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Given that that near (1, 1, 1) the curve of intersection of the surfaces x^3 + y^2 - z^4 - 3xyz =...

Question:

Given that that near (1, 1, 1) the curve of intersection of the surfaces {eq}x^3 + y^2 - z^4 - 3xyz = 0 {/eq} and {eq}xy + yz + zx - 3z^8 = 0 {/eq} has the parametric equations x = f(t), y = g(t), z = t. with f, g differentiate, the derivatives f'(1) = ?, g'(1) = ?

and the tangent line to the curve of intersection at (1, 1, 1) is?

Tangent Planes


The tangent plane to a surface {eq}\displaystyle f(x,y,z)=0, {/eq} at a point {eq}\displaystyle (x_0,y_0,z_0) {/eq} is given by {eq}\displaystyle f_x(x_0,y_0,z_0)(x-x_0)+ f_y(x_0,y_0,z_0)(y-y_0)+ f_z(x_0,y_0,z_0)(z-z_0)=0 {/eq} where

{eq}\displaystyle f_x, f_y, f_z {/eq} are the partial derivatives of {eq}\displaystyle f(x,y,z). {/eq}

To find the tangent vector to a curve of intersection of two surfaces at a given point, we will find the vector that is the cross product of the normal vectors to the surfaces, at the given point, because the tangent vector will be in both tangent planes to the given surfaces.

Answer and Explanation:


To find the tangent line to the curve of intersection of the surfaces

{eq}\displaystyle x^3 + y^2 - z^4 - 3xyz = 0, xy + yz + zx - 3z^8 = 0 {/eq} at the point {eq}\displaystyle (1,1,1) {/eq}

we will obtain the tangent vector to the curve of intersection, first.

The tangent vector is given by the cross product of the normal vectors to the two surfaces, which are

{eq}\displaystyle \begin{align} \mathbf{n}_1(x,y,z)&=\left\langle \frac{\partial }{\partial x}\left(x^3 + y^2 - z^4 - 3xyz \right), \frac{\partial }{\partial y}\left(x^3 + y^2 - z^4 - 3xyz \right), \frac{\partial }{\partial z}\left(x^3 + y^2 - z^4 - 3xyz \right)\right\rangle=\left\langle 3x^2 - 3yz , 2 y - 3xz , - 4z^3 - 3xy \right\rangle\\ \implies \mathbf{n}_1(1,1,1)&=\left\langle 0 , -1, - 7\right\rangle\\ \mathbf{n}_1(x,y,z)&=\left\langle \frac{\partial }{\partial x}\left(xy + yz + zx - 3z^8 \right), \frac{\partial }{\partial y}\left(xy + yz + zx - 3z^8 \right), \frac{\partial }{\partial z}\left(xy + yz + zx - 3z^8 \right)\right\rangle=\left\langle y+z, x+z, y+x-24z^2 \right\rangle\\ \implies \mathbf{n}_2(1,1,1)&=2\left\langle 1 , 1, - 11\right\rangle\\ \mathbf{n}_1(1,1,1)\times \mathbf{n}_2(1,1,1) &= \begin{array}{|ccc|} \mathbf{i} &\mathbf{j}&\mathbf{k}\\ 0&-1&-7\\ 1&1&-11 \end{array} =\langle 18, -7, 1\rangle. \end{align} {/eq}

Therefore the parametric curve of intersection {eq}\displaystyle x=f(t), y=g(t), z=t {/eq} has the derivatives {eq}\displaystyle \boxed{ f'(1)=18, g'(1)=-7}, {/eq}

because the derivatives represent the components of the tangent vector to the curve with the z-component of the vector, 1.

Therefore, the tangent line to the curve of intersection at {eq}\displaystyle (1,1,1) {/eq} with the direction vector {eq}\displaystyle \langle 18, -7, 1\rangle {/eq}

is {eq}\displaystyle \boxed{x=1+18t, y=1-7t, z=1+t, t\in\mathbf{R}}. {/eq}


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Parametric Equations in Applied Contexts

from Precalculus: High School

Chapter 24 / Lesson 6
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