Find an equation of the tangent plane to the given surface at the specified point. z = 3y² - 2x² + x, (2, -1, -3)

Solution:

Given, the equation is z = 3y² - 2x² + x

We have to find the equation of the plane at a given point (2, -1, -3).

We have function,

f(x, y, z) = 3y² - 2x² + x - z

In order to find the normal at any particular point point in vector space, we use gradient operator,

\(\bigtriangledown f(x, y, z)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}\)

\(\frac{\partial f}{\partial x}(3y^2-2x^2+x-z)=-4x+1\)

\(\frac{\partial f}{\partial y}(3y^2-2x^2+x-z)=6y\)

\(\frac{\partial f}{\partial z}(3y^2-2x^2+x-z)=-1\)

So, \(\bigtriangledown f(x, y, z)=-4x+1+6y-1=-4x+6y\\\bigtriangledown f(x, y, z)=-4x\hat{i}+6y\hat{j}+0\hat{k}\)

At the given point (2, -1, -3) the normal vector to the surface is given by

\(\bigtriangledown f(2, 1,-3)=-4(2)\hat{i}+6(-1)\hat{i}+0(-3)\hat{k}\\=-8\hat{i}-6\hat{j}+0\hat{k}\)

The general form of vector equation is \(\vec{r}.\vec{n}=\vec{a}.\vec{n}\)

Where, \(\vec{r}=\begin{pmatrix} x\\ y\\z \end{pmatrix}.\vec{n}=\begin{pmatrix} -8\\-6 \\0\end{pmatrix}\) is the normal vector and a is any point in the given plane.

The equation of the plane is given by

\(\begin{pmatrix} x\\y \\z \end{pmatrix}.\begin{pmatrix} -8\\-6 \\0 \end{pmatrix}=\begin{pmatrix} 2\\-1 \\-3 \end{pmatrix}.\begin{pmatrix} -8\\-6 \\0 \end{pmatrix}\)

So, x(-8) + y(-6) + z(0) = 2(-8) + -1(-6) + -3(0)

-8x - 6y + 0 = -16 + 6 + 0

-8x - 6y = -10

8x + 6y - 10 = 0

Therefore, the equation of the tangent plane is 8x + 6y - 10 = 0.

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Find an equation of the tangent plane to the given surface at the specified point. z = 3y² - 2x² + x, (2, -1, -3)

Summary:

The equation of the tangent plane to the surface z = 3y² - 2x² + x at a given point (2, -1, -3) is 8x + 6y - 10 = 0.