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Question Number 49845 by mhozhez last updated on 11/Dec/18
yxsiny+xysinx=1determinedydx
Commented by MJS last updated on 11/Dec/18
trythisforyourself.themethodissimple:ti(x,y)aretermsinxandyt1(x,y)+t2(x,y)+...=0isthesameas∑ni=1ti(x,y)=0nowderivatedx∑ni=1ddx[ti(x,y)]+dy∑ni=1ddy[ti(x,y)]=0⇒dydx=−∑ni=1ddx[ti(x,y)]∑ni=1ddy[ti(x,y)]anexampleax2+bxy+cy2+dx+ey+f=0dx(2ax+by+0+d+0+0)+dy(0+bx+2cy+0+e+0)=0dx(2ax+by+d)+dy(bx+2cy+e)=0dydx=−2ax+by+dbx+2cy+e
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Dec/18
u=yxsinylnu=xsinylny1u×dudx=xsiny×1ydydx+xlny×cosy×dydx+sinylnydudx=yxsiny[dydx(xsinyy+xcosylny)+sinylny]v=xysinxlnv=ysinxlnx1vdvdx=ysinxx+ylnxcosx+sinxlnxdydxdvdx=xysinx(ysinxx+ylnxcosx+sinxlnxdydx)u+v=1dudx+dvdx=0yxsiny[dydx(xsinyy+xcosylny)+sinylny]+xysinx[ysinxx+ylnxcosx+sinxlnxdydx]=0dydx[yxsiny(xsinyy+xcosylny)+(xysinxsinxlnx)]=−[xysinx(ysinxx+ylnxcosx)+yxsiny(sinylny)]dydx=−[xysinx(ysinxx+ylnxcosx)+yxsiny(sinylny)][yxsiny(xsinyy+xcosylny)+(xysinxsinxlnx)]itistheanswer...
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