Question #2fcf6

2 Answers
Jan 27, 2017

ans.the equation is y=x-1.

Explanation:

The equation of the curve is x+y-1=ln(x^2+y^2)
we can write it as e^(x+y-1)=x^2+y^2
or,e^(x+y)=e(x^2+y^2)
or e^x e^y=e(x^2+y^2) color (red) (1)
now differentiating eq. 1 w.r.t x we get
e^x e^y (dy/dx) +e^x e^y =2ex+2ey(dy/dx)
rearranging we get ,(dy/dx) =(2ex-e^(x+y))/(e^(x+y)-2ey)

so we get (dy/dx) at (1,0) =(2e-e)/(e-0) =e/e=1.
so the eq. of the tangent line at the point (1,0) will be ,
(y-0)=(dy/dx) at (1,0) (x-1)
or y=(x-1). (ans).

Jan 27, 2017

y=x-1

Explanation:

x+y-1=ln(x^2+y^2). Term-by-term differentiation gives

1+y'=1/(x^2+y^2)(2x+2yy'). At P(1, 0) this becomes

1+y'=2, giving y'=0, at P.

So, the equation to the tangent at P)1, 0) is

y-0=(1)(x-1)

Note in the graphs the tangent crossing the curve at the point of

inflexion P(1, 0). The scales, for the two graphs, are different

graph{(x+y-1-ln(x^2+y^2))(y-x+1)((x-1)^2+y^2-.01)=0 [-5, 5, -2.5, 2.5]}

graph{(x+y-1-ln(x^2+y^2))(y-x+1)((x-1)^2+y^2-.0001)=0 [0, 2, -.5, .5]}