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Math Help - Calculus-Please Help!

  1. #1
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    Calculus-Please Help!

    Find an equation of the tangent plane to the parametric surface x=2rcos(theta) y=-4rsin(theta) z=r at the point (2sqrt(2),-4sqrt(2),2) when r=2, theta=pi/4
    z=
    "hint: your answer should be an expression of x and y"
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  2. #2
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    Quote Originally Posted by karnold9 View Post
    Find an equation of the tangent plane to the parametric surface x=2rcos(theta) y=-4rsin(theta) z=r at the point (2sqrt(2),-4sqrt(2),2) when r=2, theta=pi/4
    z=
    "hint: your answer should be an expression of x and y"
    f(r,\theta)=<2r\cos(\theta),-4r\sin(\theta),r>

    taking some derivatives we get

    \frac{\partial f}{\partial r}=<2\cos(\theta),-4\sin(\theta),1>

    \frac{\partial f}{\partial \theta}=<-2r\sin(\theta),-4r\cos(\theta),0>

    Evaluating them at the point (2,\frac{\pi}{4})


    f(2,\frac{\pi}{4})=<2(2)\cos(\pi/4),-4(2)\sin(\pi/4),2>= <2\sqrt{2},-4\sqrt{2},2>

    \frac{\partial f}{\partial r}=<2\cos(\pi/4),-4\sin(\pi/4),1>=<\sqrt{2},-2\sqrt{2},1>

    \frac{\partial f}{\partial \theta}=<-2(2)\sin(\pi/4),-4(2)\cos(\pi/4),0>=<-2\sqrt{2},-4\sqrt{2},0>

    Now if we take the cross product of the partials we will get a normal vector to the parameteric surfact at the point.

    \begin{vmatrix}<br />
\vec i && \vec j && \vec k \\<br />
\sqrt{2} && -2\sqrt{2} && 1 \\<br />
-2\sqrt{2} && -4\sqrt{2} && 0 \\ <br />
\end{vmatrix}= (0+4\sqrt{2})\vec i -(0+2\sqrt{2}) \vec j + (-8-8) \vec k

    So the normal vector is \vec n = <4\sqrt{2},-2\sqrt{2},-16>

    Now let (x,y,z) be any point in the plane. Using the other point we calculated before (2\sqrt{2},-4\sqrt{2},2) we can create a vector in the plane  \vec v_p=<x-2\sqrt{2},y+4\sqrt{2},z-2>

    Since the above vector lies in the plane and we have a normal vector to the plane the dot product of these two vectors must be zero.(Becuase they are perpendicular)

    \vec n \cdot \vec v_p=0

    (4\sqrt{2})(x-2\sqrt{2})+(-2\sqrt{2})(y+4\sqrt{2})+(-16)(z-2)=0
    4\sqrt{2}x-16-2\sqrt{2}y-16-16z+32=0 simplifying

    2\sqrt{2}x-\sqrt{2}y-8z=0 \iff z=\frac{\sqrt{2}}{4}x-\frac{\sqrt{2}}{8}y
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    f(r,\theta)=<2r\cos(\theta),-4r\sin(\theta),r>

    taking some derivatives we get

    \frac{\partial f}{\partial r}=<2\cos(\theta),-4\sin(\theta),1>

    \frac{\partial f}{\partial \theta}=<-2r\sin(\theta),-4r\cos(\theta),0>

    Evaluating them at the point (2,\frac{\pi}{4})


    f(2,\frac{\pi}{4})=<2(2)\cos(\pi/4),-4(2)\sin(\pi/4),2>= <2\sqrt{2},-4\sqrt{2},2>

    \frac{\partial f}{\partial r}=<2\cos(\pi/4),-4\sin(\pi/4),1>=<\sqrt{2},-2\sqrt{2},1>

    \frac{\partial f}{\partial \theta}=<-2(2)\sin(\pi/4),-4(2)\cos(\pi/4),0>=<-2\sqrt{2},-4\sqrt{2},0>

    Now if we take the cross product of the partials we will get a normal vector to the parameteric surfact at the point.

    \begin{vmatrix}<br />
\vec i && \vec j && \vec k \\<br />
\sqrt{2} && -2\sqrt{2} && 1 \\<br />
-2\sqrt{2} && -4\sqrt{2} && 0 \\ <br />
\end{vmatrix}= (0+4\sqrt{2})\vec i -(0+2\sqrt{2}) \vec j + (-8-8) \vec k

    So the normal vector is \vec n = <4\sqrt{2},-2\sqrt{2},-16>

    Now let (x,y,z) be any point in the plane. Using the other point we calculated before (2\sqrt{2},-4\sqrt{2},2) we can create a vector in the plane  \vec v_p=<x-2\sqrt{2},y+4\sqrt{2},z-2>

    Since the above vector lies in the plane and we have a normal vector to the plane the dot product of these two vectors must be zero.(Becuase they are perpendicular)

    \vec n \cdot \vec v_p=0

    (4\sqrt{2})(x-2\sqrt{2})+(-2\sqrt{2})(y+4\sqrt{2})+(-16)(z-2)=0
    4\sqrt{2}x-16-2\sqrt{2}y-16-16z+32=0 simplifying

    2\sqrt{2}x-\sqrt{2}y-8z=0 \iff z=\frac{\sqrt{2}}{4}x-\frac{\sqrt{2}}{8}y
    Wow empty set...that was a lot of work.

    Couldn't ou have said that since r>0 that

    \sqrt{\frac{x^2}{4}+\frac{y^2}{16}}=z

    And then just used the fact that

    z_{\text{tangent plane}}=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)??
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