Linear Algebra and Geometry - Essay Example

Part I: Linear Algebra and Geometry Assignment: Give a detailed of how linear transformations can be used to study rotated conic sections. Your essay should include: 1) A description of rotations through an angle about the origin. Each of the conic sections which are centered at the origin can be described by the equation Ax2 + Cy2 + Dx + Ey + F = 0 This curve can be rotated through an angle about the origin by using the rotation equations: x = x’ cos - y’ sin y = x’ sin + y’ cos so that the transformation matrix is: R = cos - sin sin cos These values for x and y are substituted into the equation of the curve in order to obtain a new equation for the curve in terms of the x’-y’ coordinates: A’x’2 + B’x’y’ + C’y’2 + D’x’ + E’y’ + F’ = 0 which is the rotated conic section. The rotation of the conic section results in a cross-product term Bxy. 2) How to eliminate the xy- term from the equation of a rotated conic section. In order to eliminate the xy-term from the equation of a rotated conic section, the xy- axes can be rotated an angle so that the new axes xy are parallel to the conic sections axes. This angle can be found from the equation cot 2 = (A - C )/ B where A, B, and C are found from the conic section equation: Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 The rotation equations are then used: x = x’ cos - y’ sin y = x’ sin + y’ cos These values for x and y are substituted into the conic section equation in order to get a new conic section equation in terms of the rotated coordinate axes x’-y’: A’x’2 + C’y’2 + D’x’ + E’y’ + F’ = 0 where the cross-product term B’x’y’ = 0 since after substitution of the rotation equations in the conic section equation B’ = B(cos2 - sin2 ) + 2 (C - A) sin cos Using the trigonometric double angle formulas, this becomes: B’ = B(cos 2 ) + (C- A) sin 2 This equals 0 if cot 2 = (A - C )/ B. 3) Two examples of rotated conic sections with a picture diagram for each one. Example of a rotated ellipse: Fig. 1 Rotated ellipse. From http://www.krellinst.org/UCES/archive/resources/conics/node66.html This rotated ellipse has center at (0,0) and angle of rotation /6 = 30. The equation that describes the ellipse is: 7x2 - 6 3xy + 13 y2 -16 = 0 Using the general rotated conic section equation: Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 it can be seen that A = 7, B = 6 3, C = 13, F = -16, and D = E = 0. To check that this is the equation for an ellipse, the discriminant is calculated (see next section): B2 - 4AC = 36(3) - 4(7)(13) = -256 < 0 so that this curve fits that case of an ellipse, circle or a degenerate conic section. The angle that the coordinate axes must be rotated is cot 2 = |A - C| / B = |7 - 13|/6 3 = 1/3 so that tan 2 = 3 and = 1/2 inv tan (3 ) = 30. The rotation equations are then: x = x’ cos - y’ sin = x cos (30) - y sin(30) = .866 x’ - 1/2 y’ y = x’ sin + y’ cos = x’ sin(30) + y’cos(30) = 1/2 x’ + .866 y’ This leads to the rotated coordinates xy equation: 7 (.866 x’ -1/2 y’)2 - 6 3(.866 x’ - 1/2 y’) (1/2 x’ + .866 y’) + + 13 (1/2 x’ + .866 y’)2 -16 = 0 Dropping primes and simplifying: 4x2 + 16 y2 = 16 since the x’y’ terms cancel out. Dividing the above equation by 16: x2/4 + y2 = 1 which is the equation of an ellipse with semi-major axis a where a2 = 4, so that a = 2, and the major axis is 2a = 4 along the x-axis and semi-minor axis b where b2 = 1 so that b = 1 and the minor axis is 2b = 2 along the y-axis. Example of a rotated parabola: Fig. 2- Rotated parabola. From http://education.yahoo.com/homework_help/math_help/problem?id=miniprecalcgt_8_1_1_23_70 The equation for this rotated conic section is: x2 - 2xy + y2 -5x - 5y = 0 so that from Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 it can be seen that A = 1, B = -2, C = 1, D= E = -5, and F = 0. The discriminant here (see next section) is B2 - 4AC = 4 - 4 = 0, which fits the case of a parabola or a degenerate conic section. To determine the angle of rotation, the equation cot 2 = (A - C )/ B = (1-1)/(-2) = 0 Since cot 2 = cos 2 / sin 2 , the quantity cot 2 will be 0 whenever cos 2 = 0. This occurs at 90 = /2 so that 2 = /2 and = /4 = 45. From the graph above, it can be seen that the parabola is rotated 45 counter-clockwise from the orientation where the axis of symmetry is the x-axis. Using the rotation equations, x = x’ cos - y’ sin = 1/2 x’ - 1/2 y’ y = x’ sin + y’ cos = 1/2 x’ + 1/2 y’ Substituting this into the rotated conic section equation: x2 - 2xy + y2 -5x - 5y = 0 = (1/2 x’ - 1/2 y’ )2 - 2 (1/2 x’ - 1/2 y’)(1/2 x’ + 1/2 y’) + + (1/2 x’ + 1/2 y’ )2 - 5 (1/2 x’ - 1/2 y’) + - 5 (1/2 x’ + 1/2 y’) = 0 This simplifies to y’2 = 5/2 x’ After dropping primes, the equation becomes: y2 = 5/2 x which is the equation of a parabola centered at the origin, with the x-axis as the axis of symmetry for the parabola. With 4p = 5/2, the focal point p = 5/(42 ) = 0.88 units from the vertex of the parabola. Since p is positive, the parabola opens in the positive x-direction. 4) How to classify a rotated conic section as a parabola, ellipse, or hyperbola with a picture diagram for each one. From the equation of a conic section: Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 it can be shown that the discriminant, B2 - 4AC, is an invariant when the coordinate axes are rotated an angle . The rotated conic section is given by the equation A’x’2 + B’x’y’ + C’y’2 + D’x’ + E’y’ + F’ = 0 and the discriminant satisfies B2 - 4AC = B’2 - 4A’C’ which shows it is invariant under rotation of the coordinate axes. The discriminant can then be used to classify the conic section: If B2 - 4AC < 0, the conic section equation describes an ellipse, circle, point or no graph. If B2 - 4AC > 0, the conic section equation describes a hyperbola or pair of intersecting lines. If B2 - 4AC = 0, the conic section equation describes a parabola, line, pair of parallel lines, or no graph. Rotated conic sections can also be classified by rotating the coordinate axes by an angle where cot 2 = (A - C )/ B, by using the rotation equations, and then completing the squares for x’ and y’ until the equation is in one of the following forms: Parabola centered at (h, k) with a focus a distance p from the vertex: (y - k)2 = 4p(x - h) or (x - h) 2 = 4p(y - k) (axis of symmetry parallel to y axis or x axis, respectively) Ellipse centered at (h, k) with major axis 2a and minor axis 2b: (x - h)2 / a2 + (y - k)2 / b2 = 1 (Foci and major axis on the x axis) or (x - h)2 / b2 + (y - k)2 / a2 = 1 (Foci and major axis on the y axis) Hyperbola centered at (h, k) with 2a the distance between the vertices of the two branches, 2c the distance between the two foci, and b = (c2 - a2)1/2 (x - h)2 / a2 - (y - k)2 / b2 = 1 (Foci on the x axis) or (x - h)2 / b2 - (y - k)2 / a2 = 1 (Foci on the y axis) PART II: Linear Algebra and Calculus Assignment: The dot product (often referred to as the inner product in a Linear Algebra text) is a useful tool in developing several important techniques of multivariable calculus. Your task is to give a detailed description of why the gradient points in the direction of the maximum increase of a function. You may restrict your essay to R3. Your essay should include: 1) A definition and description of the directional derivative Given that a function f is differentiable at the point (x0, y0), a directional derivative Duf(x0, y0) of the function f at a point (x0, y0) in the direction of the unit vector u = is defined by Duf(x0, y0) = fx(x0, y0) u1 + fy(x0, y0) u2 where fx(x0, y0) and fy(x0, y0) are partial derivatives of the function f in the x and y directions, respectively, evaluated at the point (x0, y0). The unit vector components are given by u1 = dx/ds and u2 = dy/ds, where s is the distance along a line given by points in the x-y plane directly below points in a curve C that lies within the surface z = f(x, y). For instance, at the beginning of the line, the point (x0, y0) lies in the x-y plane directly below the point Q0 lying in the curve C within the surface z = f(x, y). At a distance s (in the direction of u) along the line in the x-y plane, the point (x, y) is reached, which lies directly below the point Q in the curve C. The parametric equations of the line lying in the x-y plane in the direction of u are: x = x0 + s u1 y = y0 + s u2 so that dx/ds = u1 and dy/ds = u2. Using the chain rule, dz/ds = f/x dx/ds + f/y dy/ds = fx(x, y) dx/ds + fy(x, y) dy/ds Using the components of the unit vector, this becomes dz/ds = fx(x, y) u1 + fy(x, y) u2 This quantity dz/ds is the instantaneous rate of change of the function f with s at the point (x, y). When dz/ds is evaluated at the point (x0, y0), the directional derivative Duf(x0, y0) is the result. As the point (x, y) starts from (x0, y0) and moves in the direction of the unit vector u, the directional derivative Duf(x0, y0) gives the instantaneous rate of change of the function f(x, y) in that direction. The directional derivative Duf(x0, y0) is also the slope of the tangent line at the point Q0 on the curve C within the surface f(x, y), where Q0 lies directly above the point (x0, y0). 2) How the dot product can be used to justify that the gradient points in the direction of maximum of a function. The directional derivative Duf(x0, y0) can be written as a dot product: Duf(x0, y0) = fx(x0, y0) u1 + fy(x0, y0) u2 = (fx(x0, y0) i + fy(x0, y0) j) (u1 i + u2 j) since the dot product results in the multiplied quantities in the i direction plus the multiplied quantities in the j direction. The vector (u1 i + u2 j) is just the unit vector u. The vector (fx(x0, y0) i + fy(x0, y0) j) is defined as the gradient of the function f, and is written as f(x0, y0). The directional derivative can then be written as the dot product of the gradient and the unit vector: Duf(x0, y0) = f(x0, y0) u To see that the gradient f(x0, y0) points in the direction of the maximum of the function f, first assume that f(x0, y0) is not equal to 0. If the gradient and a unit vector u make an angle , then the dot product of f(x0, y0) and u is: f(x0, y0) u = f(x0, y0) u cos by definition of the dot product. The length of the unit vector u is 1, so that f(x0, y0) u = f(x0, y0) cos so that the maximum of this quantity occurs at the maximum of cos , which is 1: max f(x0, y0) u = f(x0, y0) When cos = 1, the angle is 0, so that the gradient and the unit vector are in the same direction. Since f(x0, y0) u = Duf(x0, y0), when the dot product of the gradient and the unit vector is at a maximum, the directional derivative is also at a maximum. Therefore, max Duf(x0, y0) = f(x0, y0) and the gradient of a function at a point gives the maximum instantaneous rate of change of the function f, which is in the direction of u. Bibliography Anton, H. (1984) Calculus with analytic geometry. 2nd ed. New York: John Wiley & Sons. Edwards, B.H. and Larson, R.E. (1988) Elementary Linear Algebra. Lexington, MA: D.C. Heath and Company. Read More