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(1 point) Find an equation of the tangent plane to the parametric surface x=5rcosθx=5rcos⁡θ, y=4rsinθy=4rsin⁡θ, z=rz=r at the point (52–√,42–√,2)(52,42,2) when r=2r=2, θ=π/4θ=π/4. z = Note: Your answer should be an expression of x and y; e.g. "3x - 4y"

Answer :

LammettHash

Denote the given surface by the vector-valued function,

[tex]\vec s(r,\theta)=\langle 5r\cos\theta,4r\sin\theta,r\rangle[/tex]

Then the normal vector to this plane is

[tex]\vec n=\dfrac{\mathrm d\vec s}{\mathrm dr}\times\dfrac{\mathrm d\vec s}{\mathrm d\theta}=\langle-4r\cos\theta,-5r\sin\theta,20r\rangle[/tex]

When [tex]r=2[/tex] and [tex]\theta=\frac\pi4[/tex], the normal vector is

[tex]\vec n=\left\langle-\dfrac8{\sqrt2},-\dfrac{10}{\sqrt2},40\right\rangle[/tex]

and so the equation of the tangent plane is

[tex]\vec n\cdot\langle x-5\sqrt2,y-4\sqrt2,z-2\rangle=0[/tex]

[tex]\implies-\dfrac8{\sqrt2}(x-5\sqrt2)-\dfrac{10}{\sqrt2}(y-4\sqrt2)+40(z-2)=0[/tex]

[tex]\implies\boxed{4x+5y-20\sqrt2\,z=0}[/tex]

Judging by the "note", you need to write the equation as [tex]z(x,y)[/tex], which is trivial:

[tex]\boxed{z=\frac{4x+5y}{20\sqrt2}=\frac x{5\sqrt2}+\frac y{4\sqrt2}}[/tex]

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