Tangent planes

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We find tangent planes.

As we are hoping is becoming a habit, let’s start by looking again at the case of functions of one variable, and use what we know from our previous calculus courses to answer our questions in higher dimensions. As usual with this strategy, we proceed with care. In the past you may have learned:

Given a function $f$ and a number $a$ in the domain of $f$ , if one can “zoom in” on the graph at $(a, f(a))$ sufficiently so that it appears to be a straight line, then the function is differentiable, and that line is the tangent line to $f$ at the point $(a,f(a))$ .

We illustrate this informal definition with the following diagram:

We can give a similar informal definition for functions of two variables.

Given a function $F:\R ^2\to \R$ and a vector $\vec {a}$ in the domain of $F$ , if one can “zoom in” on the graph at $(\vec {a}, F(\vec {a}))$ sufficiently so that it appears to be a plane, then the function is differentiable, and that plane is the tangent plane to $F$ at the point $(\vec {a},F(\vec {a}))$ .

Let’s turn our attention to finding an equation for this tangent plane.

The single variable case

Given a function $f : \R \to \R$ and a point of interest $x=a$ in the domain of $f$ , we have previously found an equation for the tangent line to $f$ at $x=a$ , which we also called the linear approximation to $f$ at $x=a$ . $\l (x) = f(a) + f'(a) \cdot (x-a)$

There are two things we should notice about this linear approximation.

First, think about geometry We know that:

$f(a)$ is the “height” of $f$ at $x=a$ .
$f'(a)$ is the rate of change of $f$ at $a$ .
$(x-a)$ is the distance we have moved from the point $a$ .

Multiplying $f'(a)$ and $(x-a)$ gives us an approximate change in the value of the function, so to find the new approximate function value, we need to add on our original height, or $f(a)$ .

So we see that: $\l (x) = \underbrace {f(a)}_{\text {height}} + \overbrace {f'(a)}^{\text {rate}} \cdot \underbrace {(x-a)}_{\text {change}}$

Specifically the tangent line $\l$ to a function $f$ is a line constructed so that:

$\l (a) = f(a)$
$\l '(a) = f'(a)$

When the values of $\l$ and $f$ are equal at $x=a$ , and they are changing at the same rates at $x=a$ , then it is only reasonable that $\l$ is a good approximation of $f$ . Moreover, $\l$ is also the first-order Taylor approximation to $f$ . Recall, the degree $n$ Taylor polynomial for $f$ at $x=a$ is:

$\begin{align*} p_n(x) = f(a) &+ f'(a) (x-a) && \text{(linear approximation)} \\ &+ \frac{f''(a)(x-a)^2}{2!}\\ &+ \cdots && (\text{higher degree terms})\\ &+ \frac{f^{n}(a)(x-a)^n}{n!} \end{align*}$

This polynomial starts with the linear approximation, and then adds higher degree terms.

The two variable case

Now consider a differentiable function $F : \R ^2 \to \R$ . Here the formula for the tangent plane at $\vec {a} =\vector {a,b}$ is given by: $L(x,y) =F(\vec {a})+ F^{(1,0)}(\vec {a}) (x-a)+ F^{(0,1)}(\vec {a}) (y-b)$ We will work as we did above.

First, think about geometry Finally, let’s look at some of the geometry of the equation we just found. Working in an entirely similar way as before, we know that:

$F(\vec {a})$ is the “height” of $F$ at $\vector {x,y} = \vec {a}$ .
$F^{(1,0)}(\vec {a})$ is the rate of change of $F$ at $\vector {x,y} = \vec {a}$ in the $x$ -direction.
$F^{(0,1)}(\vec {a})$ is the rate of change of $F$ at $\vector {x,y} = \vec {a}$ in the $y$ -direction.
$(x-a)$ is the distance we have moved from the point $a$ in the $x$ -direction.
$(y-b)$ is the distance we have moved from the point $a$ in the $y$ -direction.

Multiplying $F^{(1,0)}(\vec {a})$ and $(x-a)$ gives us an approximate change in the value of the function when moving in the $x$ -direction. Multiplying $F^{(0,1)}(\vec {a})$ and $(y-b)$ gives us an approximate change in the value of the function when moving in the $y$ -direction. Adding these two together, we see that we have moved away from the point $\vec {a}$ , and we are combining these rates of change to get an approximate change in the value of $F$ . Adding this to our original function value $F(\vec {a})$ gives us our new approximate function value. In other words, $L$ approximates $F$ near $\vector {x,y} = \vec {a}$ .

So we see that: $L(x,y) = \underbrace {F(\vec {a})}_{\text {height}} + \overbrace {F^{(1,0)}(\vec {a})}^{\text {rate}} \cdot \underbrace {(x-a)}_{\text {change}} + \overbrace {F^{(0,1)}(\vec {a})}^{\text {rate}} \cdot \underbrace {(y-b)}_{\text {change}}$

Note, first that $z=L (x,y)$ is a plane. To be a bit pedantic, we can write

$\begin{align*} z &= L(x,y)\\ z &= F(\vec{a})+ F^{(1,0)}(\vec{a}) (x-a)+ F^{(0,1)}(\vec{a}) (y-b) \end{align*}$

and rearranging, we find: $a\cdot F^{(1,0)}(\vec {a}) + b\cdot F^{(0,1)}(\vec {a})- F(\vec {a}) = F^{(1,0)}(\vec {a}) x+ F^{(0,1)}(\vec {a}) y -z$ and this is the equation of a plane, as it is: $\text {number} = \text {number}\cdot x + \text {number}\cdot y + \text {number}\cdot z$ Let’s see if you get this.

Given a differentiable function $F:\R ^2\to \R$ , find a vector normal to the plane $z= L (x,y)$ . $\vec {n} = \vector {F^{(1,0)}(\vec {a}), F^{(0,1)}(\vec {a}), \answer [given]{-1}}.$

Now note that the function $L$ is cooked-up so that:

$L(a,b) = F(a,b)$
$L^{(1,0)} (a,b) = F^{(1,0)}(a,b)$
$L^{(0,1)} (a,b) = F^{(0,1)}(a,b)$

So we see that $z=L(x,y)$ is a plane that approximates $z=F(x,y)$ near the point $(a,b,F(a,b))$ .

Let $F(x,y) = 5-2(x-3)^2 -3(y-2)^2$ . Find a tangent plane at $(x,y) = (2,1)$ . $L(x,y) = \answer {4x+6y-14}$

Find a vector normal to the plane you found above. $\vec {n} = \vector {\answer {-4},\answer {-6},1}$

From above you see: $z=4x+6y-14$ So now write: $z-4x-6y = -14$

Foreshadowing a bit, our formula for our linear approximation is: $L(x,y) = F(\vec {a})+ F^{(1,0)}(\vec {a}) (x-a)+ F^{(0,1)}(\vec {a}) (y-b)$

This look like what we might want from a first-order Taylor approximation to our function $F$ as:

$L(a,b) = F(a,b)$
$L^{(1,0)} (a,b) = F^{(1,0)}(a,b)$
$L^{(0,1)} (a,b) = F^{(0,1)}(a,b)$

Said in words, the tangent plane at $\vec {a}$ touches the surface at $\vec {a}$ and the first partial derivatives of $L$ and $F$ agree at $\vec {a}$ .

Saying the tangent plane has the same partial derivatives as the function is another way of saying that tangent plane has to contain the tangent lines we can find using the partial derivatives.

Try your hand at this question:

Find an equation for the tangent plane to $F(x,y) = 3\cos (x)\sin (y)$ at $(x,y) = (\pi /3,\pi /6)$ . $z = \answer {(-3\sqrt {3}/4)\cdot (x-\pi /3) + (3\sqrt {3}/4)\cdot (y-\pi /6) + 3/4}$

Moreover, tangent planes are linear approximations of differentiable surfaces.

Let $F:\R ^2\to \R$ be a differentiable function where you know that $F(1,-3) = 5$ and that $F^{(1,0)}(1,-3) = -4$ and $F^{(0,1)}(1,-3) = 2$ . Give a plane tangent to $z =F(x,y)$ at the point $(1,-3)$ and use this tangent plane to approximate the value of $F(2,-4)$ . First we write the linear approximation, meaning the tangent plane: $L(x,y) = \answer {5 -4(x-1) + 2(y+3)}$ So we believe that $F(2,-4) \approx L\left (\answer {2},\answer {-4}\right ) = \answer {-1}$ .

Generalizing with the gradient vector

Above, given a function $F:\R ^2\to \R$ , we wrote our linear approximation as: $L(x,y) =F(\vec {a})+ F^{(1,0)}(\vec {a}) (x-a)+ F^{(0,1)}(\vec {a}) (y-b)$ We can rewrite this using the gradient vector as follows: $L(\vec {x}) = F(\vec {a})+ \grad F(\vec {a})\dotp (\vec {x}-\vec {a})$ where $\vec {x} = \vector {x,y}$ and $\vec {a} = \vector {a,b}$ . Now check this out, the formula $L(\vec {x}) = F(\vec {a})+ \grad F(\vec {a})\dotp (\vec {x}-\vec {a})$ actually works in all dimensions! So now if you have a function of three variable, we set $\vec {x} = \vector {x,y,z}$ and $\vec {a} = \vector {a,b,c}$ and we can unpack this formula as: $L(\vec {x}) = F(\vec {a})+ \grad F(\vec {a})\dotp (\vec {x}-\vec {a})$

$\begin{align*} =F(\vec{a}) &+ F^{(1,0,0)}(\vec{a}) (x-a)\\ &+ F^{(0,1,0)}(\vec{a}) (y-b)\\ &+ F^{(0,0,1)}(\vec{a}) (z-c) \end{align*}$

Moreover, if we are working with functions of a single variable, then

$\begin{align*} \l(x) &= f(a) + f'(a) \cdot (x-a) \\ &=f(\vec{a})+ \grad f(\vec{a})\dotp (\vec{x}-\vec{a}) \end{align*}$

where $\vec {a} = \vector {a}$ and $\vec {x} = \vector {x}$ . I like to say that the formula $L(\vec {x}) = F(\vec {a})+ \grad F(\vec {a})\dotp (\vec {x}-\vec {a})$ is “one formula to rule them all,” since when you know this formula, you know how to compute linear approximations in all dimensions.

An in-depth example

Our final example brings together several concepts.

Let $F(x,y) = x^2 + 3xy + 4$ .

(a): Find an equation for the plane tangent to $F$ at $(x,y) = (1,2)$ .
(b): Show that the line $y = 3x - 1$ passes through the point $(1,2)$ .
(c): Find a parametric description for the curve lying above $y = 3x-1$ on the surface $z = F(x,y)$ . Call this description $\vec {p}$ .
(d): Find a vector-valued formula for the line tangent to the curve $\vec {p}(t)$ at the point $(1, 2, F(1,2))$ . Call this $\vecl$ .
(e): Does the tangent line drawn by $\vecl (t)$ lie in the plane tangent to $F$ at $(x,y)=(1,2)$ ?

For part eg:tpa, compute $F(1,2) = \answer [given]{11}$ , $F^{(1,0)}(1,2) = \answer [given]{8}$ , $F^{(0,1)}(1,2) = \answer [given]{3}$ , and set $L(x,y) = \answer [given]{11 + 8(x-1) + 3(y-2)}$ .

For part eg:tpb, note that we are working here in . We would like to see that the point $(1,2)$ lies on the line given. To do this, we plug in $x = \answer [given]{1}$ to the equation $y = 3x - 1$ . We see that

$\begin{align*} y &= 3\left(\answer[given]{1}\right) - 1 \\ &= \answer[given]{3} - 1 \\ &= \answer[given]{2}. \end{align*}$

So we conclude that the point $(1, 2)$ on the line $y = 3x-1$ .

For part eg:tpc, we would like a parametric description $\vec {m}(t)$ of the line $y = 3x-1$ in the $(x,y)$ -plane. Since this line is already expressed in terms of $x$ , the simplest choice is to take $x = \answer [given]{t}$ . We find $\vec {m}(t) = \vector {\answer [given]{t}, \answer [given]{3t-1}}.$ The curve lying on the surface is the collection of all of the images of the points on the line $y = 3x-1$ , so we would like to evaluate $F$ on all of these points. Since we already have a parametric description $\vec {m}(t) = \vector {x(t), y(t)}$ , we plug these coordinates in for $x$ and $y$ to find our $z$ - coordinate. We find $\vec {p}(t) = F\left ( \vec {m}(t) \right ) = \vector {\answer [given]{t}, \answer [given]{3t-1}, \answer [given]{10t^2-3t+4}}$ For part eg:tpd note that since our tangent line is in $(x,y,z)$ -space, we will need to find a parametric equation for this line. Recall that the most straightforward way to write such an equation is $\vecl (t) = \vec {p} + t\vec {v},$ where $\vec {p}$ is a point on the line and $\vec {v}$ is a direction vector for the line. The direction of $\vecl (t)$ is tangent to the curve. We first find the tangent vector for any value of $t$ . $\vec {p}'(t) = \vector {\answer [given]{1}, \answer [given]{3}, \answer [given]{20t-3}}$ To find the tangent vector at $(1,2, F(1,2))$ , we plug in $t = \answer [given]{1}$ and find $\vec {p}'(1) = \vector {1, 3, 17}$ . We know that the line goes through the point $\vector {1, 2, \answer [given]{11}}$ , and so now we can write the equation of our tangent line.

$\begin{align*} \vecl(t) &= \vector{1, 2, \answer[given]{11}} + t \vector{\answer[given]{1}, \answer[given]{3}, \answer[given]{17}} \\ &= \vector{\answer[given]{1+t}, \answer[given]{2+3t}, \answer[given]{11+17t}} \end{align*}$

Does the tangent line $\vecl (t)$ lie in the tangent plane to $F$ at $(1,2)$ ?

We could solve this problem by simply plugging in the coordinate functions of $\vecl (t)$ to the equation $z = L(x,y)$ to see if the two sides of the equals sign agree. Instead, let’s use a little geometry!

We know that both the tangent plane $z=L(x,y)$ and the line $\vecl (t)$ pass through the point $(1, 2, 11)$ . Since $\vecl (t)$ is a line, its tangent vector $\vec {v}$ tells us everything we need to know about its direction. Since $L(x,y)$ is a plane, its normal vector $\vec {n}$ tells us everything we need to know about its direction. Then, if $\vec {v} \dotp \vec {n} = \answer [given]{0}$ , we know that the vector $\vec {v}$ must lie in the plane—and thus the entire line $\vecl (t)$ must line in the plane as well! First, find $\vec {v}$ . $\vec {v} = \vector {\answer [given]{1}, \answer [given]{3}, \answer [given]{17}}$ Next, use the coefficients of $z=L(x,y)$ to find $\vec {n}$ . $\vec {n} = \vector {\answer [given]{8}, \answer [given]{3}, \answer [given]{-1}}$ Finally, compute their dot product. $\vec {v} \dotp \vec {n} = \answer [given]{0}$