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I have the below system: $$u^3+v^3+x^5+4y^5=10\\\ u^2+v^2+x^9+y^8=11$$ I was asked by my students to find $u_x,v_y$ where $x,y$ are our independent variables. Please give me some hints. Thank you!

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Hint: Implicit differentiation. –  Daryl Jan 18 '13 at 6:17
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Since when students give exercises to teachers? –  Fabian Jan 18 '13 at 6:30

2 Answers 2

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The two equations $$\eqalign{F(x,y, u,v)&:=u^3+v^3+x^5+4y^5-10=0\ ,\cr G(x,y,u,v)&:= u^2+v^2+x^9+y^8-11=0\cr}$$ define a two-dimensional surface $S\subset{\mathbb R}^4$. This surface may have self-intersections, bubbles, cusps, and many other kinds of singularities. Therefore you cannot expect that there are "global" functions $$(x,y)\mapsto u(x,y)\ ,\qquad (x,y)\mapsto v(x,y)\qquad(1)$$ such that $S$ appears as graph of the function pair $(u,v)$ in the form $$S=\bigl\{(x,y,u,v)\ \bigm|\ (x,y)\in{\mathbb R}^2,\ u=u(x,y),\ v=v(x,y)\bigr\}\ .$$ But locally such a representation is possible, even if you are not able to solve the given equations for $u$ and $v$ algebraically. This means you are not able to write down $u(x,y)$, $v(x,y)$ explicitly; but such functions are guaranteed to exist; for details see below.

Assume that you have found, e.g. numerically, a point ${\bf p}=(x_0,y_0,u_0,v_0)\in S$. Then under a certain technical assumption there is a $2\times2$-dimensional box $B=Z\times W$ with center ${\bf p}$ such that the part of $S$ lying in this box can be described in the form $$S\cap B=\bigl\{(x,y,u,v)\ \bigm|\ (x,y)\in Z,\ u=u(x,y),\ v=v(x,y)\bigr\}$$ with certain differentiable functions $(1)$ defined in $Z$. It follows that for $(x,y)\in Z$ one has $$F\bigl(x,y,u(x,y),v(x,y)\bigr)=0\ ,\qquad G\bigl(x,y,u(x,y),v(x,y)\bigr)=0\ .$$ Differentiating with respect to $x$ and to $y$ we get four more identities $$\eqalign{F_x + F_u u_x+ F_v v_x&\equiv0\ , \qquad F_y + F_u u_y+ F_v v_y\equiv0\ ,\cr G_x + G_u u_x+ G_v v_x&\equiv0\ ,\qquad G_y + G_u u_y+ G_v v_y\equiv0\ .\cr}\qquad(2)$$ In particular they hold at $(x_0,y_0)$, for which $\bigl(x_0,y_0,u(x_0,y_0),v(x_0,y_0)\bigr)$ is nothing else but the point ${\bf p}=(x_0,y_0,u_0,v_0)$ we started with. The two lefthand equations $(2)$ can be solved for $u_x(x_0,y_0)$ and $v_x(y_0,y_0)$, the two righthand equations $(2)$ for $u_y(x_0,y_0)$, $v_y(x_0,y_0)$ $-$ under the crucial assumption that the determinant of the two equations, which is the same in both cases, is $\ne0$. This is the "technical assumption" referred to above. It reads $$F_u G_v -F_v G_u \bigr|_{\bf p}\ne 0\ .$$

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HINT $$u^3+v^3+x^5+4y^5=10\\\ u^2+v^2+x^9+y^8=11$$ considering $y$ as constant $$3u^2u_x+3v^2v_x+5x^4=0\\\ 2uu_x+2vv_x+9x^8=0$$ considering $x$ as constant $$3u^2u_y+3v^2v_y+20y^4=0\\\ 2uu_y+2vv_y+8y^7=0$$

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