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Course: Multivariable calculus>Unit 3

Lesson 1: Tangent planes and local linearization

Computing a tangent plane

Here you can see how to use the control over functions whose graphs are planes, as introduced in the last video, to find the tangent plane to a function graph. Created by Grant Sanderson.

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• Do the constants `a` and `b` have anything to do with the vector normal to the plane?
• Yes, the normal vector will be (a, b, -1). To see why, write the function as:

z = a(x - x0) + b(y - y0) + z0,

Rearrange, to get the plane equation in standard form:

ax + by - z = -z0 + a*x0 + b*y0.

As we know from linear algebra, the coefficients of x, y, z are the coordinates of the normal vector:

n = (a, b, -1).
• you should label the axis, otherwise it is hard to know what is x axis, y axis or z axis
• Wolfram Code:

``(* our function to analyze*)F[x_, y_] = x^2 + y^2;(* differentiate with respect to x and y *)dFx = D[F[x, y], x];dFy = D[F[x, y], y];(* function for getting rate of change with respect to x and y at a particular point *)DFx[a_] := dFx /. x -> a;DFy[b_] := dFy /. y -> b;(* the point we want to create the tangent plane to *)p1 = -20;p2 = 20;(* grab the rates *)Ma = DFx[p1];Mb = DFy[p2];(* the equation for the plane *)P[x_, y_, m1_, m2_, a_, b_, c_] := m1 *(x - a) + m2*(y - b) + c;(* wolfram function for showing the 3D plots *)Show[Plot3D[F[f, g], {f, -30, 30}, {g, -30, 30},   ColorFunction -> Hue, PlotRange -> All],  	 Plot3D[  P[x, y, Ma, Mb, p1, p2, F[p1, p2]], {x, -30, 30}, {y, -30, 30},   PlotRange -> All] ]``
• We need to have 3 coordinates in space to specify a dot, I get it. The only thing I'm having a difficult time grasping is that why do we need 2 slopes (x, y) to define a plane? Don't we need 3 slopes, including z slope?
• Assume we know L(x,y)'s partial derivatives, Lx & Ly (just 2 slopes).

We can note that they are constant values. This is because they are equal to the partial derivatives of the function you are approximating evaluated at a specific point, say (x0, y0).

Let's see how to get out a plane:

First, we can use these partial derivatives to define two direction vectors for L(x,y), d1 & d2.

In particular, L(x,y)'s partial derivative w.r.t. x, Lx, tells us that if we move 1 unit in the x-direction (no matter what input we start from, since Lx is constant), we 'stay' on the plane by moving Lx units in the z-direction. So, we can say that:

d1 = <1, 0, Lx>

And by a similar argument, for the partial derivative w.r.t. y, we can say that:

d2 = <0, 1, Ly>.

Now, using the direction vectors, notice the following:

Since {d1, d2} is linearly independent--i.e., here, d1 is not parallel to d2--the span of the two vectors, {a(d1) + b(d2): a, b are any real numbers}, fills out two dimensions--a plane!

So the slopes are enough after all.

(By the way, this is not enough to show where the plane lies in R^3. For that, we need an extra restriction, which is a point that passes through the plane.)
• Doesn't L output a scalar value? How is it a function that outputs a plane? For example, when we plug in the value for x and y, L will evaluate to a number. This number is just the function output.
(1 vote)
• The output value of L together with its input values determine the plane. The concept is similar to any single variable function that determines a curve in an x-y plane. For example, f(x)=x^2 determines a parabola in an x-y plane even though f(x) outputs a scalar value.
BTW, the topic of the video is Tangent Planes of Graphs. The functions that describe graphs are scalar-valued; i.e., function output is a scalar value.
• The practice question are mostly implicit functions and I need help with taking partial derivative for implicit functions.
for example if the function x^2+y^2+z^2=0
if you take it's partial derivative with respect to partial x
you get 2x
but if you make it explicitly z = sqrt(-x^2-y^2)
then you take it with respect to partial x
you will get -x/sqrt(-x^2-y^2).
I don't know what am I taking with respect to partial x.
• Consider that each term is positive (or zero) in:
x^2 + y^2 + z^2
due to the squaring of each variable, then:
x^2 + y^2 + z^2 = 0
implies that x = y = z = 0 (unless you want to get into complex numbers), so this is a trivial equation.
Differentiating... if:
x^2+y^2+z^2=0
Its derivative with respect to (wrt) anything will be zero, because it's a constant (zero) with respect to everything.
Say you differentiate wrt 'x' ('d' and '∂' mean the same thing, so i'm not going to think too hard about using precisely the right symbol) - remembering what you do to one side of the equation you do to the other, so:
∂(x^2+y^2+z^2)/∂x = ∂0/∂x = 0
If you like, you can put in the intermediate step:
∂(x^2+y^2+z^2)/∂x = 2x = ∂0/∂x = 0
but it doesn't change anything.
The individual terms probably (but i'm not going there in this thought experiment) don't have to equal zero if you're dealing with complex numbers with a non-zero imaginary component, so if you subtract (X^2 + y^2) and take the square root, neither side needs to be zero, so maybe you can get somewhere with differentiating: i'll leave that to you to work out: please let me know the answer in a comment below 😉.
(1 vote)
• at the function of the graph is specified, where did this function come from? I thought we were using L(x,y)=2x+1y+c
(1 vote)
• How do you draw on the grapher window?