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Multivariable Calculus Question:

Find an equation of the straight line tangent, at the given point, to the level curve of the given function passing through that point.

f(x, y) = [tex] x^{2}- y^{2} [/tex] at (2, -1)

Thanks!

Respuesta :

LammettHash
[tex]f(2,-1)=2^2-(-1)^2=3[/tex]

so we're considering the level curve

[tex]x^2-y^2=3[/tex]

The tangent line to this curve at (2, -1) will be the value of [tex]\dfrac{\mathrm dy}{\mathrm dx}[/tex] at this point. Differentiating yields

[tex]2x-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac xy[/tex]

and so the slope of the tangent would be [tex]\dfrac2{-1}=-2[/tex].

The tangent line then has equation

[tex]y-(-1)=-2(x-2)\implies y=-2x+3[/tex]

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