# Find an equation for the tangent plane to a surface (using gradient)

In summary, the equation for the tangent plane to the surface xz^2 +x^2y-z=-1 at the point (1,-3,2) is -2x+y+3z=1. To confirm the correctness of the equation, we can check that the normal to the plane is parallel to the gradient at the given point and that the plane contains the given point.

## Homework Statement

Find an equation for the tangent plane to a surface $xz^2 +x^2y-z=-1$ at the point (1,-3,2).

## Homework Equations

$(\vec{r}-\vec{r_p}) \cdot \nabla f(\vec{r_p}) = 0$

## The Attempt at a Solution

[/B]
First I found the gradient of the function
$\nabla f = (z^2+2xy)\hat{i} + x^2 \hat{j} + (2xz-1)\hat{k}$
and then evaluated at (1,-3,2)
$\nabla f(\vec{r_p}) = (2^2+2(1)(-3))\hat{i} + 1^2 \hat{j} + (2(1)(2)-1)\hat{k} = (-2,1,3)$

And then used the relation in the relevant equations section:
$(\vec{r}-\vec{r_p}) \cdot \nabla f(\vec{r_p}) = 0 \\ (x-1,y+3,z-2) \cdot (-2,1,3) = 0 \\ (-2x+2)+(y+3)+(3z-6) = 0 \\ -2x+y+3z=1$

This is the first time I've ever done this and I am just going off of a similar example that the solution was given for. It is not exactly the same question so not sure my process is even correct. Would appreciate any help :)

Last edited:

## Homework Statement

Find an equation for the tangent plane to a surface $xz^2 +x^2y-z=-1$ at the point (1,-3,2).

## Homework Equations

$(\vec{r}-\vec{r_p}) \cdot \nabla f(\vec{r_p}) = 0$

## The Attempt at a Solution

[/B]
First I found the gradient of the function
$\nabla f = (z^2+2xy)\hat{i} + x^2 \hat{j} + (2xz-1)\hat{k}$
and then evaluated at (1,-3,2)
$\nabla f(\vec{r_p}) = (2^2+2(1)(-3))\hat{i} + 1^2 \hat{j} + (2(1)(2)-1)\hat{k} = (-2,1,3)$

And then used the relation in the relevant equations section:
$(\vec{r}-\vec{r_p}) \cdot \nabla f(\vec{r_p}) = 0 \\ (x-1,y+3,z-2) \cdot (-2,1,3) = 0 \\ (-2x+2)+(y+3)+(3z-6) = 0 \\ -2x+y+3z=1$

This is the first time I've ever done this and I am just going off of a similar example that the solution was given for. It is not exactly the same question so not sure my process is even correct. Would appreciate any help :)
Looks good to me.

As a check you can do the following:
• Confirm that the normal to the plane is parallel to the gradient at the given point.
• Confirm that the plane contains the given point.

Mark44 said:
Looks good to me.

As a check you can do the following:
• Confirm that the normal to the plane is parallel to the gradient at the given point.
• Confirm that the plane contains the given point.
Thanks for taking a look, and for the advice on how to check if correct, appreciate it :)

Mark44 told you that!
"As a check you can do the following:
• Confirm that the normal to the plane is parallel to the gradient at the given point.
• Confirm that the plane contains the given point."

## 1. What is a tangent plane to a surface?

A tangent plane to a surface is a plane that touches the surface at exactly one point. It is perpendicular to the surface's normal vector at that point.

## 2. Why do we use the gradient to find the equation of a tangent plane?

The gradient is a vector that points in the direction of the steepest increase of a function. By finding the gradient at a specific point on a surface, we can determine the direction in which the surface is changing the fastest. This information is crucial in defining the orientation of the tangent plane at that point.

## 3. How do we find the gradient of a surface?

The gradient of a surface is found by taking the partial derivatives of the surface's equation with respect to each of its variables. This results in a vector with components representing the rate of change of the surface in each direction.

## 4. Can the tangent plane change at different points on a surface?

Yes, the tangent plane can change at different points on a surface. This is because the gradient and normal vector of a surface can vary at different points, resulting in a different orientation and equation for the tangent plane.

## 5. How is the equation of the tangent plane related to the equation of the surface?

The equation of the tangent plane is related to the equation of the surface by using the point-slope formula. The equation of the tangent plane will have the same variables as the surface's equation, but with the addition of a constant term to account for the specific point at which the tangent plane touches the surface.

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