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Question

Find the points on the surface

xy+yz+zxxz2=0x y + y z + z x - x - z ^ { 2 } = 0

where the tangent plane is parallel to the xy-plane.

Solution

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Answered 12 months ago
Answered 12 months ago
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1 of 2

Let

f(x,y,z)=xy+yz+zxxz2=0\begin{align} f(x,y,z)=xy+yz+zx-x-z^2=0 \end{align}

Then the gradiend f\nabla f is

f=fx(x,y,z)i+fy(x,y,z)j+fz(x,y,z)k=(y+z1)i+(x+z)j+(y+x2z)k\begin{align*} \nabla f&=f_x(x,y,z)\textbf{i}+f_y(x,y,z)\textbf{j}+f_z(x,y,z)\textbf{k}\\&= (y+z-1)\textbf{i}+(x+z)\textbf{j}+(y+x-2z)\textbf{k} \end{align*}

Since we need to find the points when the given tangent plane is parallel to the xyxy-plane we have that the gradient f\nabla f is perpendicular to the xyxy-plane.

Hence we need to solve the system:

{y+z1=0x+z=0{y=1zx=z\begin{align*} \begin{cases} y+z-1=0\\ x+z=0 \end{cases}\Rightarrow\begin{cases} y=1-z\\ x=-z \end{cases} \end{align*}

Returning this into (1)(1) gives

(z)(1z)+(1z)z+z(z)(z)z2=0z+z2+zz2z2+zz2=0z2z2=z(12z)=0z=0z=12\begin{align*} (-z)(1-z)+(1-z)z+z(-z)-(-z)-z^2=0&\Rightarrow -z+z^2+z-z^2-z^2+z-z^2=0\\&\Rightarrow z-2z^2=\\&\Rightarrow z(1-2z)=0\\&\Rightarrow\color{#4257b2} z=0\vee z=\frac{1}{2} \end{align*}

For z=0\color{#4257b2}z=0 we get

{y=10=1x=0\begin{align*} \color{#4257b2}\begin{cases} y=1-0=1\\ x=0 \end{cases} \end{align*}

and for z=12\color{#4257b2}z=\frac{1}{2} we get

{y=112=12x=12\begin{align*} \color{#4257b2}\begin{cases} y=1-\frac{1}{2}=\frac{1}{2}\\ x=-\frac{1}{2} \end{cases} \end{align*}

Hence, the wanted points are

(0,1,0),(12,12,12)\begin{align*} \color{#c34632}\boxed{(0,1,0),\left(-\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)} \end{align*}

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