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my professor didn't cover this material that well so I'm not sure how to do this one. the question is: Find an equation of the tangent plane to the parametric surface $x=2r\cos\theta$, $y=5r\sin\theta$, $z=r$, at the point $(2\sqrt2$, $5\sqrt2$, 2) where $r = 2$ and $\theta = \pi/4$

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Example 3 here might help. – Dylan Moreland Dec 18 '11 at 19:17
@Dylan I love his site; overall, the best student-centered, first/second year math courses site I've encountered on the web. – David Mitra Dec 18 '11 at 19:19

The parametric surface has the following parametric equation: $$f(r,\theta)=(x,y,z)=(2r\cos\theta,5r\sin\theta,r).$$ Then we have $$f_r(2,\pi/4)=(2\cos(\pi/4),5\sin(\pi/4),1)=(\sqrt{2},5\sqrt{2}/2,1),$$ $$f_\theta(2,\pi/4)=(-2\cdot 2\cdot\sin(\pi/4),5\cdot 2\cdot\cos(\pi/4),0)=(-2\sqrt{2},5\sqrt{2},0).$$ Therefore, the unit normal $n$ of the tangent plane at $(r,\theta)=(2,\pi/4)$ is equal to $$n=\frac{f_r(2,\pi/4)\times f_\theta(2,\pi/4)}{\|f_r(2,\pi/4)\times f_\theta(2,\pi/4)\|}=\frac{1}{\sqrt{458}}(-5\sqrt{2},-2\sqrt{2},20).$$ Since the tangent plane at $(r,\theta)=(2,\pi/4)$ passes through the point $f(2,\pi/4)=(2\sqrt{2},5\sqrt{2},2)$, the equation of the tangent plane is given by $n\cdot(x-2\sqrt{2},y-5\sqrt{2},z-2)=0$, or equivalently, $$(-5\sqrt{2},-2\sqrt{2},20)\cdot(x-2\sqrt{2},y-5\sqrt{2},z-2)=0,$$ that is $$5\sqrt{2}x+2\sqrt{2}y-20z=0.$$

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