Tangent Plane

Just like a curve has a tangent line at a given point, a surface has an intersection of tangent lines, i.e. a tangent plane.

Definition §

Tangent Plane

In cartesian coordinates, the tangent plane of a bivariate function $f(x,y)$ at a point $(a,b)$ is given by:

$$z=f(a,b)+\frac{\partial f}{\partial x}{|}_{(a,b)}\cdot (x-a)+\frac{\partial f}{\partial y}{|}_{(a,b)}\cdot (y-b)$$

$$z=f(a,b)+f_x(a,b)\cdot (x-a)+f_y(a,b)\cdot (y-b)$$

The tangent plane can also be defined by the gradient vector $\nabla f$ and the dot product:

$$z=f(a,b)+\nabla f(a,b)\cdot (x-a,y-b)$$

Note the relation to the tangent equation $y=f(a)+f^{\prime }(x)(x-a)$

Tangent Plane (Parametric Surface) ^defintion-parametric

Let $S$ be a parametric surface that arises from the parametric equation $\varphi (s,t)$. If $S$ is smooth (having a non-zero unit normal vector) at a point $\varphi (s_0,t_0)=(x_0,y_0,z_0)$, then the cartesian equation for the tangent plane at $(x_0,y_0,z_0)$ is:

$$(x-x_0,y-y_0,z-z_0)\cdot \vec{n}(s_0,t_0)=0$$

where $\vec{n}$ is the choice of unit normal vector. This equation comes from the point-normal form of a plane.

Derivation §

When graphed, the derivatives of a 2D function can be seen as the ‘direction’ of the point in both the x and y axes. Just like how a tangent line is defined as $y-y_p=f^{\prime }(x_p)\times (x-x_p)$ (Equation of a line), we can define a tangent plane using an intersection of two tangents:

$$\frac{\partial f}{\partial x}{|}_{(a,b)}\text{ gives the slope at (a,b) relative to the X-Z plane}$$

$$\frac{\partial f}{\partial y}{|}_{(a,b)}\text{ gives the slope at (a,b) relative to the Y-Z plane}$$

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If we wanted to construct a tangent of $(a,b)$ in the X-Z plane. We would use the equation:

$$y=y_1+m\cdot (x-x_1)$$

transformed to:

$$z=f(a,b)+\frac{\partial f}{\partial x}{|}_{(a,b)}\cdot (x-a)$$

And similarly for the Y-Z plane, we have:

$$z=f(a,b)+\frac{\partial f}{\partial y}{|}_{(a,b)}\cdot (y-b)$$

We can then use the vector equation of a plane, which allows use to construct a plane using the intersection of two lines:

$$z=f(a,b)+\frac{\partial f}{\partial x}{|}_{(a,b)}\cdot (x-a)+\frac{\partial f}{\partial y}{|}_{(a,b)}\cdot (y-b)$$

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Linear Approximation §

Using the tangent plane, we can obtain a linear approximation of $f$ at $P_1=(x,y)$, given we known the equation to the tangent plane at $P_0=(x_0,y_0,z_0)$ and $P_1\sim P_0$ (P1 is very close to P0):

$$f(x,y)\approx z_0+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$

Or, in different notation, where $f_x|P_0$ and $f_y|P_0$ gives the value of the partial derivatives $f_x$ and $f_y$ at point $P_0$:

$$f(x,y)\approx z_0+f_x|P_0\times (x-x_0)+f_y|P_0\times (y-y_0)$$

Approximate Change §

Using the same equation for linear approximation:

$$f(x,y)\approx z_0+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$

we can rearrange it (by moving the $z_0$):

$$f(x,y)-f(x_0,y_0)\approx f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$

$\Delta x=x-x_0$ and $\Delta y=y-y_0$. Hence, $\Delta f=\Delta x+\Delta y$ and the final equation is:

$$\Delta f=f_x(x_0,y_0)\Delta x+f_y(x_0,y_0)\Delta y$$

Examples §

1: Tangent Plane Linear Approximation

Let:

$$f(x,y)=x{e}^{xy}$$

Find the tangent plane to the surface $z=f(x,y)$ at the point where $(x,y)=(2,0)$. Hence approximate $f(2.1,-0.1)$

Solution $z_0=f(2,0)=2e^{2\times 0}=2$ $f_x(x,y)=e^{xy}+yx{e}^{xy}$ $f_y(x,y)=x^2{e}^{xy}$ $f_x(2,0)=e^0+0=1$ $f_y(2,0)=4\times e^0=4$ Tangent plane: $z=2+1(x-2)+4(y-0)=2-2+x+4y=x+4y$ Now, for the linear approximation: $f(2.1,-0.1)\approx 2+1(2.1-2)+4(-0.1-0)$ $=2+0.1-0.4=1.7$ $f(2.1,-0.1)\approx 1.7$

2: Tangent Plane to a Parametric Cone

Find the Cartesian equation of the tangent plane to parametric cone given by:

$$\vec{\Phi }:x=s\cos (t)y=s\sin (t)z=s\text{\hspace{1em}(s≥0,t∈[0,2π))}$$

at the point $(1,1,\sqrt{2})$.

Solution

The equation for the tangent plane yields:

$$(x-1,y-1,z-\sqrt{2})\cdot \vec{n}(s_0,t_0)=0$$

We need to find the values of the parameters $s,t$ at the point $(1,1,\sqrt{2})$:

• Since $z=s$, we know that $s=z=\sqrt{2}$.
• Then, using the $x$ equation we have: $1=\sqrt{2}\cos (t)⟹t=\arccos \left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}$

From a previous example, we have $\vec{n}(s,t)=(-s\cos (t),-s\sin (t),s)$. Thus:

$$\begin{array}{rl}(x-1,y-1,z-\sqrt{2})\cdot \vec{n}(\sqrt{2},\frac{\pi }{4})& =0\\ (x-1,y-1,z-\sqrt{2})\cdot (-\sqrt{2}\cos (\frac{\pi }{4}),-\sqrt{2}\sin (\frac{\pi }{4}),\sqrt{2})& =0\\ (x-1,y-1,z-\sqrt{2})\cdot (-1,-1,\sqrt{2})& =0\\ (1-x)+(1-y)+z\sqrt{2}-2& =0\\ -x-y+z\sqrt{2}& =0\\ z& =\frac{1}{\sqrt{2}}(x+y)\end{array}$$

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