Tangent Planes – Calculus Tutorials

Tangent Planes and Linear Approximation – HMC Calculus Tutorial

Just as we can visualize the line tangent to a curve at a point in 2-space, in 3-space we can picture the plane tangent to a surface at a point.

Consider the surface given by $z = f (x, y)$. Let $(x_0 , y_0 , z_0 )$ be any point on this surface. If $f (x, y)$ is differentiable at $(x_0 , y_0 )$, then the surface has a tangent plane at $(x_0 , y_0 , z_0 )$.

A function $f(x,y)$ is differentiable at the point $(x_0,y_0)$ if $f_x(x_0,y_0)$ and $f_y(x_0,y_0)$ exist and $\Delta f = f(x_0+\Delta x,y_0+\Delta y) – f(x_0,y_0)$ can be written in the form $$ \Delta f = f_x(x_0,y_0)\Delta x + f_y(x_0,y_0)\Delta y + \varepsilon_1 \Delta x + \varepsilon_2 \Delta y $$ where $\varepsilon_1$ and $\varepsilon_2$ are functions of $\Delta x$ and $\Delta y$ such that $$ \lim_{(\Delta x,\Delta y) \rightarrow (0,0)}\varepsilon_1 = \lim_{(\Delta x,\Delta y) \rightarrow (0,0)}\varepsilon_2 = 0. $$

Let $(x_0, y_0, z_0)$ be any point on the surface $z = f(x,y)$. If the tangent lines at $(x_0, y_0, z_0)$ to all smooth curves on the surface passing through $(x_0, y_0, z_0)$ lie on a common plane, then we call that plane the tangent plane to $z = f(x,y)$ at $(x_0, y_0, z_0)$.

The equation of the tangent plane at $(x_0 , y_0 , z_0 )$ is given by $$f_x (x_0 , y_0 )(x – x_0 ) + f_y (x_0 , y_0 )(y – y_0 ) – (z – z_0 ) = 0.$$

Notes

Recall that the equation of the plane containing a point $(x_0 , y_0 , z_0 )$ and normal to the vector ${\bf n} = (a, b, c)$ is $$ a(x – x_0 ) + b(y – y_0 ) + c(z – z_0 ) = 0. $$ The derivation of the equation for the tangent plane just involves showing that the tangent plane is normal to the vector ${\bf n} = (f_x (x_0 , y_0 ), f_y (x_0 , y_0 ), -1)$.

Let $(x_0,y_0,z_0)$ be any point on the surface $z=f(x,y)$ such that $f(x,y)$ is differentiable at $(x_0,y_0)$. We well show that the tangent plane is normal to the vector ${\bf n} = (f_x(x_0,y_0),f_y(x_0,y_0),-1)$.

Consider any smooth curve $C$ on the surface that passes through $(x_0,y_0,z_0)$. Parametrize the curve as \begin{eqnarray*} x & = & x(s) \\ y & = & y(s) \\ z & = & z(s). \end{eqnarray*} Let $s=s_0$ satisfy $x_0=x(s_0), \quad y_0=y(s_0), \quad z_0=z(s_0).$ Then $$ z(s) = f(x(s),y(s)) $$ for all $s$. Using the Chain Rule to differentiate both sides of this equation, $$ \frac{dz}{ds}=\frac{\partial f}{\partial x}\frac{dx}{ds} + \frac{\partial f}{\partial y}\frac{dy}{ds}. $$ Thus, $$ \frac{\partial f}{\partial x}\frac{dx}{ds} + \frac{\partial f}{\partial y}\frac{dy}{ds} – \frac{dz}{ds}=0. $$ Switching notation and writing the equation in vector form, $$ (f_x(x,y),f_y(x,y),-1) \cdot (x'(s),y'(s),z'(s))=0. $$ At $s=s_0$, $$ (f_x(x_0,y_0),f_y(x_0,y_0),-1) \cdot (x'(s_0),y'(s_0),z'(s_0))=0. $$ But $(x'(s_0),y'(s_0),z'(s_0))$ is the tangent vector to the curve $C$ at $(x_0,y_0,z_0)$. Thus, we have that the tangent vector to any smooth curve $C$ on the surface that passes through $(x_0,y_0,z_0)$ is normal to the vector ${\bf n} = (f_x(x_0,y_0),f_y(x_0,y_0),-1)$ and so is given by the equation $$ f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)-(z-z_0)=0. $$

(This proof is taken from Calculus, by Howard Anton.)

For surfaces $F (x, y, z) = 0$ that are not easily solved for $z$, the equation of the tangent plane at $(x_0 , y_0 , z_0 )$ is $$ F_x (x_0 , y_0 , z_0 )(x – x_0 ) + F_y (x_0 , y_0 , z_0 )(y – y_0 ) + F_z (x_0 , y_0 , z_0 )(z – z_0 ) = 0 $$ provided that $\nabla F (x_0 , y_0 , z_0 ) \neq 0$. Note that if we let $F (x, y, z) = f (x, y) – z$, we obtain the equation given for the tangent plane to $z = f (x, y)$ at $(x_0 , y_0 , z_0 )$.

Example

Let’s find the equation of the plane tangent to the surface $z = 4x^3 y^2 + 2y$ at the point $(1, -2, 12)$.

Since $f (x, y) = 4x^3 y^2 + 2y$, $$ f_x (x, y) = 12x^2 y^2 \textrm{ and } f_y (x, y) = 8x^3 y + 2.$$

With $x = 1$ and $y = -2$, \begin{eqnarray*} f_x (1, -2) &=& 12(1)^2 (-2)^2 = 48 \\ f_y (1, -2) &=& 8(3)^3(-2) + 2 = -14. \end{eqnarray*}

Thus, the tangent plane has normal vector $ {\bf n} = (48, -14, -1) $ at $(1, -2, 12)$ and the equation of the tangent plane is given by $$ 48(x – 1) – 14 (y – (-2)) – (z – 12) = 0.$$ Simplifying, $$ 48x – 14y – z = 64. $$

Linear Approximation

The tangent plane to a surface at a point stays close to the surface near the point. In fact, if $f (x, y)$ is differentiable at the point $(x_0 , y_0 )$, the tangent plane to the surface $z = f (x, y)$ at $(x_0 , y_0 )$ provides a good approximation to $f (x, y)$ near $(x_0 , y_0 )$:

$\qquad$ Solving $f_x (x_0 , y_0 )(x – x_0 ) + f_y (x_0 , y_0 )(y – y_0 ) – (z – z_0 ) = 0$ for $z$, $$ z = z_0 + f_x (x_0 , y_0 )(x – x_0 ) + f_y (x_0 , y_0 )(y – y_0 ). $$ $\qquad$ Since $z_0 = f (x_0 , y_0 )$, we have that $$ z = f (x_0 , y_0 ) + f_x (x_0 , y_0 )(x – x_0 ) + f_y (x_0 , y_0 )(y – y_0 ). $$ $\qquad$ Near $(x_0 , y_0 )$, the surface is close to the tangent plane. Thus, $$ f (x, y) \approx f (x_0 , y_0 ) + f_x (x_0 , y_0 )(x – x_0 ) + f_y (x_0 , y_0 )(y – y_0 ). $$ We call this the linear approximation or local linearization of $f (x, y)$ near $(x_0 , y_0 )$.

Notes

The linear approximation is really just the multivariable Taylor polynomial of degree 1 for $f (x, y)$ about $(x_0 , y_0 )$. It is only accurate near $(x_0 , y_0 )$. Better approximations can be obtained by using higher-order Taylor polynomials.
These concepts can be extended to functions of more than two variables: $$ f (x, y, z) \approx f (x_0 , y_0 , z_0 )+f_x (x_0 , y_0 , z_0 )(x-x_0 )+f_y (x_0 , y_0 , z_0 )(y-y_0 )+f_z (x_0 , y_0 , z_0 )(z-z_0 ) $$ where $f (x, y, z)$ is differentiable at $(x_0 , y_0 , z_0 )$.

Example

From our work in the previous example, the linear approximation to $f (x, y) = 4x^3 y^2 + 2y$ near $x = 1,\quad y = -2$ is $$ f (x, y) \approx 48x – 14y – 64. $$ This is, of course, exact at $x = 1, \quad y = -2$: $$ f (1, -2) = 12 = 48(1) – 14(-2) – 64. $$ At $x = 1.1$ and $y = -1.9$, according to the linear approximation, $$ f (1.1, -1.9) \approx 48(1.1) – 14(-1.9) – 64 = 15.4, $$ which is indeed very close to the exact value $f (1.1, -1.9) = 15.41964$.

Key Concepts

Tangent Plane to a Surface
Let $(x_0 , y_0 , z_0 )$ be any point on the surface $z = f (x, y)$. If $f (x, y)$ is differentiable at $(x_0 , y_0 )$, then the surface has a tangent plane at $(x_0 , y_0 , z_0 )$ given by $$ f_x (x_0 , y_0 )(x – x_0 ) + f_y (x_0 , y_0 )(y – y_0 ) – (z – z_0 ) = 0. $$
Linear Approximation to a Surface
If $f (x, y)$ is differentiable at $(x_0 , y_0 )$, then near $(x_0 , y_0 )$ $$ f (x, y) \approx f (x_0 , y_0 ) + f_x (x_0 , y_0 )(x – x_0 ) + f_y (x_0 , y_0 )(y – y_0 ). $$

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