Pure Further Mathematics Revision Notes June 6
FP JUNE 6 SDB
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... 7 Parabola... 7 Parametric differentiation... 8 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... 4 Reduction formulae... 5 Arc length... 8 Area of a surface of revolution... 5 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
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... 8 6 Matrices... 9 Basic definitions... 9 Dimension of a matri... 9 Transpose and symmetric matrices... 9 Identity and zero matrices... 9 Determinant of a matri... 9 Properties of the determinant... 4 Singular and non-singular matrices... 4 Inverse of a matri... 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... 44 Image of a plane... 44 Image of a plane... 44 Method... 44 Method, as in the book... 45 7 Eigenvalues and eigenvectors... 46 Definitions... 46 matrices... 46 Orthogonal matrices... 47 Normalised eigenvectors... 47 Orthogonal vectors... 47 Orthogonal matrices... 47 Diagonalising a matri... 48 Diagonalising symmetric matrices... 49 Eigenvectors of symmetric matrices... 49 Diagonalising a symmetric matri... 49 matrices... 5 Finding eigenvectors for matrices.... 5 Diagonalising symmetric matrices... 5 FP JUNE 6 SDB
Hyperbolic functions Definitions and graphs sinh = (e e ) cosh = (e + e ) tanh = sinh = (e e ) cosh (e +e ) y y=sinh y 4 y=cosh y y=tanh You should be able to draw the graphs of cosech, sech and coth from the above: cosech sech coth y y=cosech y y=sech y y=coth 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. Eamples: sin(a + B) = sin A cos B + cos A sin B sinh(a + B) = sinh A cosh B + cosh A sinh B but cos(a + B) = cos A cos B sin A sin B cosh(a + B) = cosh A cosh B + sinh A sinh B and + tan A = sec A no change product of two sines, so change sign tanh A = sech A tan A = sin A, product of two sines, so change sign cos A FP JUNE 6 SDB
Inverse hyperbolic functions Graphs Remember that the graph of y = f () is the reflection of y = f() in y =. y = arsinh y = arcosh y = artanh y y=sinh y= y y= y y= y=arsinh y=cosh y=arcosh y=tanh y=artanh Notice arcosh is a function defined so that arcosh. there is only one value of arcosh. However, the equation cosh z =, has two solutions, +arcosh and arcosh. Logarithmic form ) y = arsinh sinh y = (ey e y ) = e y e y = e y = ± 4 +4 = + + or + But e y > and + < e y = + + only y = arsinh = ln + + ) y = arcosh cosh y = (ey + e y ) = e y e y + = e y = ± 4 4 = ± both roots are positive y = arcosh = ln + or ln It can be shown that ln = ln + y = arcosh = ± ln + But arcosh is a function and therefore has only one value (positive) y = arcosh = ln + ( ) ) Similarly artan = ln + ( < ) 4 FP JUNE 6 SDB
Equations involving hyperbolic functions It would be possible to solve 6 sinh cosh = 7 using the R sinh( α) technique from trigonometry, but it is easier to use the eponential form. Eample: Solve 6 sinh cosh = 7 Solution: 6 sinh cosh = 7 6 (e e ) (e + e ) = 7 e 7e 4 = (e + )(e 4) = e = (not possible) or 4 = ln 4 In other cases, the trigonometric solution may be preferable Eample: Solve cosh + 5 sinh 4 = Solution: cosh + 5 sinh 4 = + sinh + 5 sinh 4 = note use of Osborn s rule sinh + 5 sinh = (sinh )(sinh + ) = sinh = or = arsinh.5 or arsinh ( ) = ln.5 +.5 + or ln( ) + ( ) + using log form of inverse = ln + 5 or ln FP JUNE 6 SDB 5
Further coordinate systems Ellipse Cartesian equation a + y b = y b ²/a²+y²/b²= Parametric equations = a cos θ, y = b sin θ a * ( ae,) * (ae,) a Foci at S (ae, ) and S ( ae, ) = a/e b =a/e Directrices at = ± a e Eccentricity e <, b = a ( e ) An ellipse can be defined as the locus of a point P which moves so that PS = e PN, where S is one of the foci, e < and N lies on the corresponding directri. y N' P N *S' * S = a/e =a/e This is true for either focus with the corresponding directri. PS = e PN and PS = e PN PS + PS = e (PN + PN ) = enn PS + PS = e a = a. e This justifies the string method of drawing an ellipse. 6 FP JUNE 6 SDB
Hyperbola Cartesian equation y= b/a y ²/a² y²/b²= a y b = Parametric equations *S' a a * S = a cosh θ, y = b sinh θ ( = a sec θ, y = b tan θ also work) y=b/a Asymptotes a = ± y b Foci at S (ae, ) and S ( ae, ) Directrices at = ± a e *S' a y ²/a² y²/b²= P a N * S Eccentricity e >, b = a (e ) = a/e =a/e 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 directri. S * y y²/c² ²/d²= y c d = is a hyperbola with foci on the y-ais, y= c/d c c y=c/d S' * Parabola N y P Cartesian equation y = 4a Parametric equations = at, y = at N * S (a,) Focus at S (a, ) = a y²=4a Directri at = a 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 directri and the eccentricity e =. FP JUNE 6 SDB 7
Parametric differentiation From the chain rule dy = dy dθ dθ dy = dy dθ dθ or dy = dy dt dt using any parameter. Tangents and normals It is now easy to find tangents and normals. Eample: Find the equation of the normal to the curve given by the parametric equations = 5 cos θ, y = 8 sin θ at the point where θ = π Solution: When θ = π, cos θ = = 5, y = 4 and sin θ = and dy dy = dθ dθ = 8 cos θ = 8 5 sin θ 5 gradient of normal is 5 8 when θ = π equation of normal is y 4 = 5 8 5 5 8y + 9 = Sometimes normal, or implicit, differentiation is (slightly) easier. Eample: Find the equation of the tangent to y = 6, or = 6t, y = 6, at the point t where t =. Solution: When t =, = 8 and y =. dy can be found in two (or more!) ways: dy = dy dt dt dy = = 6t 6 =, t when t = 9 y = 6 y = 6 dy dy = 6 = 6 =, 8 when = 8 9 equation of tangent is y = + 9y 6 = 9 ( 8) 8 FP JUNE 6 SDB
Finding a locus First find epressions for and y coordinates in terms of a parameter, t or θ, then eliminate the parameter to give an epression involving only and y, which will be the equation of the locus. Eample: The tangent to the ellipse 9 + y 6 crosses the -ais at A, and the y-ais 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 ( cos θ) y sin θ + 4 cos θ = cos θ + sin θ = Tangent crosses -ais at A when y =, = cos θ, and crosses y-ais at B when =, y = 4 sin θ mid-point of AB is, 4 (X, Y) cos θ sin θ Here X = cos θ and Y = sin θ cos θ = X equation of the locus is and sin θ = Y 9 + 4 4X Y = since cos θ + sin θ = or 9 4 + 4 y = FP JUNE 6 SDB 9
Differentiation Derivatives of hyperbolic functions y = sinh = (e e ) dy = (e + e ) = cosh and, similarly, d(cosh ) = sinh Also, y = tanh = sinh cosh dy cosh cosh sinh sinh = cosh = cosh = sech In a similar way, all the derivatives of hyperbolic functions can be found. f() sinh cosh tanh f () cosh sinh sech all positive coth cosech sech cosech cosech coth sech tanh all negative Notice: these are similar to the results for sin, cos, tan etc., but the minus signs do not always agree. The minus signs are wrong only for cosh and sech = cosh. Derivatives of inverse functions y = arsinh sinh y = cosh y dy = dy = = cosh y +sinh y d(arsinh ) = + FP JUNE 6 SDB
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 () column f() f () substitution needed for integration arcsin arccos arctan arsinh arcosh artanh sin u = cos u use = sin u cos u = sin u use = cos u + + tan u = sec u use = tan u + + sinh u = cosh u use = sinh u ln + cosh u = sinh u use = cosh u tanh u = sech u use = tanh u partial fractions, see eample below Note that = + + = + ln + c With chain rule, product rule and quotient rule you should be able to handle a large variety of combinations of functions. FP JUNE 6 SDB
4 Integration Standard techniques Recognise a standard function Eamples: sec tan = sec + c sech tanh = sech + c Using formulae to change the integrand Eamples: tan = sec = tan + c cos = + cos = + sin + c sinh = cosh = sinh + c Reverse chain rule Notice the chain rule pattern, guess an answer and differentiate to find the constant. Eample: cos sin looks like u du so try u cos dcos = cos ( sin ) = cos sin so divide by cos sin = cos + c Eample: ( 7) 4 looks like u 4 du so try u5 ( 7) 5 7 5 = 5( 7) 4 6 = ( 7) 4 so divide by ( 7) 4 = ( 7) 5 + c Eample: sech 4 tanh = sech (sech tanh ) looks like u du so try u4 = sech 4 dsech 4 = 4 sech sech tanh so divide by 4 sech 4 tanh = 4 sech4 + c FP JUNE 6 SDB
Standard substitutions a +b a +b a b b a b = a tan u better than b = a sinh u when there is no b = a sinh u better than b = a tan u when there is b = a tanh u or use partial fractions b = a cosh u better than b = a sec u when there is For more complicated integrals like or p +q+r p +q+r complete the square to give p( + a) + b and then use a substitution similar to one of the four above. Eample: 4 8 5 4 8 5 = 4( + ) 9 = 4( ) 9 = 4( ) 9 Substitute ( ) = cosh u = sinh u du = 9(cosh u ) sinh u du = du = u + c = arcosh + c Nice trick Eample: Solution: I = ( ) + 5 Substitute u =, = du I = u + 5 du = arsinh u + c, using standard formula 5 = Important tip arsinh + c 5 n, etc., is best done with the substitution a ± (i) or (ii) u (or u ) = a ±, when n is odd, a trigonometric or hyperbolic function when n is even. FP JUNE 6 SDB
Integration inverse functions and ln To integrate inverse trigonometric or hyperbolic functions and ln we use integration by parts dv with = Eample: Find arctan Solution: I = arctan take u = arctan du = + and dv = v = I = arctan + I = arctan = arctan ln ( + ) + c Eample: Find arcosh Solution: I = arcosh take u = arcosh du = and dv = v = I = arcosh I = arcosh = arcosh + c 4 FP JUNE 6 SDB
Reduction formulae The first step in finding a reduction formula is often (but not always) integration by parts (sometimes twice). The following eamples show a variety of techniques. Eample : I n = n e. (a) Find a reduction formula, (b) Find I, (c) Find I 4 Solution: (a) Integrating by parts u = n du = nn and dv = e v = e I n = n e n n e I n = n e ni n (b) I = e = e + c (c) Using the reduction formula I 4 = 4 e 4I = 4 e 4( e I ) = 4 e 4 e + ( e I ) = 4 e 4 e + e 4(e I ) = 4 e 4 e + e 4e + 4e + c since I = e + c Eample : Solution: Find a reduction formula for I n = tan n. I n = tan n = tan n tan I n = tan n (sec ) = tan n sec tan n I n = n tann I n. FP JUNE 6 SDB 5
π Eample : (i) Find a reduction formula for I n = sin n. π (ii) Use the formula to find I 6 = sin 6 Solution: (i) By splitting sin n = sin n sin, we can differentiate sin n reducing the power, and we can integrate sin I n = π sin n I n = [ cos sin n π ] π = sin n sin take u = sin n du = (n ) sin n cos and dv π = sin v = cos cos (n ) sin n cos π = + (n ) cos sin n π = (n ) sin sin n π π = (n ) sin n (n ) sin n I n = (n ) I n (n ) I n I n = n n I n. (ii) I 6 = 5 6 I 4 = I 6 = 5 π 6 5 6 4 I = = 5π. 5 6 4 I Eample 4: Find a reduction formula for I n = sec n. Solution: By splitting sec n = sec n sec, we can differentiate sec n reducing the power, and we can integrate sec I n = sec n = sec n sec take u = sec n dv du = (n ) secn sec tan and = sec v = tan I n = sec n tan tan (n ) sec n sec tan = sec n tan (n ) tan sec n = sec n tan (n ) (sec ) sec n = sec n tan (n )I n + (n )I n (n )I n = sec n tan + (n )I n 6 FP JUNE 6 SDB
Eample 5: Find a reduction formula for I n = n ( + ). Solution: We can differentiate n reducing the power, and we can integrate ( + ) I n = n ( + ) take u = n and dv = ( + ) du = nn v = ( + ) I n = n ( + ) n n ( + ). Writing ( + ) = ( + ) ( + ) allows us to write the integral in terms of I n and I n. Many reduction formulae need a fiddle like this e.g. ( ) = ( ) ( ) I n = n n I n = n n I n = n I n n I n ( + ) ( + ) ( + ) n n ( + ) n+ I n = I n = n I n n I n+ n π Eample 6: Find a reduction formula for I n = n cos Solution: We can differentiate n reducing the power, and we can integrate cos π I n = n I n = [ n π sin ] cos take u = n and dv π n sin n du n = n = cos v = sin Integrating by parts again will change the sin to cos, and reduce the power further. take u = n and I n = π n n{[ n π ( cos )] dv du n = (n ) = sin v = cos π cos (n ) n } I n = π π n n{ + (n ) n cos } I n = π n n (n ) I n FP JUNE 6 SDB 7
Eample 7: Find a reduction formula for I n = sin n sin Solution: I n = sin[(n )+] sin = sin(n ) cos +cos(n ) sin sin = sin(n ) ( sin ) + cos(n ) sin cos sin = sin(n ) sin + cos(n ) cos sin(n ) sin = I n + cos(n ) using cos(a + B) = cos A cos B sin A sin B I n = I n + n sin(n ). Arc length All the formulae you need can be remembered from this diagram Q arc PQ line segment PQ (δ s) (δ ) + (δ y) δs δ + δy δ P δ δ s δ y and as δ ds = + dy ds = + dy arc length = s = + dy 8 FP JUNE 6 SDB
Similarly and δs δy δs δt δ δy + s = dy + dy δ δt + δy δt s = dt + dy dt for parametric equations. dt or s = + y dt Eample : Find the length of the curve y =, from the point where = to the point where = 8. 8 6 y y=(/)^(/) Solution: The equation of the curve is in Cartesian form so we use s = + dy. y = dy = 4 8 6 4 8 s = + 4 6 8 s =. = ( + ) 8 = (9) (4) Eample : Find the length of one arch of the cycloid = a(t sin t), y = a( cos t). Solution: The curve is given in parametric form so we use s = + y dt. = a(t sin t), y = a( cos t) dt = a( cos t), and dy dt = a sin t t= y =a(t sint), y=a( cost) t=π + y = a ( cos t + cos t + sin t) = a ( cos t) + y = a sin t = a sin t π s = a sin t dt s = 4a cos t π = 4a 4a = 8a. FP JUNE 6 SDB 9
Area of a surface of revolution A curve is rotated about the -ais. To find the area of the surface formed between = a and = b, we consider a small section of the curve, δs, at a distance of y from the -ais. When this small section is rotated about the -ais, the shape formed is approimately a cylinder of radius y and length δs. y a y δ s b The surface area of this (cylindrical) shape 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 ) + (δ y) Eample : A sphere has radius r. Find the surface area of the sphere between the planes = a and = b. Solution: The Cartesian form is most suitable here. ²+y²=r² y b A = π y a + y = r ds + y dy = dy = y and ds = + dy =a =b b A = πy a b = πr a + y since + y = r = b π a A = [πr] a b = πr(b a) since r is constant y + Notice that the area of the whole sphere is from a = r to b = r giving surface area of a sphere is 4π r. FP JUNE 6 SDB
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 Eample : The parabola, = at, y = at, between the origin (t = ) and P (t = ) is rotated about the -ais. 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 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 JUNE 6 SDB
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 aes k i j = k, j k = i, k i = j, 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 JUNE 6 SDB
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 Eample: 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 = = 5 4 4 A B area ABC = AB = AC + 5 + = 5 FP JUNE 6 SDB
Volume of a parallelepiped In the parallelepiped, the base is parallel to b and c n is a unit vector perpendicular to the base and the height h = h n, where h = ± a cos φ = ± a. n ± because φ might be obtuse h n a φ c θ b The area of base = bc sin θ volume V = ± h bc sin θ ± V = a cos φ bc sin θ a. (b c) = a. (bc sin θ n) = a. n (bc sin θ ) a. (b c) = a cos φ bc sin θ = ± V 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 epanding 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 JUNE 6 SDB
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 n a φ c θ b volume of tetrahedron = 6 volume of parallelepiped volume of tetrahedron = {a, b, c} 6 Eample: 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 l α or y = m + λ β. z n γ a r b FP JUNE 6 SDB 5
Cartesian equation of a line in -D Eliminating λ from the above equation we obtain 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 l α = y m β and y m β = z n γ and 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. Eample: A straight line has Cartesian equation = 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 = y.5 = z the line passes through A, (,, ), and is parallel to b,.5 or 5 4 6 FP JUNE 6 SDB
(i) (ii) r = + λ 5 4 r 5 4 = r r.5 5 4 = 7 = 5 4 6 4. i j k = 5 4 7 = 6 4 Equation of a plane Scalar product form Let n be a vector perpendicular to the plane π. Let A be a fied point in the plane, and P be a general point, (, y, z), in the plane. n a r A P π Then AP 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 = a b c then r. n = y. a bc = a + b y + c z z a + b y + c z = d is the Cartesian equation of a plane perpendicular to a b. c FP JUNE 6 SDB 7
Eample: 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 6 y + z = d but A, (,, 5) lies in the plane d = 6 + 5 = Cartesian equation is 6 y + z = and scalar product equation is r. 6 =. Vector equation of a plane r = a + λ b + µ c is the equation of a plane, π, through A and parallel to the vectors b and c. n a r A b P π O c Eample: 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. 5 4 8 FP JUNE 6 SDB
Distance from a point to a plane Eample: Find the distance from the point P (,, 5) to the plane 4 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 + 6 + 44λ = λ = 69 PM = 4 69 distance = PM = 69 4 + + = 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 n + n y + n z + d = is given by s = n α + n β + n γ + d n + n + n This formula is in your formula booklets, but not in your tet books. M P n FP JUNE 6 SDB 9
Reflection of a point in a plane Eample: 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 π This meets the plane π when (+λ) ( λ) + (7+λ) = 7 P + 9λ + 4λ + 7 + λ = 7 λ = OP = + ( ) 7 6 AP = OP OA = ( ) = 4 6 OA = OA + AP = + 4 = 9 7 A the reflection of A is A, (, 9, ) FP JUNE 6 SDB
Distance between parallel planes Eample: Find the distance between the parallel planes π : 6y + z = 9 and π : 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 shortest distance s from P to the plane π is n α + n β + n γ + d n + n + n shortest distance s = 6 + 5 + 6 + = 4 7 The distance between the planes is 4 7. Shortest distance from a point to a line Eample: Find the shortest distance from the point P (,, 4) to the line l, r = + λ 6 P Solution: Any plane 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) y + 6z = ( ) + 6 4 = 6 y + 6z = 6 l X l meets plane ( + λ) ( λ) + 6(6λ) = 6 4 +4λ 9 + 9λ + 6λ = 6 λ = X is the point (,, 6) PX = = 6 4 shortest distance is PX = + + = 7 FP JUNE 6 SDB
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, B For any point, B, on π AM is the projection of AB onto the line AM AM = AB. n Eample: Find the shortest distance from the point A (,, 5) to the plane 4 y + z =. M n A Solution: By inspection B (, 7, ) lies on the plane AB = 7 = 5 5 n = 4 n = 4 + + = shortest distance = AB. n =. 5 4 = FP JUNE 6 SDB
Distance between parallel planes Eample: Find the distance between the parallel planes π π : 6y + z = 9 and π : 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 = + 6 + = 7 shortest distance = 5. 7 6 = 4 7 FP JUNE 6 SDB
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 AC 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 eample 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 JUNE 6 SDB
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. AP 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 A X and we can now find PX = AP AX, using Pythagoras Eample: 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 = + + 6 = 7, b = 7 6 and AP = 4 = 5 5 4 AX = AP. b = 5 5. 4 7 6 = +5+4 7 = 7 PX = AP AX = (5 + 5 + 4 ) 7 = 7 FP JUNE 6 SDB 5
Line of intersection of two planes Eample: Find an equation for the line of intersection of the planes + y + z = 4 I and y + z = 4 II Solution: Eliminate one variable I + II + 5z = 8 We are not epecting a unique solution, so put one variable, z say, equal to λ and find the other variables in terms of λ. z = λ = 8 5λ I or y = 4 z = 4 8 5λ y = z y = z 8 5 4 + λ 8 5 4 + 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 JUNE 6 SDB
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 Eample: Find the angle between the line + and the plane + 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 = + + + + 7 cos θ cos θ = 7 6 θ = 7. o the angle between the line and the plane, φ = 9 7. = 6.7 o FP JUNE 6 SDB 7
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 Eample: Find the angle between the planes plane plane + y + z = 5 and + 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 JUNE 6 SDB
6 Matrices Basic definitions Dimension of a matri A matri with r rows and c columns has dimension r c. Transpose and symmetric matrices The transpose, A T, of a matri, 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 matri, 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 matri. c e f Identity and zero matrices The identity matri I = and the zero matri is = Determinant of a matri The determinant of a matri, 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 JUNE 6 SDB 9
Properties of the determinant ) A determinant can be epanded 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 matri, A, is singular if its determinant is zero, det(a) = A matri, A, is non-singular if its determinant is not zero, det(a) Inverse of a matri 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 + + + + + 4 FP JUNE 6 SDB
Finding the inverse ) Find the determinant, det(a). If det(a) =, then A is singular and has no inverse. A B C ) Find the matri 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 eample 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 JUNE 6 SDB 4
Matrices and linear transformations Linear transformations T is a linear transformation on a set of vectors if (i) T ( + ) = T ( ) + T ( ) for all vectors and y (ii) T (k) = kt () for all vectors Eample: Show that T y z = + y z is a linear transformation. Solution: (i) T ( + ) = T y z + y z = T ( + ) = + + y + y = +y + +y z z z T ( + ) = T ( ) + T ( ) + y + y z + z z = T ( ) + T ( ) (ii) k k T (k) = T k y = T ky = k + ky z kz kz T (k) = kt () = k +y z = kt () 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 matri a d g b e h c f i a b c d e f g h i the first column of the matri the second column of the matri the third column of the matri This is an important result, as it allows us to find the matri for given transformations. 4 FP JUNE 6 SDB
Eample: Find the matri for a reflection in the plane y = Solution: The z-ais lies in the plane y = so the third column of the matri is Also the first column of the matri is the second column of the matri is the matri for a reflection in y = is. Eample: Find the matri of the linear transformation, T, which maps (,, ) (, 4, ), (,, ) (6,, 5) and (,, 4) (,, ). Solution: Firstly 4 first column is 4 Secondly 6 5 but = + 4 + T 6 T = 4 = 5 second column is Thirdly but 4 4 = 4 + + 4 4 + 4T = 4T T = third column is T = 4. FP JUNE 6 SDB 4
Image of a line Eample: 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 4. 9 4 9 and so a 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 a + by + cz = d, first construct a vector equation. Method Eample : Find the image of the plane y + 4z = 6 under a linear transformation, T. Solution: (i) (ii) First construct a vector equation, Put = z = y = (,, ) is a point on the plane To find vectors parallel to the plane, they must be perpendicular to n = inspection, using the top two coordinates, coordinates, 4 or parallel to the plane., and, using the bottom two 4. By must be to n (look at the scalar products), and so are 44 FP JUNE 6 SDB
The vector equation of the plane is r = The image under the matri M is M r = M + λ + µ + λm + µ M Eample : Find the image of the plane r. 5 Solution: The equation can be written as 5y = 8 = 8 under a linear transformation T. (i) Put y = = 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 = inspection, using the top two coordinates, 5 5. By, and, using the z-coordinate, must be to n (look at the scalar products), and so are parallel to the plane. Continue as in Eample. Method, as in the book Fine until the vector n has a zero coordinate, then life is a bit more complicated. Eample: Find the image of the plane y + 4z = 6 under a linear transformation, T. Solution: To construct a vector equation, put = λ, y = µ and find z in terms of λ and µ. λ µ + 4z = 6 z = 6 λ+μ 4 y = z λ μ 6 λ+μ 4 = + λ + μ 6 4 4 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 JUNE 6 SDB 45
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 matri 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 matri det(a λ I) = the solutions of the characteristic equation are the eigenvalues. matrices Eample: Find the eigenvalues and eigenvectors for the transformation with matri 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 y = y + y = = y and + 4y = y = y eigenvector e = we could use.7, but why make things nasty.7 46 FP JUNE 6 SDB
For λ = 4 y = y + y = = y and + 4y = y = y eigenvector e = choosing easy numbers. Orthogonal matrices Normalised eigenvectors A normalised eigenvector is an eigenvector of length. In the above eample, 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 matri form vectors which are (i) mutually orthogonal (or perpendicular) (ii) each of length then the matri is an orthogonal matri. Eample: 5 5 and 5 5 are both unit vectors, and 5. 5 = + 5 5 5 5 =, the vectors are orthogonal 5 M = 5 5 5 is an orthogonal matri FP JUNE 6 SDB 47
Notice that M T M = 5 5 5 5 5 5 5 5 = and so the transpose of an orthogonal matri is also its inverse. This is true for all orthogonal matrices think of any set of perpendicular unit vectors Another definition of an orthogonal matri is M is orthogonal M T M = I M = M T Diagonalising a matri Let A be a matri 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 matri whose columns are eigenvectors of A, and D as the diagonal matri, 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 matri. In this course we consider only diagonalising symmetric matrices. 48 FP JUNE 6 SDB
Diagonalising symmetric matrices Eigenvectors of symmetric matrices Preliminary result: = and y = y y The scalar product. y =. y y = y + y but ( ) y y = y + y T y =. y This result allows us to use matri multiplication for the scalar product. Theorem: Eigenvectors, for different eigenvalues, of a symmetric matri are orthogonal. Proof: Let A be a symmetric matri, 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, and (AB) T =B T A T 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 matri The above theorem makes diagonalising a symmetric matri, A, easy. ) Find eigenvalues, λ and λ, and eigenvectors, e and e ) Normalise the eigenvectors, to give e and e. ) Write down the matri P with e and e as columns. P will now be an orthogonal matri since e and e are orthogonal P = P T 4) P T A P will be the diagonal matri D = λ λ. FP JUNE 6 SDB 49
Eample: Diagonalise the symmetric matri A = 6 9. 6 λ Solution: The characteristic equation is 9 λ = (6 λ)(9 λ) 4 = λ 5λ + 5 = (λ 5)(λ ) = λ = 5 or For λ = 5 6 9 y = 5 y 6 y = 5 = y and + 9y = 5y = y e = and normalising e = For λ = 5 5 6 9 y = y 6 y = = y and + 9y = y = y e = and normalising e = 5 5 Notice that the eigenvectors are orthogonal 5 P = 5 5 5 D = P T A P = λ λ = 5. 5 FP JUNE 6 SDB
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. Eample: Given that λ = 5 is an eigenvector of the matri M = 4, find the corresponding eigenvector. Solution: Consider Me = 5e 4 y z = 5 y z y + z = 5 y + z = I + y z = 5y 4y z = II 4 y z = 5z 4 y 7z = III Now eliminate one variable, say : I II y + z = y = z We are not epecting to find unique solutions, so put z =, and then find and y. y =, and, from I, = z y = + = = 5 e = 5 or as any multiple will also be an eigenvector check in II and III, O.K. FP JUNE 6 SDB 5
Diagonalising symmetric matrices Eample: A =. Find an orthogonal matri P such that P T AP is a diagonal matri. 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 = y = y z z y = I + y + z = y II y = z III I y =, and III y = z choose = and find y and z e = and e = e = 9 = e = λ = y = y z z y = I + y + z = y II y = z III 5 FP JUNE 6 SDB
I = y, and II z = y choose y = and find and z e = and e = e = 9 = e = λ = 4 y = 4 y z z y = 4 I + y + z = 4y II y = 4z III I = y, and III y = z choose z = and find and y e = and e = e = 9 = e = ) Find orthogonal matri, P P = (e e e ) P = is required orthogonal matri 4) Find diagonal matri, 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 eaminers can test every step above! FP JUNE 6 SDB 5
Inde Differentiation hyperbolic functions, inverse hyperbolic functions, Further coordinate systems hyperbola, 7 locus problems, 9 parabola, 7 parametric differentiation, 8 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, inverse hyperbolic functions, 4 inverse trig functions, 4 ln, 4 reduction formulae, 5 standard techniques, Matrices cofactors matri, 4 determinant of matri, 9 diagonalising matrices, 48 diagonalising symmetric matrices, 49 diagonalising symmetric matrices, 5 dimension, 9 identity matri, 9 inverse of matri, 4 non-singular matri, 4 singular matri, 4 symmetric matri, 9 transpose matri, 9 zero matri, 9 Matrices - eigenvalues and eigenvectors matrices, 46 characteristic equation, 46 for matri, 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 JUNE 6 SDB