Pure Further Mathematics 3. Revision Notes

Transcription

1 Pure Further Mathematics Revision Notes February 6

2 FP FEB 6 SDB

3 Hyperbolic functions... Definitions and graphs... Addition formulae, double angle formulae etc.... Osborne s rule... Inverse hyperbolic functions... 4 Graphs... 4 Logarithmic form... 4 Equations involving hyperbolic functions... 5 Further coordinate systems... 6 Ellipse... 6 Hyperbola... 6 Parabola... 7 Parametric differentiation... 7 Tangents and normals... 8 Finding a locus... 9 Differentiation... Derivatives of hyperbolic functions... Derivatives of inverse functions... 4 Integration... Standard techniques... Recognise a standard function... Using formulae to change the integrand... Reverse chain rule... Standard substitutions... Integration inverse functions and ln x... 4 Reduction formulae... 4 Arc length... 8 Area of a surface of revolution Vectors... Vector product... The vectors i, j and k... Properties... Component form... Applications of the vector product... Volume of a parallelepiped... 4 Triple scalar product... 4 Volume of a tetrahedron... 5 Equations of straight lines... 5 Vector equation of a line... 5 Cartesian equation of a line in -D... 6 Vector product equation of a line... 6 Equation of a plane... 7 Scalar product form... 7 Cartesian form... 7 Vector equation of a plane... 8 Distance from a point to a plane... 9 Distance from any point to a plane... 9

4 Reflection of a point in a plane... Distance between parallel planes... Shortest distance from a point to a line... Projections an alternative approach... Shortest distance from a point from a plane.... Distance between parallel planes... Shortest distance between two skew lines... 4 Shortest distance from a point to a line... 5 Line of intersection of two planes... 6 Angle between line and plane... 7 Angle between two planes Matrices... 9 Basic definitions... 9 Dimension of a matrix... 9 Transpose and symmetric matrices... 9 Identity and zero matrices... 9 Determinant of a matrix... 9 Properties of the determinant... 4 Singular and non-singular matrices... 4 Inverse of a matrix... 4 Cofactors... 4 Finding the inverse... 4 Properties of the inverse... 4 Matrices and linear transformations... 4 Linear transformations... 4 Base vectors i, j, k... 4 Image of a line Image of a plane Image of a plane Method Method, as in the book Eigenvalues and eigenvectors Definitions matrices Orthogonal matrices Normalised eigenvectors Orthogonal vectors Orthogonal matrices Diagonalising a matrix Diagonalising symmetric matrices Eigenvectors of symmetric matrices Diagonalising a symmetric matrix matrices... 5 Finding eigenvectors for matrices Diagonalising symmetric matrices... 5 FP FEB 6 SDB

5 Hyperbolic functions Definitions and graphs sinh x = (ex e x ) cosh x = (ex + e x ) tanh x = sinh x = (ex e x ) cosh x (e x +e x ) 4 You should be able to draw the graphs of cosech x, sech x and coth x from the above: cosech x sech x coth x Addition formulae, double angle formulae etc. The standard trigonometric formulae are very similar to the hyperbolic formulae. Osborne s rule If a trigonometric identity involves the product of two sines, then we change the sign to write down the corresponding hyperbolic identity. Examples: sin(a + B) = sin A cos B + cos A sin B sinh(a + B) = sinh A cosh B + cosh A sinh B no change but cos(a + B) = cos A cos B sin A sin B cosh(a + B) = cosh A cosh B + sinh A sinh B product of two sines, so change sign and + tan A = sec A tanh A = sech A tan A = sin A, product of two sines, so change sign cos A FP FEB 6 SDB

6 Inverse hyperbolic functions Graphs Remember that the graph of y = f (x) is the reflection of y = f(x) in y = x. y = arsinh x y = arcosh x y = artanh x 4 y sinh x arsinh x y cosh x arcosh x x y artanh x tanh x Notice arcosh x is a function defined so that arcosh x. there is only one value of arcosh x. However, the equation cosh z =, has two solutions, +arcosh and arcosh. Logarithmic form ) y = arsinh x sinh y = (ey e y ) = x e y xe y = e y = x± 4x +4 = x + x + or x x + But e y > and x x + < e y = x + x + only y = arsinh x = lnx + x + ) y = arcosh x cosh y = (ey + e y ) = x e y xe y + = e y = x± 4x 4 = x ± x both roots are positive y = arcosh x = lnx + x or lnx x It can be shown that lnx x = lnx + x y = arcosh x = ± lnx + x But arcosh x is a function and therefore has only one value (positive) y = arcosh x = lnx + x ( x ) ) Similarly artan x = ln +x ( x < ) x 4 FP FEB 6 SDB

7 Equations involving hyperbolic functions It would be possible to solve 6 sinh x cosh x = 7 using the R sinh(x α) technique from trigonometry, but it is easier to use the exponential form. Example: Solve 6 sinh x cosh x = 7 Solution: 6 sinh x cosh x = 7 6 (ex e x ) (ex + e x ) = 7 e x 7e x 4 = (e x + )(e x 4) = e x = (not possible) or 4 x = ln 4 In other cases, the trigonometric solution may be preferable Example: Solve cosh x + 5 sinh x 4 = Solution: cosh x + 5 sinh x 4 = + sinh x + 5 sinh x 4 = note use of Osborn s rule sinh x + 5 sinh x = (sinh x )(sinh x + ) = sinh x = or x = arsinh.5 or arsinh ( ) x = ln or ln( ) + ( ) + using log form of inverse x = ln + 5 or ln FP FEB 6 SDB 5

8 Further coordinate systems Ellipse Cartesian equation x a + y b = a b y a x Parametric equations b x = a cos θ, y = b sin θ y Foci at S (ae, ) and S ( ae, ) Directrices at x = ± a e ae ae x Eccentricity e <, b = a ( e ) x = a e x = a e An ellipse can be defined as the locus of a point P which moves so that PS = e PN, where S is the focus, e < and N lies on the directrix. y S P N x Hyperbola x = a e Cartesian equation x a y b = y y = b a x Parametric equations x = a cosh θ, y = b sinh θ (x = a sec θ, y = b tan θ also work) a a x Asymptotes x a = ± y b y = b a x Foci at S (ae, ) and S ( ae, ) Directrices at x = ± a e Eccentricity e >, b = a (e ) N P A hyperbola can be defined as the locus of a point P which moves so that PS = e PN, where S is the focus, e > and N lies on the directrix. S 6 FP FEB 6 SDB

9 y c x d = is a hyperbola with foci on the y-axis, x = a e Parabola Cartesian equation y = 4ax Parametric equations N P x = at, y = at Focus at S (a, ) Directrix at x = a Ṡ A parabola can be defined as the locus of a point P which moves so that PS = PN, where S is the focus, N lies on the directrix and eccentricity e =. x = a Parametric differentiation From the chain rule dy = dy dθ dθ dy = dy dθ dθ or dy = dy dt dt using any parameter FP FEB 6 SDB 7

10 Tangents and normals It is now easy to find tangents and normals. Example: Find the equation of the normal to the curve given by the parametric equations x = 5 cos θ, y = 8 sin θ at the point where θ = π Solution: When θ = π, cos θ = and sin θ = x = 5, y = 4 and dy dy = dθ dθ = 8 cos θ = 8 5 sin θ 5 when θ = π gradient of normal is 5 8 equation of normal is y 4 = 5 8 x 5 5 x 8y = 7 Sometimes normal, or implicit, differentiation is (slightly) easier. Example: Find the equation of the tangent to xy = 6, or x = 6t, y = 6, at the point t where t =. Solution: When t =, x = 8 and y =. dy can be found in two (or more!) ways: dy = dy dt dt dy = = 6t 6 =, t when t = 9 xy = 6 y = 6 dy dy = 6 x = 6 =, 8 when x = 8 9 x equation of tangent is y = 9 (x 8) x + 9y 6 = 8 FP FEB 6 SDB

11 Finding a locus First find expressions for x and y coordinates in terms of a parameter, t or θ, then eliminate the parameter to give an expression involving only x and y, which will be the equation of the locus. Example: The tangent to the ellipse x 9 + y 6 crosses the x-axis at A, and the y-axis at B. =, at the point P, ( cos θ, 4 sin θ ), Find an equation for the locus of the mid-point of AB as P moves round the ellipse, or as θ varies. Solution: dy dy = dθ dθ = 4 cos θ sin θ equation of tangent is y 4 sin θ = 4 cos θ sin θ (x cos θ) y sin θ + 4x cos θ = cos θ + sin θ = Tangent crosses x-axis at A when y =, x = cos θ, and crosses y-axis at B when x =, y = 4 sin θ mid-point of AB is cos θ, 4 sin θ Here x = cos θ and y = sin θ cos θ = x and sin θ = y equation of the locus is 9 4x + 4 y = since cos θ + sin θ = FP FEB 6 SDB 9

12 Differentiation Derivatives of hyperbolic functions y = sinh x = (ex e x ) dy = (ex + e x ) = cosh x and, similarly, d(cosh x) = sinh x Also, y = tanh x = sinh x cosh x dy cosh x cosh x sinh x sinh x = cosh x = cosh x = sech x In a similar way, all the derivatives of hyperbolic functions can be found. f(x) sinh x cosh x tanh x f (x) cosh x sinh x sech x all positive coth x cosech x sech x cosech x cosech x coth x sech x tanh x all negative Notice: these are similar to the results for sin x, cos x, tan x etc., but the minus signs do not always agree. The minus signs are wrong only for cosh x and sech x = cosh x. Derivatives of inverse functions y = arsinh x sinh y = x cosh y dy = dy = = cosh y +sinh y d(arsinh x) = +x FP FEB 6 SDB

13 The derivatives for other inverse hyperbolic functions can be found in a similar way. You can also use integration by substitution to find the integrals of the f (x) column f(x) f (x) substitution needed for integration arcsin x arccos x arctan x arsinh x arcosh x artanh x x sin u = cos u use x = sin u x cos u = sin u use x = cos u +x + tan u = sec u use x = tan u +x + sinh u = cosh u use x = sinh u x x ln +x x x cosh u = sinh u use x = cosh u tanh u = sech u use x = tanh u partial fractions, see example below Note that x = +x + = x ln +x x + c With chain rule, product rule and quotient rule you should be able to handle a large variety of combinations of functions. FP FEB 6 SDB

14 4 Integration Standard techniques Recognise a standard function Examples: sec x tan x = sec x + c sech x tanh x = sech x + c Using formulae to change the integrand Examples: tan x = sec x = tan x x + c cos x = + cos x = x + sin x + c sinh x = cosh x = sinh x x + c Reverse chain rule Notice the chain rule pattern, guess an answer and differentiate to find the constant. Example: cos x sin x looks like u u x so try u cos x dcos x = cos x ( sin x) = cos x sin x so divide by cos x sin x = cos x + c Example: x (x 7) 4 looks like u 4 u x so try u5 (x 7) = 5(x 7) 4 6x = (x 7) 4 so divide by x (x 7) 4 = (x 7) 5 + c Example: sech 4 x tanh x = sech x (sech x tanh x) looks like u u x so try u4 = sech 4 x dsech 4 x = 4 sech x sech x tanh x so divide by 4 sech 4 x tanh x = 4 sech4 x + c FP FEB 6 SDB

15 Standard substitutions a +b x a +b x a b x b x a bx = a tan u better than bx = a sinh u when there is no bx = a sinh u better than bx = a tan u when there is bx = a tanh u or use partial fractions bx = a cosh u better than bx = a sec u when there is For more complicated integrals like or px +qx+r px +qx+r complete the square to give p(x + a) + b and then use a substitution similar to one of the four above. Example: 4x 8x 5 4x 8x 5 = 4(x x + ) 9 = 4(x ) 9 = 4(x ) 9 Substitute (x ) = cosh u = sinh u du = 9(cosh u ) sinh u du = du = u + c = arcosh x + c Important tip xn a ±x is best done with the substitution u (or u ) = a ± x, when n is odd, or a trigonometric or hyperbolic function when n is even. FP FEB 6 SDB

16 Integration inverse functions and ln x To integrate inverse trigonometric or hyperbolic functions and ln x we use integration by parts dv with = Example: Find arctan x Solution: I = arctan x take u = arctan x du = +x and dv = v = x I = x arctan x x +x Example: I = arctan x = x arctan x Find arcosh x ln ( + x ) + c Solution: I = arcosh x take u = arcosh x du = x and dv = v = x I = x arcosh x x x I = arcosh x = x arcosh x x + c Reduction formulae The first step in finding a reduction formula is usually often integration by parts (sometimes twice). The following examples show a variety of techniques. Example : I n = x n e x. (a) Find a reduction formula, (b) Find I, and (c) find I 4 Solution: (a) Integrating by parts u = x n du = nxn and dv = ex v = e x I n = x n e x nx n e x I n = x n e x ni n 4 FP FEB 6 SDB

17 (b) (c) I = e x = e x + c Using the reduction formula I 4 = x 4 e x 4I = x 4 e x 4(x e x I ) = x 4 e x 4x e x + (x e x I ) = x 4 e x 4x e x + x e x 4(xe x I ) = x 4 e x 4x e x + x e x 4xe x + 4e x + c since I = e x + c Example : Find a reduction formula for I n = π sin n x. Use the formula to find I 6 = π sin 6 x π Solution: I n = sin n x π = sin n x sin x take u = sin n x du = (n ) sin n x cos x and dv = sin x v = cos x I n = [ cos x sin n π x] π cos x (n ) sin n x cos x π = + (n ) cos x sin n x π = (n ) sin x sin n x π π = (n ) sin n x (n ) sin n x I n = (n ) I n (n ) I n I n = n n I n. Now I 6 = 5 6 I 4 = I = I I 6 = 5 π 6 = 5π. FP FEB 6 SDB 5

18 Example : Find a reduction formula for I n = sec n x. Solution: I n = sec n x = sec n x sec x take u = sec n x and dv du = sec x v = tan x = (n ) secn x sec x tan x I n = sec n x tan x tan x (n ) sec n x sec x tan x = sec n x tan x (n ) tan x sec n x = sec n x tan x (n ) (sec x ) sec n x = sec n x tan x (n )I n + (n )I n (n )I n = sec n x tan x + (n )I n Example 4: Solution: Find a reduction formula for I n = tan n x. I n = tan n x = tan n x tan x I n = tan n x (sec x ) = tan n x sec x tan n x I n = n tann x I n. Example 5: Find a reduction formula for I n = x n ( + x). Solution: I n = x n ( + x) take u = x n du = nxn and dv = ( + x) v = ( + x) I n = x n ( + x) nx n I n = n xn I n = n xn I n = n I n n I n ( + x) ( + x) ( + x) n xn ( + x). ( + x) n+ I n = I n = n I n n I n+ n 6 FP FEB 6 SDB

19 π Example 6: Find a reduction formula for I n = x n cos x π Solution: I n = x n cos x take u = x n du = nxn I n = [x n π sin x] and π n sin x x n dv = cos x v = sin x take u = x n du = (n ) xn and dv = sin x v = cos x I n = π n n{[ x n π ( cos x)] π cos x (n )x n } I n = π π n n{ + (n ) x n cos x } I n = π n n (n ) I n Example 7: Find a reduction formula for I n = sin nx sin x Solution: I n = sin[(n )x+x] sin x = sin(n )x cos x +cos(n )x sin x sin x = sin(n )x ( sin x) + cos(n )x sin x cos x sin x = sin(n )x sin x + cos(n )x cos x sin(n )x sin x = I n + cos(n )x using cos(a + B) = cos A cos B sin A sin B I n = I n + n sin(n )x. FP FEB 6 SDB 7

20 Arc length All the formulae you need can be remembered from this diagram arc PQ line segment PQ Q (δ s) (δ x) + (δ y) δs δx + δy δx P δ x δ s δ y and as δ x ds = + dy ds = + dy arc length = s = + dy Similarly δs δy δx δy + s = dy + dy and δs δt δx δt + δy δt s = dt + dy dt dt or s = x + y dt for parametric equations. Example : Find the length of the curve y = x, from the point where x = to the point where x = 8. 5 Solution: The equation of the curve is in Cartesian form so we use s = + dy. y = x 8 s = + x dy = x 5 5 = ( + x) 8 = (9) (4) s =. 8 FP FEB 6 SDB

21 Example : Find the length of one arch of the cycloid x = a(t sin t), y = a( cos t). Solution: The curve is given in parametric form so we use s = x + y dt. x = a(t sin t), y = a( cos t) t= t=π dt = a( cos t), and dy dt = a sin t x + y = a ( cos t + cos t + sin t) = a ( cos t) x + y = a sin t = a sin t π s = a sin t dt s = 4a cos t π = 4a 4a = 8a. Area of a surface of revolution A curve is rotated about the x-axis. To find the area of the surface formed between x = a and x = b, we consider a small section of the curve, δs, at a distance of y from the x-axis. y y δ s x a b When this small section is rotated about the x-axis, the shape formed is approximately a cylinder of radius y and length δs. The surface area of this (cylindrical) shape π rl π yδs The total surface area b a π yδs b and, as δs, the area of the surface is A = π y a ds. And so b A = π y a ds b or A = π y a ds dt dt We can use ds = + dy or ds dt = dt + dy dt, as appropriate, remembering that (δ s) (δ x) + (δ y) FP FEB 6 SDB 9

22 Example : Find the surface area of a sphere with radius r,, between the planes x = a and x = b. Solution: The Cartesian form is most suitable here. b A = π y a x + y = r ds x=a x=b x + y dy = dy = x y and ds = + dy b A = πy a + x y = b π a y + x b = πr a since x + y = r A = [πrx] a b = πr(b a) since r is constant Notice that the area of the whole sphere is from a = r to b = r giving surface area of a sphere is 4π r. Historical note. Archimedes showed that the area of a sphere is equal to the area of the curved surface of the surrounding cylinder. Thus the area of the sphere is h = r A = π rh = 4π r since h = r. r FP FEB 6 SDB

23 Example : The parabola, x = at, y = at, between the origin (t = ) and P (t = ) is rotated about the x-axis. Find the surface area of the shape formed. y P, (t = ) Solution: The parametric form is suitable here. b A = π y a ds dt dt x and ds dt = dt + dy dt = at and dy = a dt dt ds dt = (at) + (a) = a t + A = π at a t + dt = 8πa (t + ) A = 8πa 5 FP FEB 6 SDB

24 5 Vectors Vector product The vector, or cross, product of a and b is a b = ab sinθ n where n is a unit (length ) vector which is perpendicular to both a and b, and θ is the angle between a and b. n θ b a The direction of n is that in which a right hand corkscrew would move when turned through the angle θ from a to b. Notice that b a = ab sinθ n, where n is in the opposite direction to n, since the corkscrew would move in the opposite direction when moving from b to a. Thus b a = a b. n θ b a The vectors i, j and k For unit vectors, i, j and k, in the directions of the axes i j = k, j k = i, k i = j, k i k = j, j i = k, k j = i. j i Properties a a = since θ = a b = a is parallel to b since sin θ = θ = or π or a or b = a (b + c) = a b + a c a b is perpendicular to both a and b remember the brilliant demo with the straws! from the definition FP FEB 6 SDB

25 Component form Using the above we can show that a b a b a b i j k a b = a b = a b + a b = a a a a b a b a b b b b Applications of the vector product Area of triangle OAB = ab sin θ b B area of triangle OAB = a b O θ a A Area of parallelogram OADB is twice the area of the triangle OAB area of parallelogram OADB = a b O θ b a B A D Example: A is (,, ), B is (,, ) and C is (, 4, ). Find the area of the triangle ABC. C Solution: The area of the triangle ABC = AB AC AB = b a = and AC = c a = i j k AB AC = = A B area ABC = = 5 FP FEB 6 SDB

26 Volume of a parallelepiped n In the parallelepiped h = a cos φ and area of base = bc sin θ volume = h bc sin θ φ h a θ c b = a bc sin θ cos φ and (i) (ii) φ is the angle between n and a bc sin θ is the magnitude of b c a. (b c) = a bc sin θ cos φ volume of parallelepiped = a. (b c) Triple scalar product a b c b c a. (b c) = a. b c + b c a b c b c = a (b c b c ) + a ( b c + b c ) + a (b c b c ) a a a = b c b c b c By expanding the determinants we can show that a. (b c) = (a b). c keep the order of a, b, c but change the order of the and. For this reason the triple scalar product is written as {a, b, c} {a, b, c} = a. (b c) = (a b). c It can also be shown that a cyclic change of the order of a, b, c does not change the value, but interchanging two of the vectors multiplies the value by. {a, b, c} = {c, a, b} = {b, c, a } = {a, c, b} = {c, b, a} = {b, a, c} 4 FP FEB 6 SDB

27 Volume of a tetrahedron The volume of a tetrahedron is Area of base h The height of the tetrahedron is the same as the height of the parallelepiped, but its base has half the area h a c b volume of tetrahedron = 6 volume of parallelepiped volume of tetrahedron = 6 {a, b, c} Example: Find the volume of the tetrahedron ABCD, given that A is (,, ), B is (,, ), C is (,, ) and D is (4,, ). Solution: Volume = AD, AC, AB 6 AD = d a =, AC =, 5 AB = AD,, AC AB = 5 = + = volume of tetrahedron is 6 = 5 Equations of straight lines Vector equation of a line r = a + λ b is the equation of a line through the point A and parallel to the vector b, A P l x l α or y = m + λ β. z n γ a r b FP FEB 6 SDB 5

28 Cartesian equation of a line in -D Eliminating λ from the above equation we obtain x l α = y m β = z n γ (= λ) α is the equation of a line through the point (l, m, n) and parallel to the vector β. γ This strange form of equation is really the intersection of the planes x l α = y m β and y m β = z n γ and x l α = z n γ. Vector product equation of a line AP = r a and is parallel to the vector b A P l AP b = (r a) b = is the equation of a line through A and parallel to b. a r or r b = a b = c is the equation of a line parallel to b. b Notice that all three forms of equation refer to a line through the point A and parallel to the vector b. Example: A straight line has Cartesian equation x = y+4 5 = z. Find its equation (i) in the form r = a + λ b, (ii) in the form r b = c. Solution: First re-write the equation in the standard manner x = y.5 = z the line passes through A, (,, ), and is parallel to b,.5 or FP FEB 6 SDB

29 (i) (ii) r = + λ 5 4 r 5 4 = r r = 7 = i j k = = 6 4 Equation of a plane Scalar product form n A P Let n be a vector perpendicular to the plane π. π a Let A be a fixed point in the plane, and P be a general point, (x, y, z), in the plane. r Then is parallel to the plane, and therefore perpendicular to n O AP. n = (r a). n = r. n = a. n = a constant, d r. n = d is the equation of a plane perpendicular to the vector n. Cartesian form If n = α β γ x α then r. n = y. β = α x + β y + γ z z γ α x + β y + γ z = d is the Cartesian equation of a plane perpendicular to α β. γ FP FEB 6 SDB 7

30 Example: Find the scalar product form and the Cartesian equation of the plane through the points A, (,, 5), B, (,, ) and C, (,, ). Solution: We first need a vector perpendicular to the plane. A, (,, 5), B, (,, ) and C, (,, ) lie in the plane AB = 4 and AC = 7 AB AC is perpendicular to the plane are parallel to the plane AB AC = i j k 4 = 6 7 = 6 using smaller numbers 6x y + z = d but A, (,, 5) lies in the plane d = = Cartesian equation is 6x y + z = and scalar product equation is r. 6 =. Vector equation of a plane A P r = a + λ b + µ c is the equation of a plane, π, through A and parallel to the vectors b and c. π a r b O c Example: Find the vector equation of the plane through the points A, (, 4, ), B, (, 5, ) and C, (4, 7, ). Solution: We want the plane through A, (, 4, ), parallel to AB = and AC = 5 4 vector equation is r = 4 + λ + μ FP FEB 6 SDB

31 Distance from a point to a plane Example: Find the distance from the point P (,, 5) to the plane 4x y + z =. Solution: Let M be the foot of the perpendicular from P to the plane. The distance of the origin from the plane is PM. We must first find the intersection of the line PM with the plane. PM is perpendicular to the plane and so is parallel to n = 4. M P n the line PM is r = 5 + λ 4 + 4λ = λ, 5 + λ and the point of intersection of PM with the plane is given by 4( +4λ) ( λ) + (5 + λ) = 8 + 6λ 9 + 9λ λ = λ = 69 PM = 4 69 distance = PM = = The distance of the P from the plane is. Distance from any point to a plane The above technique can be used to find the formula:- distance, s, from the point P (α, β, γ) to the plane M ax + by + cz d = is given by aα + bβ + cγ d s = a + b + c P n FP FEB 6 SDB 9

32 Reflection of a point in a plane Example: Find the reflection of the point A (,, 7) in the plane π, r. = 7. Solution: Find the point of intersection, P, of the line through A and perpendicular to π with the plane π. Then find AP, to give OA = OA + AP. Line through A perpendicular to π is r = + λ 7 A This meets the plane π when (+λ) ( λ) + (7+λ) = 7 + 9λ + 4λ λ = 7 π P λ = OP = + ( ) 7 6 AP = OP OA = ( ) = 4 6 OA = OA + AP = + 4 = 9 7 A the reflection of A is A, (, 9, ) FP FEB 6 SDB

33 Distance between parallel planes Example: Find the distance between the parallel planes π : x 6y + z = 9 and π : x 6y + z = 5 Solution: Take any point, P, on one of the planes, and then use the above formula for the shortest distance, PQ, between the planes. P Q π π By inspection the point P (,, ) lies on π aα + bβ + cγ d shortest distance s from P to the plane π is a + b + c shortest distance s = = 4 7 The distance between the planes is 4 7. Shortest distance from a point to a line Example: Find the shortest distance from the point P (,, 4) to the line l, r = + λ 6 P Solution: Any plane x y + 6z = d must be perpendicular to the line l. If we make this plane pass through P and if it meets the line l in the point X, then PX must be perpendicular to the line l, and so PX is the shortest distance from P to the line l. Plane passes through P (,, 4) x y + 6z = ( ) = 6 x y + 6z = 6 l X l meets plane ( + λ) ( λ) + 6(6λ) = λ 9 + 9λ + 6λ = 6 λ = X is the point (,, 6) PX = = 6 4 shortest distance is PX = + + = 7 FP FEB 6 SDB

34 Projections an alternative approach Imagine a light bulb causing a rod, AB, to make a shadow, A B, on the line l. If the light bulb is far enough away, we can think of all the light rays as parallel, and, if the rays are all perpendicular to the line l, the shadow is the projection of the rod onto l (strictly speaking an orthogonal projection). l A A θ B n B The length of the shadow, B A, is BA cos θ = BA. n, where n is a unit vector parallel to the line l. Modulus signs are needed in case n is in the opposite direction. Shortest distance from a point from a plane. π To find AM, the shortest distance from A to the plane π, For any point, B, on π AM is the projection of AB onto the line AM AM = AB. n Example: Find the shortest distance from the point A (,, 5) to the plane 4x y + z =. B M n A Solution: By inspection B (, 7, ) lies on the plane AB = 7 = n = n = = shortest distance = AB. n =. 5 4 = FP FEB 6 SDB

35 Distance between parallel planes Example: Find the distance between the parallel planes π π : x 6y + z = 9 and π : x 6y + z = 5 A X Solution: Take any point, B, on one of the planes, π, and then consider the line BX perpendicular to both planes; BX is then the shortest distance between the planes. Then choose any point, A, on π, and BX is now the projection of AB onto BX B n π shortest distance = BX = AB. n or shortest distance = b a. n, for any two points A and B, one on each plane, where n is a unit vector perpendicular to both planes. By inspection the point A (,, ) lies on π, and the point B ( 5,, ) lies on π 5 5 AB = = n = 6 n = = 7 5 shortest distance =. 7 6 = 4 7 FP FEB 6 SDB

36 Shortest distance between two skew lines It can be shown that there must be a line joining two skew lines which is perpendicular to both lines. This line is XY and is the shortest distance between the lines. Y A l r = a + λ b The vector n = b d is perpendicular to both lines X C l r = c + µ d the unit vector n = b d b d Now imagine two parallel planes π and π, both perpendicular to n, one containing the line l and the other containing the line l. A and C are points on l and l, and therefore on π and π. We now have two parallel planes with two points, A and C, one on each plane, and the planes are both perpendicular to n. As in the example for the distance between parallel planes, the shortest distance d = AC. n d = (c a). b d b d This result is not in your formula booklet, SO LEARN IT please 4 FP FEB 6 SDB

37 Shortest distance from a point to a line In trying to find the shortest distance from a point P to a line l, r = a + λb, we do not know n, the direction of the line through P perpendicular to l. Some lateral thinking is needed. P l We do know A, a point on the line, and b, the direction of the line l AP. b = AX, the projection of AP onto l and we can now find PX = AP AX A X using Pythagoras Example: Find the shortest distance from the point P (,, 4) to the line l, r = + λ 6 Solution: If l is r = a + λb, then a = and b = 6 b = = 7, b = 5 and AP = = AX = AP. b = = +5+4 = PX = AP AX = ( ) 7 = 7 FP FEB 6 SDB 5

38 Line of intersection of two planes Example: Find an equation for the line of intersection of the planes x + y + z = 4 I and x y + z = 4 II Solution: Eliminate one variable I + II x + 5z = 8 We are not expecting a unique solution, so put one variable, z say, equal to λ and find the other variables in terms of λ. z = λ x = 8 5λ I or y = 4 x z = 4 8 5λ x y = z x y = z λ λ λ = 4 λ making the numbers nicer in the direction vector only which is the equation of a line through 8, 4, and parallel to 5. 6 FP FEB 6 SDB

39 Angle between line and plane Let the acute angle between the line and the plane be φ. First find the angle between the line and the normal vector, θ. There are two possibilities as shown below: l l n θ φ θ φ n (i) n and the angle φ are on the same side of the plane φ = 9 θ (ii) n and the angle φ are on opposite sides of the plane φ = θ 9 Example: Find the angle between the line x+ and the plane x + y 7z = 5. = y = z Solution: The line is parallel to, and the normal vector to the plane is. 7 a. b = ab cos θ = cos θ cos θ = 7 6 θ = 7. o the angle between the line and the plane, φ = 9 7. = 6.7 o FP FEB 6 SDB 7

40 Angle between two planes n n If we look end-on at the two planes, we can see that the angle between the planes, θ, equals the angle between the normal vectors. θ θ Example: Find the angle between the planes plane plane x + y + z = 5 and x + y + z = 7 Solution: The normal vectors are and a. b = ab cos θ = cos θ cos θ = 4 θ = 44.4 o 8 FP FEB 6 SDB

41 6 Matrices Basic definitions Dimension of a matrix A matrix with r rows and c columns has dimension r c. Transpose and symmetric matrices The transpose, A T, of a matrix, A, is found by interchanging rows and columns a b c a d g A = d e f A T = b e h g h i c f i (AB) T = B T A T - note the change of order of A and B. A matrix, S, is symmetric if the elements are symmetrically placed about the leading diagonal, or if S = S T. a b c Thus, S = b d e is a symmetric matrix. c e f Identity and zero matrices The identity matrix I = and the zero matrix is = Determinant of a matrix The determinant of a matrix, A, is a b c det (A) = = d e f = a e f f e b d + c d g h i h i g i g h = aei afh bdi + bfg + cdh ceg FP FEB 6 SDB 9

42 Properties of the determinant ) A determinant can be expanded by any row or column using a b c e.g = d e f = d b c c b + e a h i g i f a g h. g h i ) Interchanging two rows changes the sign of the determinant using the middle row and leaving the value unchanged a b c d e f d e f = a b c g h i g h i which can be shown by evaluating both determinants ) A determinant with two identical rows (or columns) has value. a b c = a b c interchanging the two identical rows gives = = g h i 4) det(ab) = det(a) det(b) this can be shown by multiplying out Singular and non-singular matrices A matrix, A, is singular if its determinant is zero, det(a) = A matrix, A, is non-singular if its determinant is not zero, det(a) Inverse of a matrix This is tedious, but no reason to make a mistake if you are careful. Cofactors a b c In d e f the cofactors of a, b, c, etc. are A, B, C etc., where g h i A = + e f f e, B = d, C = +d h i g i g h, D = b c c b, E = + a h i g i F = a g h, G = + b c c, H = a e f d f, I = + a b d e These are the matrices used in finding the determinant, together with the correct sign from FP FEB 6 SDB

43 Finding the inverse ) Find the determinant, det(a). If det(a) =, then A is singular and has no inverse. A B C ) Find the matrix of cofactors C = D E F G H I A D G ) Find the transpose of C, C T = B E H C F I 4) Divide C T by det(a) to give A = See example on page 48. A D G det (A) B E H C F I Properties of the inverse ) A A = AA = I ) (AB) = B A - note the change of order of A and B. Proof (AB) AB = I from definition of inverse (AB) AB (B A ) = I (B A ) (AB) A (BB )A = B A (AB) A I A = B A (AB) A A = B A (AB) = B A ) det(a ) = det (A) FP FEB 6 SDB 4

44 Matrices and linear transformations Linear transformations T is a linear transformation on a set of vectors if (i) T (x + x ) = T (x ) + T (x ) for all vectors x and y (ii) T (kx) = kt (x) for all vectors x Example: x Show that T y z x = x + y z is a linear transformation. x Solution: (i) T (x + x ) = T y z + x y z = T (x + x ) x x = x + x + y + y = x +y + x +y z z z T (x + x ) = T (x ) + T (x ) x + x y + y z + z z = T (x ) + T (x ) (ii) x kx kx T (kx) = T k y = T ky = kx + ky z kz kz T (kx) = kt (x) = k x x +y z = kt (x) Both (i) and (ii) are satisfied, and so T is a linear transformation. All matrices can represent linear transformations. Base vectors i, j, k i =, j =, k = Under the transformation with matrix a d g b e h c f i a b c d e f g h i the first column of the matrix the second column of the matrix the third column of the matrix This is an important result, as it allows us to find the matrix for given transformations. 4 FP FEB 6 SDB

45 Example: Find the matrix for a reflection in the plane y = x Solution: The z-axis lies in the plane y = x so the third column of the matrix is Also the first column of the matrix is the second column of the matrix is the matrix for a reflection in y = x is. Example: Find the matrix of the linear transformation, T, which maps (,, ) (, 4, ), (,, ) (6,, 5) and (,, 4) (,, ). Solution: Firstly 4 first column is 4 Secondly 6 5 but = T 6 T = 4 = 5 second column is Thirdly but 4 4 = T = 4T T = third column is T = 4. FP FEB 6 SDB 4

46 Image of a line Example: Find the image of the line r = where T = 4. + λ under T, Solution: As T is a linear transformation, we can find T (r) = T + λ = T + λt T (r) = 4 + λ 4 T (r) = r = + λ + λ and so the vector equation of the new line is Image of a plane Similarly the image of a plane r = a + λ b + µ c, under a linear transformation, T, is T (r) = T (a + λ b + µ c) = T (a) + λ T(b) + µ T (c). Image of a plane To find the image of a plane with equation of the form ax + by + cz = d, first construct a vector equation. Method Example : Find the image of the plane x y + 4z = 6 under a linear transformation, T. Solution: (i) (ii) First construct a vector equation, Put x = z = y = (,, ) is a point on the plane To find vectors parallel to the plane, they must be perpendicular to n =. By 4 inspection, using the top two coordinates,, and, using the bottom two 44 FP FEB 6 SDB

47 coordinates, 4 or must be to n (look at the scalar products), and so are parallel to the plane. The vector equation of the plane is r = + λ + µ The image under the matrix M is M r = M + λm + µ M Example : Find the image of the plane r. 5 = 8 under a linear transformation T. Solution: The equation can be written as x 5y = 8 (i) Put y = x = 4 (4,, ) is a point on the plane z could be anything (ii) To find vectors parallel to the plane, they must be perpendicular to n = 5. By 5 inspection, using the top two coordinates,, and, using the z-coordinate, must be to n (look at the scalar products), and so are parallel to the plane. Continue as in Example. Method, as in the book Fine until the vector n has a zero coordinate, then life is a bit more complicated. Example: Find the image of the plane x y + 4z = 6 under a linear transformation, T. Solution: To construct a vector equation, put x = λ, y = µ and find z in terms of λ and µ. λ µ + 4z = 6 z = 6 λ+μ 4 x y = z λ μ 6 λ+μ 4 = + λ + μ x 4 y = + λ + μ z 6 4 making the numbers nicer in the parallel vectors and now continue as in method. NOTE that M (b c) is not equal to M (b) M (c), since this does not follow the conditions of a linear transformation, so you must use one of the methods above. FP FEB 6 SDB 45

48 7 Eigenvalues and eigenvectors Definitions ) An eigenvector of a linear transformation, T, is a non-zero vector whose direction is unchanged by T. So, if e is an eigenvector of T then its image e is parallel to e, or e = λ e e = T (e) = λ e. e defines a line which maps onto itself and so is invariant as a whole line. If λ = each point on the line remains in the same place, and we have a line of invariant points. ) The characteristic equation of a matrix A is det(a λ I) = Ae = λ e (A λ I) e = has non-zero solutions eigenvectors are non-zero A λ I is a singular matrix det(a λ I) = the solutions of the characteristic equation are the eigenvalues. matrices Example: Find the eigenvalues and eigenvectors for the transformation with matrix A = 4. Solution: The characteristic equation is det(a λ I) = λ 4 λ = ( λ)(4 λ) + = λ 5λ + 6 = λ = and λ = For λ = 4 x y = x y x + y = x x = y and x + 4y = y x = y eigenvector e = we could use.7, but why make things nasty.7 46 FP FEB 6 SDB

49 For λ = 4 x y = x y x + y = x x = y and x + 4y = y x = y eigenvector e = choosing easy numbers. Orthogonal matrices Normalised eigenvectors A normalised eigenvector is an eigenvector of length. In the above example, the normalized eigenvectors are e =, and e = 5. 5 Orthogonal vectors A posh way of saying perpendicular, scalar product will be zero. Orthogonal matrices If the columns of a matrix form vectors which are (i) mutually orthogonal (or perpendicular) (ii) each of length then the matrix is an orthogonal matrix. Example: 5 5 and 5 5 are both unit vectors, and 5. 5 = =, the vectors are orthogonal 5 M = is an orthogonal matrix FP FEB 6 SDB 47

50 Notice that M T M = = and so the transpose of an orthogonal matrix is also its inverse. This is true for all orthogonal matrices think of any set of perpendicular unit vectors Another definition of an orthogonal matrix is M is orthogonal M T M = I M = M T Diagonalising a matrix Let A be a matrix with eigenvalues λ and λ, and eigenvectors e = u v and e = u v then A e = u v = λ u λ v and A e = u v = λ u λ v A u v u v = λ u λ v λ u λ v = u v u v λ λ. I Define P as the matrix whose columns are eigenvectors of A, and D as the diagonal matrix, whose entries are the eigenvalues of A I P = u v u v and D = λ λ AP = PD P AP = D The above is the general case for diagonalising any matrix. In this course we consider only diagonalising symmetric matrices. 48 FP FEB 6 SDB

51 Diagonalising symmetric matrices Eigenvectors of symmetric matrices Preliminary result: x = x x and y = y y The scalar product x. y = x x. y y = x y + x y but (x x ) y y = x y + x y x T y = x. y This result allows us to use matrix multiplication for the scalar product. Theorem: Eigenvectors, for different eigenvalues, of a symmetric matrix are orthogonal. Proof: Let A be a symmetric matrix, then A T = A Let A e = λ e, and A e = λ e, λ λ. λ e T = (λ e ) T = (A e ) T = e T A T = e T A since A T = A λ e T = e T A λ e T e = e T A e = e T λ e = λ e T e λ e T e = λ e T e (λ λ ) e T e = But λ λ e T e = e. e = the eigenvectors are orthogonal or perpendicular Diagonalising a symmetric matrix The above theorem makes diagonalising a symmetric matrix, A, easy. ) Find eigenvalues, λ and λ, and eigenvectors, e and e ) Normalise the eigenvectors, to give e and e. ) Write down the matrix P with e and e as columns. P will now be an orthogonal matrix since e and e are orthogonal P = P T 4) P T A P will be the diagonal matrix D = λ λ. FP FEB 6 SDB 49

52 Example: Diagonalise the symmetric matrix A = λ Solution: The characteristic equation is 9 λ = (6 λ)(9 λ) 4 = λ 5λ + 5 = (λ 5)(λ ) = λ = 5 or For λ = x y = 5 x y 6x y = 5x x = y and x + 9y = 5y x = y e = and normalising e = For λ = x y = x y 6x y = x x = y and x + 9y = y x = y e = and normalising e = 5 5 Notice that the eigenvectors are orthogonal 5 P = D = P T A P = λ λ = 5. 5 FP FEB 6 SDB

53 matrices All the results for matrices are also true for matrices (or n n matrices). The proofs are either the same, or similar in a higher number of dimensions. Finding eigenvectors for matrices. Example: Given that λ = 5 is an eigenvector of the matrix M =, find the corresponding eigenvector. 4 Solution: Consider Me = 5e x x y = 5 y 4 z z x y + z = 5x x y + z = I x + y z = 5y x 4y z = II 4x y z = 5z 4x y 7z = III Now eliminate one variable, say x: I II y + z = y = z We are not expecting to find unique solutions, so put z = t, and then find x and y in terms of t. from I, y = t, and, x = z y = t + t = t x = 5t 5t e = t or choosing t = t check in II and III O.K. FP FEB 6 SDB 5

54 Diagonalising symmetric matrices Example: A =. Find an orthogonal matrix P such that P T AP is a diagonal matrix. Solution: ) Find eigenvalues The characteristic equation is det(a λi) = λ λ = λ ( λ)[ λ ( λ) 4] + [λ ] + = λ λ 6λ + 8 = By inspection λ = is a root (λ + ) is a factor (λ + )(λ 5λ + 4) = (λ + ) (λ ) (λ 4) = λ =, or 4. ) Find normalized eigenvectors λ = x x y = y z z x y = x I x + y + z = y II y = z III I y = x, and III y = z choose x = and find y and z e = and e = e = 9 = e = λ = x x y = y z z x y = x I x + y + z = y II y = z III 5 FP FEB 6 SDB

55 I x = y, and II z = y choose y = and find x and z e = and e = e = 9 = e = λ = 4 x x y = 4 y z z x y = 4x I x + y + z = 4y II y = 4z III I x = y, and III y = z choose z = and find x and y e = and e = e = 9 = e = ) Find orthogonal matrix, P P = (e e e ) P = is required orthogonal matrix 4) Find diagonal matrix, D λ P T AP = D = λ = λ 4 A nice long question! But, although you will not be asked to do a complete problem, the examiners can test every step above! FP FEB 6 SDB 5

56 Index Differentiation hyperbolic functions, inverse hyperbolic functions, Further coordinate systems hyperbola, 6 locus problems, 9 parabola, 7 parametric differentiation, 7 tangents and normals, 8 Hyperbolic functions addition formulae, definitions and graphs, double angle formulae, inverse functions, 4 inverse functions, logarithmic form, 4 Osborne s rule, solving equations, 5 Integration area of a surface, 9 inverse hyperbolic functions, 4 inverse trig functions, 4 ln x, 4 reduction formulae, 4 standard techniques, Matrices cofactors matrix, 4 determinant of matrix, 9 diagonalising matrices, 48 diagonalising symmetric matrices, 49 diagonalising symmetric matrices, 5 dimension, 9 identity matrix, 9 inverse of matrix, 4 non-singular matrix, 4 singular matrix, 4 symmetric matrix, 9 transpose matrix, 9 zero matrix, 9 Matrices - eigenvalues and eigenvectors matrices, 46 characteristic equation, 46 for x matrix, 5 normalised eigenvectors, 47 orthogonal matrices, 47 orthogonal vectors, 47 Matrices - linear transformations base vectors, 4 image of a line, 44 image of a plane, 44 Vectors area of triangle, triple scalar product, 4 vector product, volume of parallelepiped, 4 volume of tetrahedron, 5 Vectors - lines angle between line and plane, 7 cartesian equation of line in -D, 6 distance between skew lines, 4 distance from point to line,, 5 line of intersection of two planes, 6 vector equation of a line, 5 vector product equation of a line, 6 Vectors - planes angle between line and plane, 7 angle between two planes, 8 Cartesian equation, 7 distance between parallel planes,, distance of a point from a plane, 9, line of intersection of two planes, 6 reflection of point in plane, vector equation, 8 54 FP FEB 6 SDB