MATH20132 Calculus of Several Variables. 2018

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1 MATH013 Calculus of Several Variables 018 Solutions to Problems 7 Tangent Spaces & Planes Full rank Jacobian matrices 1 Are the following parametrically defined sets surfaces? Give your reasons Look at their Jacobian matrices i ii iii iv { } +y, y, 3y T :,y R, { } +y, y, 3 3y T :,y R \{0}, { } +y, y, 3 3y T : > 0,y > 0, { } ye, e y, 1 T :,y R Solution For the image set to be a surface the Jacobian JF has to be of full-rank at all points i Here the set given is the image set of the function F : R R 3, +y F = y 3y, which has the Jacobian matri JF = y y 3 A quick observation shows that when = 0 the columns 0,0, T and 0,0,3 T are not linearly independent Hence ImF = {F : R } is not a surface ii The Jacobian matri is JF = y y 6 6y 1

2 The columns are obviously linearly dependent if = 0 but that point has been omitted We look, instead, for 0 for which there eists λ R such that y y = λ 6 6y That is, = λy, y = λ and = λy The first two combine as = λ in which case either = 0 or λ = 1 If = 0 then y = λ = 0 but 0,0 has been omitted This leaves λ = 1, ie either λ = 1 or λ = 1 If λ = 1 then = y and = y = Since 0 we have = 1 and thus y = 1 That is, at 1,1 T the Jacobian is not of full rank Hence the given set is not a surface We should complete the study and consider λ = 1 Then = y and = y = This implies = 1 and y = 1 So 1,1 T is another and the last point at which the Jacobian matri if not of full rank iii We have the same Jacobian matri as in part ii We saw there that if, and only if,,y = 0, 1,1 or 1, 1 then JF is not of full-rank These three points do not lie in the region > 0,y > 0 and so {F : > 0,y > 0} is a surface iv In this eample we have ye e JF = e y e y 0 0 Does there eist and λ such that ye = λe and e y = λe y? Since e,e y 0 divide to get y = λ and 1 = λ Substitute in to get that the parametric set fails to be a surface at all points on the two hyperbola given by y = 1 Are the following level sets surfaces? Look at the Jacobian matrices of the level sets i { R 3 : +3y +z = 9}, ii The set of R 3 satisfying +y z = 1, +3y +z = 9

3 iii The set of R 3 satisfying +y z = 11, +3y +z = 9 iv The set of R 4 satisfying 3+y +u +v = 13, 3 y 3 +u 3 v 3 = 0, y +5u v = 4 Hint the point p = 1,1,, T may be of interest Solution i With f = +3y +z 9 the Jacobian matri is Jf =, 6y, 4z Is this full-rank for : f = 0? To talk of one vector being linear independent is that you cannot find a non-zero coefficient which you multiply your vector to get the zero vector You can only do this if you start with the zero vector Is our Jacobian ever 0? Yes, when = 0 But f0 = 9 0 so 0 is not in the level set Thus for all points in the level set the Jacobian matri is non-zero and so the level set is a surface ii With the Jacobian matri is f = Jf = +y z 1 +3y +z 9 y z 6y 4z Is this full-rank, ie the two rows linearly independent, for : f = 0? Remember, two vectors are linearly dependent iff one is a scalar multiple of the other Thus Jf will not be of full rank for for which there eists λ R such that 6y 4z = λ y z, That is, = λ, 3y = λy and z = λz From = λ either = 0 or λ = 1 3

4 If λ = 1 then the last two equations become y = 3y and z = z, ie y = z = 0 So the two rows are linearly dependent if y = z = 0 Yet, for : y = z = 0 we have f = 1, 9 T 0 for all R If = 0 then look at 3y = λy For this, either y = 0 or λ = 3 If = 0 and λ = 3 then, from z = λz = 3z we get z = 0 So the two rows are linearly dependent if = z = 0 Yet, for : = z = 0 we have f = y 1, 3y 9 T 0 for all y R If = 0 and y = 0 we can see the two rows are linearly dependent with λ = Yet, for : = y = 0 we have f = z 1, z 9 T 0 for all z R We have found many points at which Jf is not of full-rank but none satisfy f = 0 Hence the level set is a surface The set of points +y z = 1 is an Hyperboloid: 4

5 The set of points +3y +z = 9 is an ellipsoid: There intersection is a disjoint union of two closed lines in R 3, here shown in blue iii Be Careful! You can go through the argument of part ii but in this case the level set is empty! Subtract equation 1 from to find that any points on the level set must satisfy y +z =, impossible 5

6 iv With the Jacobian matri is f = Jf = 3+y +u +v 13 3 y 3 +u 3 v y +5u v 4 3 4y u v 3 3y 3u 3v u v The point p = 1,1,, T is of interest because fp = 0, so p is in the level set But also Jfp = This is not of full rank because r 1 +r = r 3 Hence the level set is not a surface at p Planes in R n 3 i A plane in R 3 is given parametrically by +4y 5 +y : R y 3y Epress this plane as a a graph { φ of some function φ : R R, b a level set, for some f : R 3 R : R }, f 1 0 = { R 3 : f = 0 } 6

7 ii Repeat for the parametric set +y +y 1 3y : y R iii Repeat for 4 4y +8 +y 1 3 4y +6 4y 4 : y R, this time epressing this as a graph of some function φ : R R, and then as a level set Solution i a In the hope that +4y 5 +y : R y = 3y s t φs write s = +4y 5 and t = +y and solve: y = 1 3 s t+3 and = 1 6 4t s+3 s : s = R t, 1 Then φs = 3y = 1 3 7t 4s 6 Thus the given plane is the graph of φs = 7t 4s 6/3 b By the result in part a, as a level set the points s,t,u T R 3 on the graph satisfy u = φ s,t T, ie 4s 7t+3u = 6 So f s = 4s 7t+3u+6, where s = s,t,u T Alternative Approach Write +4y 5 +y = 3y y 4 1 3

8 Then v 1 =,, T and v = 4,1, 3 T span the plane Thus v 1 v = 4, 7,3 T is normal to the plane, and for this reason relabeled as n A definition of the plane is that s = s,t,u T is in the plane iff n s p = 0 where p = 5,,0 T This again leads to ii a There is no hope that +y +y 1 : R y = 3y for this would imply s t φs s = +y = +y 1 = t, s : s = R t, whereas s and t should be independent variables Instead we could choose, for eample, s = +y 1 and t = 3y to get +y +y 1 s : R y = s s : R t 3y t This is a vertical plane in R 3, and it is a graph but where the -coordinate is a function of the y, z-coordinates b As a level set this is the set of points v,s,t T R 3 such that v s = 0; equivalently it is the set f 1 0 for f v = v s where v = v,s,t T iii We hope to write the given surface as a graph s { } s : s R t s = φs φ 1 s : s = R, t φ s where φ : R R Hoping that 4 4y +8 +y 1 3 4y +6 = 4y 4 first solve 4 4y +8 = s +y 1 = t 8 s t φ 1 s φ s, 3

9 The solution is = s 4t+4 4 and y = s t+6 Then, from 3, s 4t+4 s t+6 φ 1 s = 3 4y +6 = and similarly = 5s+4t 1 4 φ s = 4y 4 = s 4t+8 Thus the given plane is the graph of φ : R R, 5s+4t 1/4 φs = s 4t+8 The level set is those s,t,u,v T R 4 satisfying 5s+4t 4u = 1, s+4t+v = 8 Equivalently, it is the set f 1 0 where f : R 4 R, 5s+4t 4u 1 f s =, s+4t+v 8 where s = s,t,u,v T 4 i Let M be an n r real valued matri Prove that {Mt : t R r }, the image of the linear map M, is a vector subspace of R n spanned by the columns of M ii Let N be an m n real valued matri Prove that { R n : N = 0}, the kernel of the linear map N, is a vector subspace of R n These two results are used in the proof not given in lectures concerning planes This is relegated to an additional question at the end of this sheet SolutionV isavectorsubspaceofr n iffforallu,v V wehaveαu+βv V for all α,β R 9

10 i Let u,v {Mt : t R r } Then there eist k,l R r such that u = Mk and v = Ml Then given α,β R we have αu+βv = αmk+βml = M αk+βl {Mt : t R r }, since αk+βl R r Hence {Mt : t R r } is a vector subspace of R n Further, by matri multiplication, Mt = r C i t t, i=1 ie every element in the set is a linear combination of the columns C i of M ii Let u,v { R n : N = 0} Given α,β R consider αu+βv : N αu+βv = αnu+βnv = α0+β0 = 0 Hence αu + βv { R n : N = 0} and thus { R n : N = 0} is a vector subspace of R n Best Affine Approimations Recall that the Best Affine Approimation to a function f at a point a is given by f a+df a a = f a+jf a a 5 Write down the Best Affine Approimation to i f = +y at a =, 1 T, and what value does the approimation give at a = 1, 09 T? ii f = y + yz + z at a = 1, 1,4 T, and what value does the approimation give at a = 09, 11,41 T? iii f = y y at a =, 3 T, and what value does the approimation give at a = 19, 31 T? 10

11 Hint Look back at Sheet 4 for appropriate results Solution i As seen in Question 3 Sheet 4, the Fréchet derivative is df a t = a+bs + at for general a = a,b T and t = s,t T With the specific a =, 1 T this becomes df a t = 3s+t Then, with t = a, df a a = 3 +y +1 = 3+y 4, where =,y T Thus, the Best Affine Approimation is f a+3+y 4 = 3+y At a = 1, 09 T the approimation is 5 The actual value is 5 ii As seen in Question 4 Sheet 4, the derivative is df a t = b+cs + a+ct + a+bu for general a = a,b,c T and t = s,t,u T With the specific a = 1, 1,4 T this becomes df a t = 3s+3t u Thus, the Best Affine Approimation is fa+3+1+3y +1 z 4 = 3+3y z +7 At a = 09, 11,41 T the approimation is 7 The actual value is 71 iii As seen in Question 5 Sheet 4, the derivative is b s+abt df a t = abs+a t With the specific a =, 3 T this becomes 9s 1t df a t = 1s+4t Thus, the Best Affine Approimation is 9 1y +3 f a+ = 1 +4y y y +4 At a = 19, 31 T the approimation is T The actual value is T There is a form of Taylor s Theorem for scalar-valued functions of several variables This can be used to estimate the error between the Best Affine Approimation and the original function See my web site for non-eaminable notes on this 11

12 6 Define the function f : R R by f,y T = 3 y, + y T Show that f locally invertible at a = 1, 1 T What is the Best Affine Approimation to the inverse function near f a? What approimation does this give to f 1 09,01 T? Solution The Jacobian matri is 3 Jfa = y 4y 1 1 1, 1 T = The matri is of full-rank and so, by the Inverse Function Theorem, f is invertible in some open set containing a ie it is locally invertible From the Chain Rule applied to f f 1 = id we deduce that the Jacobian of the inverse is the inverse of the Jacobian That is, with b = f a, the Jacobian of the inverse is Jf 1 b = Jfa 1 = Then the Best Affine Approimation to f 1 near b = fa = 1,0 T is, as a function of u = u,v T R, = 1 3 f 1 b+jf 1 bu b = a+jfa 1 u fa 1 = u+4v +u v u+1 v The approimation this gives to f 09,01 1 T is 11, 1 T To get someideahowgoodorbadthisis,notethatf 11, 1 T = 0869,01 T The Inverse Function Theorem tells us when an inverse function eists but not what it is It is an eample of an eistence result Nonetheless we can find the Best Affine Approimation to the inverse function Tangent Spaces for Graphs 7 For each of the following scalar-valued functions φ, find both a basis for the Tangent Space and the equation of the Tangent Plane to the graph of ϕ for the given point on the graph For the latter give your answer in the form the Tangent plane to ϕ at q is the graph of the function g = 1

13 i ϕ = 4 +y, q = 1, 1 T R, ii ϕ = 9 y, q =,1 T R, iii ϕ = 9 y, p =,,1 T G ϕ, iv ϕ = 5/1+ +3y, p = 1, 1,1 T G ϕ Solution First I recollect a bit of the Theory: From the notes: the graph of ϕ is the image of F := ϕ and a basis for the Tangent Space at p = Fq is given by the columns of JFq The Jacobian matri of F is where I is the identity matri JF = I Jϕ The Tangent Plane to graph of ϕ at p is the image of the Best Affine Approimation to F at q The Best Affine Approimation to F at q F is q I Fq+JFq q = + q ϕq Jϕq = = q ϕq, +, q Jϕq q ϕq+jϕq q That is, the Best Affine Approimation to F at q is the graph of the Best Affine Approimation to ϕ at q i First note that, given q = 1, 1 T R, we are looking for the Tangent Space and plane at the point q 1 p = = 1 ϕq 5 13

14 on the graph Net, Jϕ = 8,y so Jϕq = 8, Thus I 1 0 JFq = = 0 1, Jϕq 8 and hence the columns of this matri, the vectors 1 0 0, 1 8 form a basis for the Tangent Space at p Also, since ϕq = 5, the Best Affine Approimation to ϕ at q is ϕq+jϕq q = = 8 y 5 y +1 Hence the Tangent Plane to the graph of ϕ at p is the graph of the function g = 8 y 5 z p q y 14

15 ii This time p =, 1, T while ϕq = Net! # y Jϕ = p,, p 9 y 9 y so Jϕq = 1, 1/ Thus JFq =! I Jϕq # 1 0 1, = 0 1 1/ and so 1, 0, 1T and 0, 1, 1/T form a basis for the Tangent Space at p The Best Affine Approimation to ϕ at q is 1 9 = y + ϕq + Jϕq q = + 1 1/ y 1 Hence the Tangent plane to the graph of ϕ at p is the graph of the function g = y/ + 9/ Figure for Question 7i The boundary of the surface is jagged since it is plotted using rectangular coordinates rather than spherical ones z p q y 15

16 In the net two parts we are given a point p on the graph Being on the graph means that q p =, ϕq for some q R iii Since p =,,1 T then q =, T and ϕq = 1 Also Jϕ = y in which case Jϕq = 4 4 Thus the columns of I Jϕ = , namely 1,0, 4 T and 0,1,4 T, form a basis for the Tangent Space at p The Best Affine Approimation to ϕ at q is = 4+4y +17 y + Hence the Tangent Plane to the graph of ϕ at p G ϕ is the graph of the function g = 4+4y +17 iv Since p = 1, 1,1 T then q = 1, 1 T and ϕq = 1 Also 5 10 ϕ = implies Jϕ = 1+ +3y 1+ +3y, 30y 1+ +3y So Jϕq = /5,6/5 Thus the columns of I 1 0 = 0 1 Jϕ /5 6/5 form a basis for the Tangent Space to G ϕ at p We can scale these and give the basis as 5,0, T and 0,5,6 T The Best Affine Approimation to ϕ at q is y +13 = 5 5 y

17 z p y q Hence the Tangent Plane to the graph of ϕ at p Gϕ is the graph of the function g = + 6y + 13 /5 8 Repeat Question 7 for the vector-valued function y φ = at p =, 1,, 5T Gφ, +y with =, yt R T with q =, 1T Then Solution First note that p = qt, φ qt Jφ = y y so Jφ q = 1 4 This matri is of full rank so the column of I 1, = 1 Jφ q 4 that is 1, 0, 1, 4T and 0, 1,, T, form a basis to the Tangent Space to Gϕ at p The Best Affine Approimation to ϕ at q is 1 ϕq + Jϕq q = y+1 + y + = 4 y 5 17

18 Hence the Tangent Plane to the graph of ϕ at p G ϕ is the graph of the function +y + g = 4 y 5 Image or Parametric sets are locally graphs 9 Show that the following Image sets are graphs around the point given { } i 4+costcoss,4+costsins,sint T : s,t R withq = 3π/4,π/4 T, ii iii { } y, +y, 3 y, y T :,y R, with q = 1, T, { } cost, sint, t T : t R with q = 3π Solution To show that the image set of F : R r R n is locally a graph we apply a Corollary of the Inverse Function This says that if the rows 1 to r of JFq are linearly independent then the image set is the graph of some function φ : W R r R n r with q W i JFq = 4+costsins sintcoss 4+costcoss sintsins 0 cost s=q 4+ / 1 = 4+ / 1 0 The first two rows are linearly independent and so the image set can be written as a graph of the first two variables in some open set around q ii JFq = y y 1 3 y 0 y =q = The first two rows are linearly independent and so the image set can be written as a graph of the first two variables in some open set around q iii This is already a graph Usually a graph is written with the variables first followed by the functions of the variables Here we have a permutation of this, functions first variable last Tangent Spaces for Image sets 18

19 10 Let the surface S be given parametrically as S = { Fu : u U and JFu is of full-rank }, of a C 1 -function F : U R r R n Let p S, so p = Fq for some q U, at which the Jacobian matri JFq is of full rank Prove that where T p S is the Tangent Space to S at p {JFqw : w R r } T p S, 4 Hint Look at the definition of T p S and consider γt = Fq+tw Solution By definition, v T p S if, and only if, there eists a curve in S, ie differentiable γ : η,η S, such that γ0 = p and γ 0 = v Let v {JFqw : w R r } so v = JFqw for some w R r Define γt = Fq+tw Note that q U and U is an open set so q+tw U for t in some interval η,η with η > 0 For such t, Fq+tw and thus γt is defined Checking we see that γ0 = Fq = p And by the Chain Rule, γ 0 = JFqw = v Hence v T p S and the result 4 follows Note this is the easier half of a major result in the course that says there is equality in 4 The result is proven in the lectures when F is the image of a graph The proof of 4 in general is in the appendices to the notes and requires noting that a surface given parametrically is locally a graph 11 In each case, find parametric equations for the Tangent Plane passing through the point Fq on the parametric surfaces given by the following functions i F,y T = +y, y, 3y T, at q = 1, T, ii F,y T = y, +y, 3 y, y T, at q = 1, T, iii Ft = cost, sint, t T at q = 3π Solution From the Theory: A result from the notes states that if JFq is of full rank then the tangent plane to the image set of a function is the image set of the Best Affine Approimation to the function 19

20 i The Jacobian matri is y JF = y so JFq = The Jacobian matri JFq is of full rank so the Best Affine Approimation and thus the Tangent Plane is the image set of 5 4 Fq+JFq q = y y 5 = +y, 3y for R Note that the last coordinate function for the Tangent plane is identical to the last one in the definition of F This should be no surprise since 3y is linear ii As already seen in Question 9 the Jacobian matri is y y JFq = 3 = y y 0 4 =q It is quickly seen from the last row that the two columns are linearly independent make sure you understand this and so JFq is of full-rank Then the Best Affine Approimation and thus the Tangent Plane is the image set of Fq+JFq q = = y +8 +y 1 3 4y +6 4y , +1 y

21 with R iii The Best Affine Approimation and thus the Tangent Plane is the image set of 1 0 Fq+JFqt q = 0 + t 3π 3π 1 = 1 t+6π t Though called a tangent plane this is geometrically a line Figure for Question 11iii: z p y Level sets are locally sets 1 Show that the following level sets are graphs around the point given i,y,z T R 3 : y z 3 y z + 3 y = 18 with p =,3, 1 T ii,y,z T R 3 : +3y +z = 9, yz =, with p =, 1,1 T 1

22 Solution To show that the level set : f = 0 for some C 1 -function f : U R n R m is locally a graph we apply the Implicit Function Theorem ThissaysthatifthefinalmcolumnsoftheJacobianmatriJfparelinearly independentthenthelevelsetisthegraphofsomefunctionφ : V R n m R m i With f = y z 3 y z + 3 y 1 we have Jf p = f,3, 1 = y z 3 y z +3 y,yz 3 yz + 3 y,3y z y z =p = 63,1,16 The last column is non-zerothe equivalence of linearly independent when only one term So by the Implicit Function Theorem the last variable, z, can be given as a function of the first two, and y, in a neighbourhood of,3 T Note you can also solve for y as a function of and z in a neighbourhood of, 1 T or for in terms of y and z in a neighbourhood of 3, 1 T ii The Jacobian of the level set at p is 6y 4z Jfp = yz z y =p = 1 The last two columns are linearly independent so the system can be solved with y and z functions of in a neighbourhood of Tangent Planes to level sets 13 Let the surface S be given as a level set S = { U : f = 0 and Jf is of full-rank }, 5 for some C 1 -function f : U R n R m Let p S Prove that T p S { R n : Jfp = 0} Solution Let v T p S This means there eists a differentiable curve γ : η,η S such that γ0 = p and γ 0 = v Since S is a level set the requirement γt S means that fγt = 0 for all η < t < η

23 Differentiate this using the Chain Rule to find Jfγtγ t = 0 Choose t = 0 to get Jfpv = 0 Hence v { R n : Jfp = 0} and 5 follows Note this is the easier half of a fundamental result in the course which states that we have equality in 5 See the appendices for a proof of the general result It proceeds by using the Implicit Function Theorem to say that S is, locally near p, a graph A proof of 5 is given in the notes for a graph; this can simply be applied, the complication comes from epressing the result in the form seen in the RHS of 5 14 For each of the following level sets find the tangent plane to the surface at the given point p and give your answer as a level set i,y,z T R 3 : +y +z = 14 with p =,1, 3 T, ii,y,z T R 3 : with p =, 1,1 T, iii,y,u,v R 4 : with p = 1,, 1, T, +3y +z = 9, yz =, 3 3yu+u +v = 1 v +y 3u 3yv = 3 Solution Theory: From the notes the Tangent Plane to a point p on a level set f 1 0 is { : Jfp p = 0} i With f = +y +z 14 and p =,1, 3 T the Jacobian matri is Jfp = 4,, 6 This is non-zero and so of full-rank and thus the Tangent Plane is those R 3 satisfying that is 4+y 6z 8 = 0 4,, 6 y 1 z +3 = 0, 3

24 FigureforPartishowingthespherewiththeTangentPlanez = 4+y 8/6 : z y p ii The Jacobian of the level set at p is 6y 4z Jfp = yz z y =p = 1 The rows are not linear multiples of each other so Jfp is of full rank Hence the Tangent Plane is those R 3 satisfying y +1 = 0, 1 z 1 that is 9 3y +z = 0, +6+y z = 0 iii The Jacobian of the level set at p is 3 +v 3u 3y +u v 4y 3v 6u v 3y =p = The last two columns are linearly independent so Jfp is of full rank Hence the Tangent Plane is those R 4 satisfying y 4 6 u+1 = 0, v 4

25 which is the level set 7+3y 8u+v = 5, 4+y +6u v = 15 Return to your answers of Question 14 and write them as graphs instead of level sets Then give a basis for the Tangent Space Solution i The Tangent Plane was previously given, in Question 14, as those R 3 satisfying 4+y 6z 8 = 0 This can be written as the graph Since y 4+y 8/6 y 4+y 8/6 = 0 0 8/6 : y + R 1 0 /3 +y 0 1 1/3 a basis for the Tangent Space will be 1,0,/3 T and 0,1,1/3 T which we could re-scale as 3,0, T and 0,3,1 T ii The Tangent Plane was previously given as those R 3 satisfying 9 3y +z = 0, +6+y z = 0 These simultaneous equations can be solved for y and z as functions of, giving y = 3 and z = / Thus the plane is given by the graph 3 : R / This is the graph of the vector valued function 3 φ = / Though called a plane it is geometrically a line 5,

26 Writing 3 / = 1 1 1/ , we see that,,1 T is a basis vector for the Tangent Space iii The Tangent Plane was previously given as those R 4 satisfying 7+3y 8u+v = 5 4+y +6u v = Equivalent to solving for u and v is to start with the augmented matri Then apply row operations r 1 r 1 +r r r +3r We could continue to get the identity matri in the columns corresponding to u and v, but instead we translate the matri back into equations u = 11+5y 3 v = 37+17y 67 Thus the level set is the graph of the function φ : R R, 11+5y 3/ φ =, 37+17y 67/ where =,y T R Looking at the columns in T, φ T T we find that the vectors,0,11,37 T and 0,,5,17 T span the Tangent Space 16 Return to Question 3 on Sheet 6 You were asked to show, by using the Implicit Function Theorem, that the following equations +y +uv = y 3 +u 3 v 3 = 0, 6

27 determine u and v as functions of and y for,y T in an open subset of R containing the point q = 1,1 T R The implicit function theorem is an eistence result, it does not say what u and v are as functions of and y Nonetheless it is possible to find their partial derivatives and you were asked to do this The answer was u v q = 0, q = 1, u y v q = 1 and q = 0 y Use these partial derivatives to find a basis for the tangent space at p = 1,1,1,1 T Solution The Implicit Function Theorem says that, for,y T restricted to some set V containing q, the points in the level set lie in the image set of F = y u,y, v,y where =,y T V Yet the Tangent space at a point p S on a surface given parametrically as the image of F is spanned by the columns of JFq In our case the Jacobian at q is uq/ uq/ y vq/ vq/ y = Hence the Tangent Space is spanned by 1,0,0,1 T and 0,1, 1,0 T To double check The Jacobian matri of the system at p is y v u Jfp = 3 3y 3u 3v = The rows of Jfp span T p S so we need vectors orthogonal to the rows of Jfp It is easily checked that both 1,0,0,1 T and 0,1, 1,0 T are orthogonal to all rows of Jfp In addition they are linearly independent which means they form a basis for T p S =p 17 Let Su = cosusinv, sinusinv, cosv T, where u = u,v T, with 0 v π, 0 u π This is the surface of the unit ball in R 3 in standard spherical coordinates 7

28 i Show that the tangent space of S at q = π, π/ T is T p S = Spane, e 3, where p = Sq ii Determine also the tangent space at q = 0, π/4 T iii a Letw = 1,, 1 T / 6 Showthatw T p S wherep = S 0, π/4 T b But the definition of T p S is that w T p S only if there eists a curve a : I S such that α0 = p and α 0 = w Find a α in this case Hint In the notes we prove that T p S = {JFq} when S = ImF Look at that proof which constructs a curve within a surface Solution For a parametrically defined set the tangent space is spanned by the columns of the Jacobian matri In this case sinusinv cosucosv JSu = cosusinv sinucosv 0 sinv i With q = π, π/ T, JSq = The columns are e and e 3 but Spane, e 3 = Spane, e 3 so result follows z y p 8

29 ii With q = 0, π/4t, JSq = So we can choose 1, 0, 1T and 0, 1, 0T as a basis for Tp S z p y iii a The first part follows since w= = + Tp S, It might not look obvious from the following figure that w lies in the plane spanned by v1 and v : z p y 9

30 But it we change our viewpoint to sideways on to the plane it is more believable z p y b With v R to be chosen our curve will be for 1 t 1, say α : R R 3, t Sq+tv, By its definition the image of α lies in the surface of the sphere and α0 = Sq = p as required The function α is a composition of S and f t = q+tv so, to find the tangent vector α t, we need apply the Chain Rule For a vector-valued function of one variable the derivative equals the Jacobian matri, so α t = Jαt = J S ft = JSftJft = JSftf t = JSftv Putting t = 0 gives α 0 = JSf0v = JSqv We require α 0 = w so choose v = u,v T such that JSqv = w, ie u v =

31 Solve this to find v = / 3,1/ 3 T Thus the required curve is That is t S 0 π/4 t Figure for Question 17 iiib + t 3 1 = S cos t/ 3 sin π/4+t/ 3 sin t/ 3 sin π/4+t/ 3 cos π/4+t/ 3 z t/ 3 π/4+t/ 3 p w y 31