# Tangent Plane Examples | Multivariable Calculus

In two-variable calculus, the tangent line on a curve y = f(x) at a point (x_{0}, y_{0}) is given by the equation

y - y_{0} = f'(x_{0})*(x - x_{0})

where f'(x_{0}) is the derivative of f(x) evaluated at x_{0}. For an implicitly defined curve g(x, y) = 0, the equation of the tangent line at (x_{0}, y_{0}) is

0 = [∂g/∂x(x_{0}, y_{0})]*(x - x_{0}) + [∂g/∂y(x_{0}, y_{0})]*(y - y_{0})

where ∂g/∂x(x_{0}, y_{0}) is the partial derivative of g with respect to x evaluated at (x_{0}, y_{0}) and ∂g/∂y(x_{0}, y_{0}) is the partial derivative of g with respect to y evaluated at (x_{0}, y_{0}).

In 3-dimensional space, the object we start with is a surface or rather than a curve in 2-dimensional space, and the tangent object is a plane rather than a line. If we start with a surface defined by z = f(x, y), the tangent plane at the point (x_{0}, y_{0}, z_{0}) is

z - z_{0} = f_{x}*(x - x_{0}) + f_{y}*(y - y_{0})

where f_{x} = ∂f/∂x and f_{y} = ∂f/∂y, the partial derivatives evaluated at x = x_{0} and y = y_{0}. The two notations for partial derivatives are interchangeable. Similarly, if a surface is given by an implicit equation g(x, y, z) = 0, the tangent plane at the point (x_{0}, y_{0}, z_{0}) is

0 = g_{x}*(x - x_{0}) + g_{y}*(y - y_{0}) + g_{z}*(z - z_{0})

where g_{x}, g_{y}, and g_{z} are the partial derivatives of g with respect to x, y, an z evaluated at (x_{0}, y_{0}, z_{0}). Below are several examples of how to compute the tangent planes of explicit surfaces z = f(x, y) and implicit surfaces g(x, y, z) = 0.

## Example 1: z = Ln |x^2 + xy^2 - y| at (-2, 1, 0)

For our first example we analyze the surface z = Ln |x^2 + xy^2 - y|, the natural logarithm of the absolute value of x^2 + xy^2 - y. To use the formula z - z_{0} = f_{x}(x - x_{0}) + f_{y}(y - y_{0}) we need to find the partial derivatives f_{x} and f_{y} at the point (-2, 1, 0), i.e., {x = -2, y = 1, z = 0}.

f_{x} = ∂f/∂x = (2x + y^2)/(x^2 + xy^2 - y)

∂f/∂x(-2, 1) = (-4 + 1)/(4 - 2 - 1) = -3

f_{y} = ∂f/∂y = (x - 1)/(x^2 + xy^2 - y)

∂f/∂y(-2, 1) = (-2 - 1)/(4 - 2 - 1) = -3

The equation of the tangent plane is then

z - 0 = -3(x - (-2)) + (-3)(y - 1)**z = -3x - 3y - 3**

Below is a 3-D plot and contour plot of this surface over the square bounded by -3 ≤ x ≤ 3 and -3 ≤ y ≤ 3. This surface is defined everywhere except where x^2 + xy^2 - y = 0. The graph of this implicitly defined curve is also shown below.

## Example 2: x^2 + y^2 + z^2 = 2xyz + 2

The surface defined by x^2 + y^2 + z^2 = 2xyz + 2 is an example of an implicitly defined surface, one that is not written in the form z = f(x, y) but instead in the form g(x, y, z) = 0. In this case, g is be the expression x^2 + y^2 + z^2 - 2xyz - 2. Because this fucntion is symmetric on the three variables x, y, and z, this strange tube-like surface has beautiful tetrahedral symmetry. The equation of the tangent plane at a point (x_{0}, y_{0}, z_{0}) is

0 = g_{x}(x - x_{0}) = g_{y}(y - y_{0}) + g_{z}(z - z_{0})

Let's find the equations of the tangent planes at the points (4, 1, 3) and (2, -1, -1). The partial derivatives g_{x}, g_{y}, and g_{z} are

g_{x} = ∂g/∂x = 2x - 2yz

g_{y} = ∂g/∂x = 2y - 2xz

g_{z} = ∂g/∂x = 2z - 2xy

Evaluated at the point (4, 1, 3) we have

g_{x}(4, 1, 3) = 2, g_{y}(4, 1, 3) = -22, g_{z}(4, 1, 3) = -2

Therefore the tangent plane at the point (4, 1, 3) is

0 = 2(x - 4) + (-22)(y - 1) + (-2)(z - 3)**2x - 22y - 2z + 20 = 0**

And evaluating the partial derivatives at (2, -1, -1) we have

g_{x}(2, -1, -1) = 2, g_{y}(2, -1, -1) = 2, g_{z}(2, -1, -1) = 2

Therefore the tangent plane at (2, -1, -1) is

0 = 2(x - 2) + 2(y - (-1)) + 2(z - (-1))

2x + 2y + 2z = 0**x + y + z = 0**

Several views of the surface x^2 + y^2 + z^2 = 2xyz - 2 are shown in the figure below, in the bounding box {-4 ≤ x ≤ 4, -4 ≤ y ≤ 4, -4 ≤ z ≤ 4}

## Example 3: z = (yx^3 + xy^4)/20 at (1, 4, 13)

Like Example 1, this function is in the form z = f(x, y), so we can use the tangent plane equation z - z_{0} = f_{x}(x - x_{0}) + f_{y}(y - y_{0}). The partial derivatives fx and fy are

fx = (3yx^2 + y^4)/20

fy = (x^3 + 4xy^3)/20

Evaluated at the point x = 1 and y = 4, the partial derivatives are

fx(1, 4) = (12 + 256)/20 = 13.4

fy(1, 4) = (1 + 256)/20 = 12.85

The equation of the tangent plane is

z - 13 = 13.4(x - 1) + 12.85(y - 4)**z = 13.4x + 12.85y - 51.8**

Some views of the graph of z = (yx^3 + xy^4)/20 are shown below in the bounding box {-6 ≤ x ≤ 6, -6 ≤ y ≤ 6, -3 ≤ z ≤ 3}.

## Example 4: x/(y^2 + 1) + 9z/(x^2 + y^2 + 1) + 1 = 0

Here is another implicitly defined surface, however, this one can easily be solved for z to put it in the explicit form z = f(x, y).

f(x, y) = z = -x(x^2 + y^2 + 1)/(9y^2 + 9) - (x^2 + y^2 + 1)/9

We can find the partial derivatives of the surface using either the implicit form or the explicit form. Using the explicit form we have

fx = -(3x^2 + y^2 + 1)/(9y^2 + 9) - 2x/9

fy = (2yx^3)/[9(y^2 + 1)^2)] - 2y/9

If we want to find the equation of the tangent plane at the point (2, 2, -1.4), we have

fx(2, 2) = -37/45

fy(2, 2) = -64/225

Therefore the equation of the tangent plane is

z - (-1.4) = (-37/45)(x - 2) + (-64/225)(y - 2)**z = -37x/45 - 64y/225 + 61/75**

Several views of the graph of x/(y^2 + 1) + 9z/(x^2 + y^2 + 1) + 1 = 0 are shown below. The bounding box is {-8 ≤ x ≤ 8, -8 ≤ y ≤ 8, -8 ≤ z ≤ 8}.

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