13. Implicit functions Consider the curve y2 = x in the plane R2 , C = { (x,y) ∈ R2 | y 2 = x }. This is not the graph of a function, and yet it is quite close to the graph of a function. Given any point on the graph, let’s say the point (2, 4), we can always find open intervals U containing 2 and V containing 4 and a smooth function f : U −→ V such that C ∩ (U × V ) is the graph of f. Indeed, take U = (0, ∞), V = (0, ∞) and f(x)= √x. In fact, we can do this for any point on the graph, apart from the origin. If it is above the x-axis, the function above works. If the point we are interested in is below the x-axis, replace V by (0, −∞) and f(x)= √x, by g(x)= −√x. How can we tell that the origin is a point where we cannot define an implicit function? Well away from the origin, the tangent line is not vertical but at the origin the tangent line is vertical. In other words, if we consider F : R2 −→ R, given by F (x,y)= y2 − x, so that C is the set of points where F is zero, then DF (x,y)=(−1, 2y). The locus where we run into trouble, is where 2y = 0. Somewhat amazingly this works in general: Theorem 13.1 (Implicit Function Theorem). Let A ⊂ Rn+m be an open subset and let F : A −→ Rm be a C1-function. Suppose that (�a,�b) ∈ S = { (�x,�y) ∈ A | F (�x,�y)= �0 }. Assume that �� ∂Fidet =0. ∂yj �Then we may find open subsets �a ∈ U ⊂ Rn and �b ∈ V ⊂ Rm, where U × V ⊂ A and a function f : U −→ V such that S ∩ (U × V ) is the graph of f, that is, F (�x,�y)= �0 if and only if �y = f(�x). where �x ∈ U and �y ∈ V . Let’s look at an example. Let F : R3 −→ R, 1 ���� ���� be the function F (x1,x2,y)= x 31x2 − x2y 2 + y 5 +1. Let S = { (x1,x2,y) ∈ R3 | F (x1,x2,y)=0 }. Then (1, 3, −1) ∈ S. Let’s compute the partial derivatives of F , ∂F 2(1, 3, −1) = 3x ∂x1 1x2 =9(1,3,−1) ∂F 32(131)=()−−xy,, 1∂x2(131)−,,����=0∂F ∂y (1, 3, −1) = (−2x2y +5y 4) = 11.(1,3,−1) So DF (1, 3, −1) = (9, 0, 11). Now what is important is that the last entry is non-zero (so that the 1 × 1 matrix (1) is invertible). It follows that we may find open subsets (1, 3) ∈ U ⊂ R2 and −1 ∈ V ⊂ R and a C1 function f : U −→ V such that F (x1,x2,f(x1,x2)) = 0. It is not possible to write down an explicit formula for f, but we can calculate the partial derivatives of f. Define a function G: U −→ R, by the rule G(x1,x2)= F (x1,x2,f(x1,x2)) = 0. On the one hand, ∂G ∂G =0 and =0. ∂x1 ∂x2 On the other hand, by the chain rule, ∂G = ∂F ∂x1 + ∂F ∂x2 + ∂F ∂f ∂x1 ∂x1 ∂x1 ∂x2 ∂x1 ∂x3 ∂x1 Now ∂x1 = 1 and ∂x2 = 0. ∂x1 ∂x1 So ∂F ∂f ∂x1 . ∂x1 = − ∂F ∂x3 2 � � Similarly ∂F ∂f ∂x2 . ∂x2 = − ∂F ∂x3 So ∂F ∂f (1, 3) = − ∂x1 (1, 3, −1) 9 ,∂x1 ∂F (1, 3, −1) = −11∂x3 and ∂f ∂F (1, 3, −1) 0∂x2 ∂x2 (1, 3) = − ∂F (1, 3, −1) = −11 =0. ∂x3 Definition 13.2. Let A ⊂ Rn be an open subset and let f : Rn −→ R be a function. The directional derivative of f in the direction of the unit vector uˆis Duˆf(P ) = lim f(P + huˆ) − f(P ) . h0 h→If ˆu =ˆei then,∂fDeˆi f(P )= (P ),∂xi the usual partial derivative. Proposition 13.3. If f is differentiable at P then Duˆf(P )= Df(P ) u.ˆ· Proof. Since A is open, we may find δ> 0 such that the parametrised line r :(−δ,δ) −→ A, given by r(h)= f(P )+ huˆis entirely contained in A. Consider the composition of r and f, f ◦ r : R −→ R. Then Duˆf(P )= d(f ◦ r) (0)dh = D(r(0)) Dr(0)· = Df(P )ˆ�u.· Note that we can also write Duˆf(P )= �f(P ) u.ˆ· 3 Note that the directional derivative is largest when �f(P ) uˆ= , ��f(P )� so that the gradient always points in the direction of maximal change (and in fact the magnitude of the gradient, gives the maximum change). Note also that the directional derivative is zero if ˆu is orthogonal to the gradient and that the directional derivative is smallest when u = −�f(P )ˆ. ��f(P )� Proposition 13.4. If �f(P )=�0 then the tangent hyperplane Π to the hypersurface S = { Q ∈ Rn | f(Q) − f(P )=0 }, is the set of all points Q which satisfy the equation �f(P ) −→PQ =0.· Remark 13.5. If f is C1, then f is the graph of some function, locally about P . Proof. By definition, the point Q belongs to the tangent hyperplane if and only if there is a curve r :(−δ,δ) −→ S, such that r(0) = P and PQ. r�(0) = −→Now, since r(h) ∈ S for all h ∈ (−δ,δ), we have F (r(h))=0. So dF (r(h))0 = (0)dh = �F (r(0)) r�(0)· = �F (P ) −→�PQ. · 4 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.