The Implicit Function Theorem - UCLA Mathematics

1 Math 32A Week 10 NotesNovember 29 and December 1, 2016 Austin ChristianThe Implicit Function TheoremSuppose we have a Function of two variables,F(x, y), and we re interested in its height-clevelcurve; that is, solutions to the equationF(x, y) =c. For instance, perhapsF(x, y) =x2+y2andc= 1, in which case the level curve we care about is the familiar unit circle. It wouldbe nice if choosing a value forxin the equationF(x, y) =cwould immediately determinethe value ofy that is, ifF(x, y) =cdeterminedyas a Function ofx. But we know thatthis isn t generally true. In the case of the unit circle, fixing a valuexleaves both 1 x2and 1 x2as possibilities for the value ofy. Graphically, this obstruction is representedby the fact thatx2+y2= 1 fails the familiar vertical line test, as can be seen in Figure 1: The level curvex2+y2= 1. The green segment represents a neighborhood of thered point on whichyis determined else that can be seen in Figure 1, though, is that our graph does pass thevertical line test locally.

2 That is, at most points on the circle we can choose a small neigh-borhood where our curve satisfies the vertical line test, and thus determinesyas a functionofx. The two points where we cannot choose such a neighborhood are (0,1) and (0, 1).Indeed, these are precisely the points exempted from the following important Implicit Function Theorem for a continuously differentiablefunctionF:R2 Rand a point (x0, y0) R2so thatF(x0, y0) =c. If F y(x0, y0)6= 0,then there is a neighborhood of (x0, y0) so that wheneverxis sufficiently close tox0thereis a uniqueyso thatF(x, y) =c. Moreover, this assignment is makesya continuousfunction Theorem says that we can makeya Function ofx except when F y= 0. In our caseFy= 2yvanishes whenevery= 0, and this happens at two points: the two we ve alreadyidentified as problems. The Theorem also holds in three dimensions:The Implicit Function Theorem for a continuously differentiablefunctionF:R3 Rand a point (x0, y0, z0) R3so thatF(x0, y0, z0) =c.

3 If F z(x0, y0, z0)6= 0, then there is a neighborhood of (x0, y0, z0) so that whenever (x, y) issufficiently close to (x0, y0) there is a uniquezso thatF(x, y, z) =c. Moreover, thisassignment is makesza continuous Function , the upshot of the Implicit Function Theorem is that for sufficiently nice pointson a surface, we can (locally) pretend this surface is the graph of a Function . The primaryuse for the Implicit Function Theorem in this course is forimplicit differentiation. You vealready seen the two-variable version of this in your first calculus class. In particular, youprobably did the unit circle example we saw above. The usual way to go about this is toconsider the equationF(x, y) =cand to differentiate both sides with respect tox. For theunit circle this yields2x+ 2ydydx= 0,from which we solve to find thatdydx= x/y. Of course, this expression doesn t make anysense aty= 0, but we ve already excepted the points where this happens.

4 Now that wehave the multivariable chain rule at our disposal, we can actually address this problem moregenerally. Differentiating both sides ofF(x, y) =cwith respect toxgives F x 1 + F y y x= fory (x) leads us toy (x) = FxFywheneverFy6= 0. While we re here we can also address the three-variable case. Supposewe haveF:R3 Rand a point (x0, y0, z0) where the Implicit Function Theorem (x, y), we re interested in the partial derivatives z xand z y. As before, we ll dothis by differentiating the equationF(x, y, z) =cwith respect tox, and then with respecttoy. An important point here is that we re consideringzas a Function ofxandy, butwe re not consideringxandyas depending on each other. We could actually say that we reinterested in thex- ory-derivative of the equationF(x, y, z(x, y)) = respect toxwe have F x 1 + F y 0 + F z z x= for z xgives z x= F/ x F/ z= , z y= Fy/Fz. Here s an example (due to Lincoln Chayes) to test our two three-variable functionsH(x, y, z) andK(x, y, z) and the associatedlevel surfacesH(x, y, z) =aandK(x, y, z) = assume that these surfaces intersect along a curve which contains the point (x0, y0, z0),and that on some neighborhood of this point, the curve determinesyas a functiony(x) ofx.

5 Derive a formula fory (x) nearx0in terms of the partial derivatives ofHandK. (Weassume that the denominators involved in this derivation do not vanish.)(Solution) First we apply the Implicit Function Theorem toHat the point (x0, y0, z0). Thisgives us a functionzH(x, y) on some neighborhood of (x0, y0) so thatH(x, y, zH(x, y)) = we care about the intersection of the two level surfaces, we may substitute thisfunction in place ofzin the formulaK(x, y, z) =b, giving usK(x, y, zH(x, y)) = left side is now a Function of two variables xany and we may apply the implicitfunction Theorem to writeyas a Function ofxfor values ofxnearx0. According to ourformula above fory (x) we havey (x) = x(K(x, y, zH(x, y))) y(K(x, y, zH(x, y))).We apply the chain rule to the numerator to obtain x(K(x, y, zH(x, y))) = K x 1 + K y 0 + K z zH according to the three-variable Implicit Function Theorem we have zH x= HxHz,so x(K(x, y, zH(x, y))) =Kx , y(K(x, y, zH(x, y))) =Ky KzHyHz,soy (x) = Kx KzHx/HzKy KzHy/Hz= KxHz KzHxKyHz see our formula in action, considerH(x, y, z) =x2+y2andK(x, y, z) =x2+y2+ level surfacesH(x, y, z) = 1 andK(x, y, z) = 1 intersect in the unit circle in the plane,where we know thaty (x) = x/yfrom our earlier work.

6 Our new formula givesy (x) = (2x)(0) (2z)(2x)(2y)(0) (2z)(2y)= xy,a welcome reality check. 4