We know that if F:R3→Rm is a smooth function, and S is a surface, then F|S:S→Rm is differentiable. Moreover, if p∈S, then (dF|S)p=(dF)p|TpM. In the right side, we have the differential of the restriction of F and in the right side, we have the derivative of F which is only viewed in the tangent plane TpS. Denote f=F|S. From now, and to simplify, we suppose m=1.
It is usual that we work with the Jacobian matrix of dF, which is noting the matrix expression of the linear map dF with respect to the usual basis of Euclidean spaces, here, of R3 and R. By definition, the elements are the partial derivatives of f, namely, Jac F(p)=M(dFp,Bu,B′u)=(fx fy fz)(x,y,z), where p=(x,y,z).
If we want to write the matrix expression of the linear map dfp, we need to fix basis in both spaces, that is, in TpS and in R. Although for R we choose the basis B′u={1}, there is not a natural basis on TpS. This only appears when we fix previously a parametrization X=X(u,v) around p. Then the basis could be B={Xu(q),Xv(q)}, with X(q)=p and the matrix would be M(dfp,B,B′u). Therefore, there is not a relation between this matrix and Jac F(p).
We show an example. In order to avoid 'simple numbers', consider the elliptic paraboloid z=x2+y2 and the function F(x,y,z)=z. Then Jac F(x,y,z)=(0 0 0).
Consider now f=F|S. Then (df)p(v)=α′3(0)=v3, where v=(v1,v2,v3). Let p=(x,y,z)∈S and the parametrization X(x,y)=(x,y,x2+y2). Then Xx=(1,0,2x) and Xy=(0,1,2y). Moreover, using the chain-rule, we have
(df)X(x,y)(Xx)=(f∘X)x=2x, (df)X(x,y)(Xy)=(f∘X)y=2y.
If B={Xx,Xy}, we have M((df)X(x,y),B,B′u)=(2x 2y),
a matrix completely different than JacF(p).
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