Objectives
This section will cover how to...
Find equations of tangent lines and tangent planes to surfaces.
Section7.3Tangent Planes
¶Review7.3.1
If you know that a plane passes through the point \((1,2,3)\) and has normal vector \((4,5,6)\text{,}\) then give an equation of the plane. See 1 An equation of the plane is \(4(x-1)+5(y-2)+6(y-3)=0\text{.}\) If \((x,y,z)\) is any point in the plane, then the vector \((x-1,y-2,z-3)\) is a vector in the plane, and hence orthogonal to \((4,5,6)\text{.}\) The dot product of these two vectors should be equal to zero, which is why the plane's equation is \((4,5,6)\cdot (x-1,y-2,z-3)=0\text{.}\)for an answer.
Exercise7.3.2
Consider again the parametric surface \(\vec r(a,t) = (a\cos t, a\sin t, t)\) for \(2\leq a\leq 4\) and \(0\leq t\leq 4\pi\text{.}\)
We'd like to obtain an equation of the tangent plane to this surface at the point \(\vec r(3,\pi)\text{.}\) To find a plane we need a normal (orthogonal) vector to the [tangent] plane and a point.
(a)
State the coordinates for \(\vec r (3,\pi)\)
(b)
Since \(\vec r_a (3,\pi)\) and \(\vec r_t (3, \pi)\) (the two partial derivatives) are tangent to the surface they must also be in our tangent plane. So a vector orthogonal to \(\vec r_a\) and \(\vec r_t\) will be normal to the tangent plane.
Find this normal vector.
Hint
How do I obtain a vector orthogonal to both partial derivatives? Is it the scalar, dot, or cross product?
(c)
Give an equation for the tangent plane.
I promised earlier in this chapter that you can obtain most of the results in multivariate calculus by replacing the \(x\) and \(y\) in \(dy=f'dx\) with \(\vec x\) and \(\vec y\text{.}\) The last exercise asked you to obtain a tangent plane to a parametric surface. You've never had to find a tangent plane to a function of the form \(z=f(x,y)\text{.}\) Let's review how to do it for functions of the form \(y=f(x)\text{,}\) and then generalize.
Review7.3.3Tangent Lines
Consider the function \(y=f(x)=x^2\text{.}\)
(a)
The derivative is \(f'(x) = ?\text{.}\)
(i)
At the point \(x=3\) the derivative is \(f'(3)=?\)
(ii)
and the output \(y\) is \(y=f(3)=?\text{.}\)
(b)
If we move from the point \((3,f(3))\) to the point \((x,y)\) along the tangent line, then a small change in \(x\) is \(dx=x-3\text{.}\)
(i)
What is \(dy\) in terms of \(y\text{?}\)
(c)
Differential notation states that a change in the output \(dy\) equals the derivative times a change in the input \(dx\text{,}\) which gives us the equation \(dy=f'(3)dx\text{.}\)
(i)
Replace \(dx\text{,}\) \(dy\text{,}\) and \(f'(3)\) with what we know they equal, to obtain an equation \(y-?=?(x-?)\text{.}\)
(ii)
What does this equation represent?
(iii)
Draw both \(f\) and the equation from the previous part on the same axes.
Subsection7.3.1Tangent Planes with Differentials
In first semester calculus, differential notation says \(dy=f' dx\text{.}\) A change in the output equals the derivative times a change in the inputs. For the next exercise, the output is \(z\text{,}\) and input is \((x,y)\text{,}\) which means differential notation says \(dz = Df \begin{bmatrix}dx\\dy \end{bmatrix}\text{.}\)
Exercise7.3.4Tangent Planes
Consider the function \(z=f(x,y)=9-x^2-y^2\text{.}\)
See Sage for a picture.
(a)
The derivative is \(Df(x,y) = \begin{bmatrix}-2x\amp ? \end{bmatrix}\text{.}\)
(b)
At the point \((x,y)=(2,1)\text{,}\) the derivative is \(Df(2,1) = \begin{bmatrix}-4\amp ? \end{bmatrix}\) and the output \(z\) is \(z=f(2,1)=?\text{.}\)
(c)
If we move from the point \((2,1,f(2,1))\) to the point \((x,y,z)\) along the tangent plane, then a small change in \(x\) is \(dx=x-2\text{.}\) What are \(dy\) and \(dz\) in terms of \(y\) and \(z\text{?}\)
(d)
Explain why an equation of the tangent plane is
\begin{equation*} z-4=\begin{bmatrix}-4 \amp -2 \end{bmatrix} \begin{bmatrix}x-2\\y-1 \end{bmatrix} \text{ or } z-4=-4(x-2)-2(y-1). \end{equation*}We'll construct a graph of \(f\) and it's tangent plane in class.
Hint
What does differential notation tell us?
Look back at the previous two exercises. The first semester calculus tangent line equation, with differential notation, generalized immediately to the tangent plane equation for functions of the form \(z=f(x,y)\text{.}\) We just used the differential notation \(dy=f'dx\) in 2D, and generalized to \(dz = Df \begin{bmatrix}dx\\dy \end{bmatrix}\text{.}\) Let's repeat this on another exercise.
Exercise7.3.5
See Sage.
Let \(f(x,y)=x^2+4xy+y^2\text{.}\) Give an equation of the tangent plane at \((3,-1)\text{.}\)
Hint
Just as in the previous exercise, find \(Df(x,y)\text{,}\) \(dx\text{,}\) \(dy\text{,}\) and \(dz\text{.}\) Then use differential notation.
Let's look again at the function \(z=9-x^2-y^2\text{,}\) and show how parametric surfaces can add more light to unlocking the derivative and its geometric meaning. With a parametrization, partial derivatives are vectors, instead of just numbers. Once we have vectors, we can describe motion. This makes it easier to visualize.
Exercise7.3.6
Let \(z=f(x,y)=9-x^2-y^2\text{.}\) We'll parameterize this function by writing \(x=x, y=y, z=9-x^2-y^2\text{,}\) or in vector notation we'd write
\begin{equation*} \vec r(x,y) = (x,y,f(x,y)). \end{equation*}See Sage for a picture.
(a)
Compute \(\ds \frac{\partial f}{\partial x}\) and \(\ds \frac{\partial f}{\partial y}\text{.}\)
(b)
Next evaluate these partials at \((x,y)=(2,1)\text{.}\) You should have \(f_x=-4\text{.}\)
(c)
What does the number \(-4\) mean? What does the number \(f_y\) mean?
(d)
Compute \(\ds \frac{\partial \vec r}{\partial x}\) and \(\ds \frac{\partial \vec r}{\partial y}\text{.}\)
(e)
Now evaluate these partials at \((x,y)=(2,1)\text{.}\)
(f)
What do these vectors mean?
Hint
Draw the surface, and at the point \((2,1,4)\text{,}\) draw these vectors. See the Sage plot.
(g)
The vectors above are tangent to the surface. Use them to obtain a normal vector to the tangent plane, and then given an equation of the tangent plane. (You should compare it to your equation from Exercise 4.)
Subsection7.3.2Generalizing the Tangent Plane Formulas
The next exercise generalizes the tangent plane and normal vector calculations above to work for any parametric surface \(\vec r(u,v)\text{.}\)
Exercise7.3.7
Consider the cone parametrized by \(\vec r(u,v)=(u\cos v, u\sin v,u)\text{.}\)
See Sage.
(a)
Give vector equations of two tangent lines to the surface at \(\vec r(2,\pi/2)\) (so \(u=2\) and \(v=\pi/2\)).
(b)
Give a normal vector to the surface at \(\vec r(2,\pi/2)\text{.}\)
(c)
Give an equation of the tangent plane at \(\vec r(2,\pi/2)\text{.}\)
We now have two different ways to compute tangent planes. One way generalizes differential notation \(dy=f'dx\) to \(dz = Df \begin{bmatrix}dx\\dy \end{bmatrix}\) and then uses matrix multiplication. This way will extend to tangent objects in EVERY dimension. It's the key idea needed to work on really large exercises.
The other way requires that we parametrize the surface \(z=f(x,y)\) as \(\vec r(x,y)=(x,y,f(x,y))\) and then use the cross product on the partial derivatives. Both give the same answer. The next exercise has you give a general formula for a tangent plane. To tackle this exercise, you'll need to make sure you can use symbolic notation. The review exercise should help with this.
Review7.3.8
Joe wants to to find the tangent line to \(y=x^3\) at \(x=2\text{.}\) He knows the derivative is \(y=3x^2\text{,}\) and when \(x=2\) the curve passes through \(8\text{.}\) So he writes an equation of the tangent line as \(y-8=3x^2(x-2)\text{.}\)
What's wrong?
What part of the general formula \(y-f(c) = f'(c) (x-c)\) did Joe forget?
See 2 Joe forgot to replace \(x\) with \(2\) in the derivative. The equation should be \(y-8=12(x-2)\text{.}\) The notation \(f'(c)\) is the part he forgot. He used \(f'(x)=3x^2\) instead of \(f'(2)=8\text{.}\)for an answer.
Exercise7.3.9Tangent Plane General Formula
Consider the function \(z=f(x,y)\text{.}\) Explain why an equation of the tangent plane to \(f\) at \((x,y)=(a,b)\) is given by
\begin{equation*} z-f(a,b) = \frac{\partial f}{\partial x}(a,b) (x-a) + \frac{\partial f}{\partial y}(a,b) (y-a). \end{equation*}Then give an equation of the tangent plane to \(f(x,y) = x^2+3xy\) at \((3,-1)\text{.}\)
Hint
Use either differential notation or a parametrization (see Task 6.5.5.c). Try both ways.
Subsection7.3.3Computational Practice
These are provided to help you achieve better skills in basic computational answers.