and F (x, y ), and the vertices as V (x, y ) and V (x, y ). If we look at one of these vertices, say V, we see that the difference of the distances from V to the foci is ( (c a) + a ) (c a) = a. (39) Since the difference of the distances from the foci to every point is the same, it is always a. If the centre of the hyperbola is at the origin and the foci are on either the x- or y-axis, the hyperbola is in one of the standard positions, with the resulting standard equations. The parabola are shown in Figure 9 and the equations that you should recognise as standard hyperbola at the origin are x a y b =, or y a x b =, with asymptotes y = ± b a x, or y = ±a b x. (4) 3
y F c, a x 5 b b a y a b F c, 5 5 x 5 y a b x b F,c 5 a a F, c b 5 5 5 y y a b x x a b Figure 9: Standard hyperbolas The asymptotes are given the limit x, y of which gives the ratios y b = x a, or y a = + x b, (4) or y x = b a, or y x = a b, (4) y = ± b a x, or y = ±a b x. (43) To draw a hyperbola from a standard equation, locate the foci and thus the focal axis. Next, locate the vertices and the asymptotes. If necessary, plot some extra points and then draw the hyperbola passing through the vertices, any extra points and approaching the asymptotes. More generally, we want to consider hyperbolas with centres that are not the origin. In this case, the equations look like Hyperbola with centre (h, k), and focal axis parallel to x-axis (x h) (y k) =, a b (44) 3
Hyperbola with centre (h, k), and focal axis parallel to y-axis (y k) (x h) (45) =, a b Please note that when we write the hyperbolas this way, the asymptotes for the first case are the lines with slope ± b passing through (h, k) and for the a second first case the asymptotes are the lines with slope ± a passing through b (h, k). Example: Determine if x 4y + x + 8y 7 = represents a hyperbola, and if show, sketch the resulting curve. Solution: We have x 4y + x + 8y 7 = (x + ) 4(y ) + 4 7 = (x + ) 4(y ) = 4 (x + ) (y ) 4 =. This is clearly a hyperbola centred on (, ). Its focal axis is parallel to the x-axis. The vertices are length a way from the centre, and since a = by comparing with equation (44), the vertices are at ( 3, ) and (, ). To find the foci, recall that they are c = a + b away from the centre. Here c = + = 5, and so the foci are at ( ± 5, ). Finally, the asymptotes can be found by finding the lines that have slope ± that pass through (, ). Using 3 F 5,, F 5, 6 4 4 Figure 3: The hyperbola (x+) 4 (y ) = y = mx + b, 33
we find that for m =, b = 3 and for m =, b =. Therefore the asymptotes are This parabola is shown in Figure 3. y = x + 3, y = x +..3.5 Reflection properties for conic sections Reflection property of parabolas: Consider a tangent line at a point P on the parabola. The angles between the tangent line and the line through P parallel to the focal axis and between the tangent line and the line connecting P to the focus are equal. Reflection property of ellipses: Consider a tangent line at a point P on the ellipse. The angles between the tangent line and the lines joining P to the foci are equal. Reflection property of hyperbolas: Consider a tangent line at a point P on the hyperabola. The angles between the tangent line and the lines joining P to the foci are equal. These reflection properties are drawn in Figure 3..3.6 Applications of conic sections Some applications of conic sections include: Parabola: Parabolic reflector - used in headlights on cars, etc. to project a focussed beam of light. Ellipse: Planetary orbits - Kepler s problem. More on this later. Hyperbola: Scattering of atomic particles - Rutherford s experiment..4 Rotated conic sections.4. Quadratic equations All conic sections are special cases of the general quadratic equation Ax + Bxy + Cy + Dx + Ey + F =. (46) If A, B, C it is quadratic in x and y, and if they vanish, it reduces to a linear equation in x and y. Obviously B = gives rise to the conic sections 34
P Α Α y 4 3 F x.5. F..5 P Α Α P.5. F.5 Α F Α F 4 4.5..5 Figure 3: Reflection properties we have already met when appropriate constants are chosen. However, if B, then we have a cross-product term, which is new. The effect of this term is to rotate the axes of the conic section so that they no longer align with the x and y axes. To understand why this is true, consider the ellipse described by the foci F (, ), F (, ) and a = 3, which is drawn in Figure 3. Based on the information we have, the point in this figure should satisfy (x ) + (y ) + (x + ) + (y + ) = a = 6, (47) which we square to give x + (x ) + (y ) (x + ) + (y + ) + y + = 36. (48) If we now write ( (x ) + (y ) (x + ) + (y + ) ) = (6 y x ), (49) when we simply we get 8x 4xy + 5y = 36, (5) 35
y P x,y, 3 3 x, Figure 3: Rotated ellipse which has the xy term we expected. A useful tool to help identify conic sections is the discriminant, δ = B 4AC. δ < represents an ellipse, with A = C and B = being the special case of a circle. δ = represents a parabola. Finally, δ > represents a hyperbola..5 Rotation of the x and y axes How do we relate these rotated conic sections to those we are familiar with? To do this, we must understand how to perform a rotation of the plane. If we begin with a coordinate system in polar coordinates x = r cos(θ + α), y = r sin(θ + α), (5) for θ constant such that the polar axis at θ +α = coincides with the x-axis, we can easily imagine that we could equally construct a coordinate system which has it s polar axis at α =, which would be x = r cos α, y = r sin α. (5) 36
y. y'..8 P.6.4 r x'..5. x. Figure 33: Rotated coordinate axes These two systems are shown in Figure 33. We can use trigonometric identities to rewrite the unprimed coordinates as x = r cos θ cos α r sin θ sin α, y = r sin θ cos α + r cos θ sin α, (53) which means that we can relate the coordinate systems by x = x cos θ y sin θ, y = x sin θ + y cos θ. (54) These are known as the rotation equations. We can also invert these to get x = x cos θ + y sin θ, y = x sin θ + y cos θ. (55) We can eliminate a cross-product term by choosing a convenient coordinate system x y. The trick is to find the correct angle to rotate by. Theorem: For the equation Ax + Bxy + Cy + Dx + Ey + F =, (56) 37
such that B, the x y -coordinate system obtained by rotating the xy-axes through an angle θ satisfying then the equation in x y -coordinates becomes cot θ = A C B, (57) A x + C y + D x + E y + F =. (58) Proof: If we substitute equation (55) into equation (56), we get A(x cos θ y sin θ) + B(x cos θ y sin θ)(x sin θ + y cos θ) which means If we have then as required. + C(x sin θ + y cos θ) + D(x cos θ y sin θ) + E(x sin θ + y cos θ) + F =, A = A cos θ + B cos θ sin θ + C sin θ B = B(cos θ sin θ) + (C A) sin θ cos θ C = A sin θ B sin θ cos θ + C cos θ D = D cos θ + E sin θ E = D sin θ + E cos θ F = F. cos θ sin θ (59) (6) = cot θ = A C B, (6) B = B cos θ + (C A) sin θ cos θ = (6) Example: Rotate to the coordinates that coincide with the axes of the conic section x + 4xy y 6 =. What type of conic section is it? Solution: We need to rotate by θ, where cot θ = ( ) = 3 4 4 θ =.463648 rad = 6.565 o sin θ =.4474, cos θ =.89447. 38
Using these, we can substitute into equation (6) to get A =, B =, C = 3, F = 6, where of course, D = E = since there are no linear terms initially. The equation in the new variables is therefore which is a hyperbola. x 3y = 6 x 3 y =,.6 Conic Sections in Polar Coordinates To find how to represent conic sections in polar coordinates, first we must state a theorem. Focus-Directrix Property of Conics: Suppose that a point P moves in the plane, with the motion determined by a fixed point (the focus) and a fixed line (the directrix) such that the focus does not lie on the directrix. Then, if the point moves in such a way that its distance to the focus divided by its distance to the directrix is a constant e, called the eccentricity, then the curve the point traces is a conic section. The eccentricity determines the conic section as follows: (a) a parabola if e =, (b) an ellipse if < e <, (c) a hyperbola if e >. (63) In the case of a parabola, this is fairly easy to see. For a parabola y = 4px curving to the right, we take the usual focus and directrix x = p. However, for an ellipse or a hyperbola we must define the directrix. We will take the directrix to be the line x = a /c if we consider the standard shapes in Figure 34. Below we will call the point P, the focus F and the directrix D. In this notation, we require P F DP = e. (64) 39
D x p y F x,y 3 x e F c,.5. y e y..5 P x,y D x P x,y D.5..5 F c,.5 4 4 x.5 e..5 x a c Figure 34: Conic sections defined and eccentricities For a parabola, we already know that the distance P F to the focus is equal to the distance to the directrix P D, so e =. For the ellipse, Recalling the fifth line of equation 34, we have (x c) + y = c ( ) a a c x, (65) where the left-hand side is P F, and the bracketed term on the right is P D. Therefore, for an ellipse, e = c a, (66) which given < c < a gives < e <, unless a = or c =. What would the conic section be in these situations? Finally, for the hyperbola, we know that the distance to the further focus minus the distance to the nearer focus is a. Therefore, in analogy with equation (34), we get (x + c) + y (x c) + y = a, (67) so performing the same procedures, we arrive at (x c) + y = c ) (x a, (68) a c 4
where the left-hand side is again P F and the right-hand side is P D since the directrix is the opposite side to the point compared with the case of the ellipse. Therefore, we also have for a hyperbola. e = c a, (69) Also note that for an ellipse, eccentricity can be regarded as a measure of the flatness of the ellipse: as e approaches, the ellipse becomes circular; as e approaches the ellipse flattens out. Earth s orbit has e =.7. See Figure 35 for plot of ellipses of different eccentricites...5 F..5.5..5..5. e e. e.4 e.6 e.8 Figure 35: Ellipses with same focus and same semi-major axis, but different eccentricities.6. Polar equations of conics In order to derive the polar equations of conics, it is convenient to chose the pole to coincide with the focus, and the polar axis to be either parallel or perpendicular to the directrix. In addition, there are two choices for the direction the polar axis points. As a result there are four possibilities that one might consider. We will choose the directrix to be perpendicular to the polar axis, which will point towards the directrix. Figure 36 shows the construction. The distance between the pole and directrix is d, see Figure 36. d is constant is set by our choice of pole. We know that P F = ep D. (7) 4
P x,y D r F Pole Θ r cos Θ d Figure 36: Polar coordinates defined for conics For the situation in the diagram, we have P F = r and P D = d r cos θ, and therefore ed r = + e cos θ. (7) This can be used to represent a parabola, ellipse or hyperbola. In summary, for the different cases, we have (a) Directrix right of pole ed r = + e cos θ (c) Directrix above pole ed r = + e sin θ (b) Directrix left of pole ed r = e cos θ (d) Directrix below pole ed r = e sin θ (7) These are sketched in Figure 37. To sketch conic sections from their polar coordinates, we need to identify the relevant parameters. For a parabola, d tell us the distance from the directrix to the focus, and so we can immediately sketch it. Example: Sketch r = cos θ. Solution: This looks like (b) of (7) with d = and e =. 4
a F D b D F D F F D Figure 37: Choices of polar coordinates for conics For an ellipse, if we consider the first diagram in Figure 38, we see that r = a c, r = a + c, (73) and so a = (r + r ), c = (r r ). (74) We also find r r = a c = b b = r r. (75) For an hyperbola, if we consider the second diagram in Figure 38, we see that r = c a, r = c + a, (76) and so We also find Finally let s do a quick example. Example: Sketch r = + sin θ. a = (r r ), c = (r + r ). (77) r r = c a = b b = r r. (78) 43
5 a r b a r b c 5 5 3 3 c 5 a a r r Figure 38: Ellipses and hyperbolas in polar coordinates Solution: This looks like (c) of (7). We have d = and e =. This is a hyperbola with directrix length above the pole. Because e >, the denominator causes complicated behavior. At θ = 7π/6, sin θ = /, causing the denominator to be, and a switch from r being positive to r being negative. At θ = 3π/, sin θ = and is the turning point of the upper branch of the curve. At θ = π/6, r returns to being positive. In 5 Π Π 6 3Π 3 4 5 6 3 y y 4 3 y 3 5 3 3 Π 6 Figure 39: Rough sketch of a hyperbola in polar coordinates short, we get r by setting θ = π/ and r from θ = 3π as these represent the vertices of the hyperbola, i.e. the turning points of the branches. Note that the upper branch appears for r < because in the plot of the function 44
it is below the axis. The function and the hyperbola are drawn in Figure 39. Then, r = + sin π/ = 3, r = + sin 3π/ = =. (79) Therefore a = (r r ) = 3, b = r r = 3 3, c = (r + r ) = 4 3. (8).6. Planetary motion While you might see more of this in a mechanics course, we will here state Kepler s Laws of Planetary Motion:. Law of Orbits: The motion of each planet traces an ellipse with the Sun at one of the foci.. Law of Areas: The line joining the Sun to the centre of the planet sketches out equal areas in equal times. 3. Law of Periods: The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. Kepler s second law is shown graphically in Figure.6.. These laws are valid..5 Earth.5 Sun. Figure 4: The Law of Areas. The two red areas represent two areas traced out during equal time periods. Of course, this is not to scale. for any celestial body under the influence of a single gravitational force. As 45
a result, these are only ever approximations as all the planets also exert gravitational forces on each other, but it is a very good approximation. The closest point between the bodies during the orbit is called the perigee (i.e. r ), and the farthest point is called the apogee (i.e. r ), which are also known as the perihelion and aphelion when the Sun is orbited, from the Greek word for the Sun, Helios. Kepler s third law may be denoted T = a 3, (8) provided a is measured in astronomical units (AU=.5 8 km), and T is Apogee..5 Perigee D.5..5 ae a a e Figure 4: The length parameters for an orbit measured in Earth years so that T = is the period of the Earth s orbit and a = is the distance to the sun. Normally, orbits are specified by eccentricity and the semi-major axis, so using c = ea and we get and d = a e c = a e ea = a( e ) e (a) r = ( e ) + e cos θ (c) r = ( e ) + e sin θ (b) r = ( e ) e cos θ (d) r = ( e ) e sin θ,, (8) (83) r = a( e), r = a( + e). (84) The various length parameters are shown in Figure.6.. If there is time, we will consider Kepler s laws in greater detail later. 46