Notes for Advanced Level Further Mathematics. (iii) =1, hence are parametric

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1 Hyperbolic Functions We define the hyperbolic cosine, sine tangent by also of course Notes for Advanced Level Further Mathematics, The following give some justification for the 'invention' of these functions. (i) a number of integrals which otherwise cannot be obtained, are expressible in terms of hyperbolic functions, (ii) A uniform chain hanging freely betweeen two fixed points takes the form of a CATENARY with equation where is a constant, referred to suitable axes. (iii) =1, hence are parametric equations for the hyperbola This last result indicates the close analogy with the circular functions In fact there is a hyperbolic identity corresponding to each trigonometric one. We have already seen that Consider now = = = So we have but OSBORN'S RULE enables us to write down any hyperbolic identity from the corresponding trigonometric one: Change cos to cosh, sin to sinh change the sign of any term involving the product of two sines. Example translates to Note that translates to Other hyperbolic identities are usually proved using the exponential definitions. Equations involving hyperbolic functions may be solved either by using similar methods to those used for trigonometric equations or by using the exponential definitions as in the following example. Solve the equation = but cannot be negative so the only solution is The Inverse Functions Consider the graphs of the hyperbolic functions which are readily obtained from the graphs of Red Blue Purple Green We can see from these graphs that is one-to-one with range domain both equal to We therefore define the inverse hyperbolic sine,, for all values of to be that value of such that. Thus, Since we must have it follows that only the positive root is acceptable so 1

2 for is not one-to-one unless we restrict the domain, the usual restriction being to allow only can then define to be the positive value of such that Thus. Now The domain of is for we must have so we cannot have. Hence is clearly one-to-one for all real so we define to be that value of such that for Graphs. it is easily shown that. We 2

3 Coordinate Geometry-The Ellipse In FP1 we said that with where is the focus is the directrix, defines an ellipse. Taking the focus to be the point the directrix to be the line (see diagram) we have Notes for Advanced Level Further Mathematics writing we have the usual stard equation of the ellipse with eccentricity given by Since this equation is symmetrical about the -axis it follows that there is another focus at directrix at another Summarising we have with, is the equation of an ellipse with major axis along the -axis, foci at directrices If then the major axis is along the -axis, foci are at directrices with The parametric equations of the ellipse are Hyperbola so equation of tangent at the normal will be by If then similar reasoning to that above leads to the equation with p[aramtric equations are It is not worth trying to memorise the equations of tangent normal but you must be able to work them out. is 3

4 Differentiation /Integration of Inverse Hyperbolic Trigonometric Functions We first consider the derivatives of the inverse trigonometric functions. (by implicit differentiation) so but, from the graph of we see that has a positive gradient throughout so we only require the positive root so Similarly, This time we require the negative root since the gradient of Finally is always negative. Reversing these results we have, More generally, by making the substitution we can show that putting we have Examples Differentiate (i) (i) (ii) (ii) (Chain rule) Evaluate (i) (ii) (i) (ii) The derivatives of the inverse hyperbolic functions are established in the same way are: So the general integral results are, with an added arbitrary constant in each case of course. Note however that for the last case it is usually preferable to use the logarithmic result obtained by using partial fractions. Harder Integrals We can now consider integrals of functions of the form where is a polynomial in. If is of equal or higher degree than the denominator then we perform division until we have a remainder of lower degree so we need only consider the case where If then the numerator is the derivative of the denominator we have a logarithmic integral. Otherwise we write giving a logarithmic integral an integral of the form. We therefore only have to consider. We do this by completing the square in the denominator to produce one of the following three cases 4

5 , Notes for Advanced Level Further Mathematics The first third may be split into partial fractions resulting in a pair of logarithmic integrals whilst the second gives an inverse tangent. Integrals of the form. Again by similar techniques as before we can reduce the problem to two types of integral. again, by completing the square inside the root we have or The first of these gives an inverse sine integral. For the second, by substituting so the integral becomes we have The third case gives an inverse hyperbolic cosine. =, Integration of inverse functions Some functions such as the inverse functions the logarithmic function can be integrated by a special application of integration by parts. Example Find We take so Similarly to integrate an inverse trig or hyperbolic function, take the function Reduction Formulae Some integrals, usually involving a power of, may be obtained by deriving a reduction formua, expressing the integral in terms of the integral with a lower power of, usually by one or more applications of integration by parts. Example Given that show that Using integration by parts with we have as required. Length of Arc Surface Area Consider the curve defined by If is a small element of the curve then the length of the curve between is given by or surface area of revolution by or For a rotation about the -axis you just interchange 5

6 Further Vector Geometry-The Vector Product The vector product a x b of two vectors at an angle to each other is defined to be the vector of magnitude in a direction perpendicular to the plane containing a b, such that a,b a x b form a right-hed set of vectors. an ordinary (right-hed) screw rotated from the direction of a to the direction of b would move in the direction of a x b. a x b where is a unit vector in the direction of a x b It follows that a x b measures the area of a parallelogram with adjacent sides at an angle of to each other. Hence the area of triangle ABC is given by a x b or b x c or c x a An immediate consequence of the definition is that a x b = b x a so vector multiplication is anticommutative.= a x b = 0 a = 0, b = 0 or a is parallel to b. In prticular a x a = 0, hence i x i = j x j = k x k = 0 while i x j = k, j x k = i, k x i = j, j x i = k, k x j = i k x i = j The distributive law a x (b + c) = a x b +a x c can be shown to hold so, in component form, with usual notation, a x b =( i + j k) x ( i j k) = ( i j k The following diagram is a convenient aid to memory The three triple products sloping upwards from left to right are positive the three sloping downwards are negative. Examples (i) Given p = i + j q = 2i j+2k, find the components of the vector p x q in the directions of i,j k. Find also the magnitude of the resolved part of p x q in the direction of i + j + k p x q = i + j + k =2i 2j k the components are thus 2i, j k The magnitude of the resolved part in the direction of is (p x q). hence resolved part of 2i 2j k in direction of i + j + k is (ii) Find a unit vector perpendicular to both a = 2i +2j k b = 4i +j k a x b i + j + k Thus are both perpendicular to a b so a unit vector is Applications of the Vector Product Line of intersection of two planes. The vector product of the normals to the planes gives us the direction vector for the line of intersection. Finding the coordinates of any one common point of the planes then enables us to write down the equation of the line of intersection. Distance of a point from a line To find the distance of the point P (x,y,z) from the line r =a + b is the direction vector of the line. Let be any point on the line, the required distance is PN The vector produce AP x AN = AP AN so AP x = PN Special Case (Two dimensions) Consider a point a line in the plane If the equation of the line is P is the point then we may take A to be b so AP x b = Hence, 6

7 Equation of a Plane Just as any straight line in a plane has an equation of the form so any plane in 3-dimensional space has an equation of the form with respect to mutually perpendicular axes. There are two vector forms for the equation of a plane. If A is any point in the plane with position vector a, b,c are the direction vectors of any two non-parallel lines in the plane then the position vector of any point of the plane is given by which is thus the equation of the plane. Alternatively, if is any point in the plane with position vector r = i+ j+ k, n = ai+bj+ck is any vector not in the plane then r.n=, so the equation of the plane may be written as r.n=constant. If is any point of the plane with position vector a, then putting r=a, the equation may be written r.n=a (r a).n. Now r a = AP, a vector in the plane, hence, n must be perpendicular to the plane. Thus the equation of any plane may be written in the form r.n = constant, where n is any vector perpendicular to the plane. If, in particular, n is a unit vector then we write r.n, we see that p is the perpendicular distance of the plane from the origin. It is usual to specify n so that p is positive. A consequence of the above reasoning is that we can easily write down a vector perpendicular to a plane from its cartesian equation. If the equation is then the vector ai j k is perpendicular to the plane. Conversely, if ai j k is perpendicular to a plane then the equation of the plane must be constant the constant is easily found if we know any one point in the plane. Example Find the distances of the points from the plane. Let be the foot of the perpendicular from to the plane let be any other point in the plane. Then = AP.n where n is a unit normal to the plane. Now i j k is a normal to the plane we may take to be (1,1,0) so AP i, Distance of from the plane is thus Similarly BP so distance of from the plane is The different signs tell us that are on opposite sides of the plane, with on the same side as the origin. The angle between two planes This will be the same as the angle between the normals to the planes. e.g the angle between the planes will be given by Example Find the coordinates of the foot of the perpendicular from the point to the plane The direction of the perpendicular is the direction of the normal to the plane, so the equation of the perpendicular is ) we simply have to solve this equation simultaneously with that of the plane. From the equation of the perpendicular we have so substituting in the equation of the plane, so the point is (6,1,1) Triple Products There are two types of triple product, a scalar valued one (a x b).c a vector valued one a x (b x c) We consider the first type. Consider a parallelepiped with edges a, b, c. Taking the plane defined by a b as base, the area of this base is where is the angle between a b. area of base = a x b n = a x b is perpendicular to the base so the height of the parallelepiped is c. so volume of parallelepiped is (a x b).c similarly, it is given by (b x c).a or (c x a).b Applications An immediate consequence is that if the scalar triple product is zero, then the three vectors are coplanar. 7

8 Note that the scalar product is zero if any two of the vectors are equal. Shortest Distance between two skew lines. ( lines which do not intersect) Let be any points on each of the lines. The shortest distance is the length of the common perpendicular, hence, since are right angles, is the projection of on this common perpendicular. The direction of is given by r x s where r s are the direction vectors of the two liines, hence, Distance of a point from a plane To find the distance of the point from the plane r = a has direction a x b (the normal to the plane) Note is positive if P is on the same side of the plane as the origin negative if on opposite side. If the equation of the plane is is the point a normal to the plane is so taking to be the point (0, 0, ) we have AP.PN hence,. Note similarity to result for distance of a point from a line in two dimensions. Alternative Vector Equation of a Line Consider the line through the point with position vector a in the direction of b, then if r is the position vector of any point on the line, r a will be parallel to b, hence (r a) x b which is an alternative to r = a b 8

9 Matrices Again Determinant of a Matrix Consider M = We form the determinant as follows: Form all products consisting of exactly one element from each row each column. There are 6 such products, each of the form where the 6 triads of suffixes are the 6 arrangements of (1 2 3) To each product we prefix a sign; this is to be when is a cyclic arrangement of (1 2 3), one of (1 2 3), (2 3 1), (3,1,2) is to be when is one of the remaining arrangements, one of (1 3 2), (2,1,3), (3,2,1). The sum of the 6 signed products is the required determinant. The following diagram will help you remember this As with the vector product the downward sloping diagonals are positive the upward sloping ones negative. The determinant may also be written as each element of the first row is multiplied by the determinant of the matrix obtained by eliminating the row column containing the top row element. The determinants obtained in theis way are called the MINORS of the corresponding elements. is the minor of, is the minor of etc. We may denote a minor by where it is the minor obtained by deleting the element in the row column. We can then write the matrix of minors if we now attach a or sign to each minor according to the pattern then we produce the COFACTORS., are called the CO-FACTORS of respectively, denoted by The transpose of the matrix of cofactors, namely is called the ADJUGATE of A denoted by adja. The inverse matrix is now given by adj Example Find the inverse of the matrix Matrix of minors is so Matrix of cofactors is the adjugate is the determinant is so inverse is Transformations of 3 dimensional space. Just as a matrix can be used to represent a transformation of the plane, so a matrix can be used to represent a transformation of space. We saw how, in the case the columns of the matrix were simply the images of the points (1,0) (0,1), in the same way the matrix for a 3-dimensional transformation has columns 9

10 which are the images of the points ) respectively. Eigenvalues Eigenvectors An EIGENVECTOR of a matrix A is a non-zero column vector x that satisfies Ax = x for some scalar is called an EIGENVALUE of the matrix corresponding to the eigenvector x. Eigenvectors are also known as characteristic vectors, or latent vectors. Immediately from the definition we have Ax x Ix (A I)x As by definition x is non-zero, we must have det(a I)=0 which is called the CHARACTERISTIC EQUATION of the matrix A. its solutions for are the eigenvalues. Example Find the eigenvalues of the matrix find the corresponding eigenvectors. The characteristic equation is (i) So equating corresponding elements we require Let then so a suitable eigenvector is (ii) so whilst can take any value so an eigenvector is (iii) Taking we have so an eigenvector is Orthogonal, Symmetric Diagonal Matrices A matrix M is orthogonal if it follows immediately that M A matrix M is symmetric if M A matrix M is diagonal if all elements not on the leading diagonal are zero. Two eigenvectors are said to be orthogonal if their scalar product is zero. A symmetric matrix A can be diagonalised as follows: (a) Find normalised ( unit) eigenvectors of A (b) Form a matrix P with thgese eigenvectors as its columns. (c) Obtain the transpose of P. AP = D is the required diagonal matrix. Example Reduce the matrix A to diagonal form We find the eigenvalues, characteristic equation is so an eigenvectort is so an eigenvector is normalised eigenvectors are thus where 10

11 So P = so D Notes for Advanced Level Further Mathematics Note that the diagonal elements are the eigenvalues. 11