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Lecture Comments (2)

1 answer

Last reply by: Professor Hovasapian
Sat Jan 19, 2013 3:59 PM

Post by mateusz marciniak on January 19, 2013

hello again professor for example 2 my cross product was (2t^2 cos pheta + 2t^2 sin pheta - t) when i crossed the last row i took out the negative which left me with just a t making the last term a negative t once i distributed the negative instead of a positive t which was what you got professor. did i factor this incorrectly?

Tangent Plane & Normal Vector to a Surface

Find the normal vector N to the surface described by P(t,u) = (u2,t2,tu).
  • To find the normal vector N to a surface P(t,u) we compute N(t,u) = [dP/du] ×[dP/dt]. Note that this is a cross product.
  • Now, [dP/du] = (2u,0,t) and [dP/dt] = (0,2t,u).
  • So [dP/du] ×[dP/dt] = (2u,0,t) ×(0,2t,u) = det(
    i
    j
    k
    2u
    0
    t
    0
    2t
    u
    ) = (0 − 2t2)i − (2u2 − 0)j + (4tu − 0)k = − 2t2i − 2u2j + 4tuk.
Hence N(t,u) = ( − 2t2, − 2u2,4tu).
Find the normal vector N to the surface described by P(t,u) = ( − u + 3t2,u3 − 2tu + t2,1 − tu).
  • To find the normal vector N to a surface P(t,u) we compute N(t,u) = [dP/du] ×[dP/dt]. Note that this is a cross product.
  • Now, [dP/du] = ( − 1,3u2 − 2t, − t) and [dP/dt] = (6t, − 2u + 2t, − u).
  • So [dP/du] ×[dP/dt] = ( − 1,3u2 − 2t, − t) ×(6t, − 2u + 2t, − u) = det(
    i
    j
    k
    − 1
    3u2 − 2t
    − t
    6t
    − 2u + 2t
    − u
    ) = (( − 3u3 + 2tu) − (2tu − 2t2))i − (u + 6t2)j + ((2u − 2t) − (18tu2 − 12t2))k
  • Simplifying yields [dP/du] ×[dP/dt] = ( − 3u3 + 2tu − 2tu + 2t2)i + ( − u − 6t2)j + (2u − 2t − 18tu2 + 12t2)k = (2t2 − 3u3)i + ( − 6t2 − u)j + (12t2 − 18tu2 − 2t + 2u)k
Hence N(t,u) = (2t2 − 3u3, − 6t2 − u,12t2 − 18tu2 − 2t + 2u).
Find the normal vector N to the surface described by P(ϕ,θ) = (sinϕ,cosθ,tan(f + θ)).
  • To find the normal vector N to a surface P(ϕ,θ) we compute N(ϕ,θ) = [dP/(dθ)] ×[dP/(dϕ)]. Note that this is a cross product.
  • Now, [dP/(dθ)] = (0, − sinθ,sec2(ϕ+ θ)) and [dP/(dϕ)] = (cosϕ,0,sec2(ϕ+ θ)).
  • So [dP/(dθ)] ×[dP/(dϕ)] = (0, − sinθ,sec2(ϕ+ θ)) ×(cosϕ,0,sec2(ϕ+ θ)) = det(
    i
    j
    k
    0
    − sinθ
    sec2(ϕ+ θ)
    cosϕ
    0
    sec2(ϕ+ θ)
    ) = ( − sinθsec2(ϕ+ θ) − 0)i − (0 − cosϕsec2(ϕ+ θ))j + (0 − ( − sinθcosϕ))k
  • Simplifying yields [dP/(dθ)] ×[dP/(dϕ)] = ( − sinθsec2(ϕ+ θ))i + (cosϕsec2(ϕ+ θ))j + (sinθcosϕ)k
Hence N(ϕ,θ) = ( − sinθsec2(ϕ+ θ),cosϕsec2(ϕ+ θ),sinθcosϕ).
Find the unit vector n for or N(t,u) = (5t, − 2u,t + u).
  • To normalize our vector N we compute n = [N/(|| N ||)].
Hence n = [((5t, − 2u,t + u))/(|| (5t, − 2u,t + u) ||)] = [((5t, − 2u,t + u))/(√{25t2 + 4u2 + (t + u)2} )] = [((5t, − 2u,t + u))/(√{26t2 + 2tu + 5u2} )]
Find the unit vector n for N(ϕ,θ) = (sinϕ,1,cosθ).
  • To normalize our vector N we compute n = [N/(|| N ||)].
Hence n = [((sinϕ,1,cosθ))/(|| (sinϕ,1,cosθ) ||)] = [((sinϕ,1,cosθ))/(√{sin2ϕ+ 1 + cos2θ} )] = [((sinϕ,1,cosθ))/(√{1 + sin2ϕ+ cos2θ} )]
Find the normal vector N for a sphere of radius 1 (centered at the origin) at ( 1,0,0 ).
  • The normal vector N for a sphere of radius R is N(ϕ,θ) = (R2sin2ϕcosθ,R2sin2ϕsinθ,R2sinϕcosϕ) for 0 ≤ ϕ ≤ π and 0 ≤ θ ≤ 2π.
  • At (1,0,0) we have not moved any degrees in the θ direction but have moved ϕ = 90o upward. Hence we compute N( [(π)/2],0 ).
So N( [(π)/2],0 ) = ( 12sin2( [(π)/2] )cos(0),12sin2( [(π)/2] )sin(0),12sin( [(π)/2] )cos( [(π)/2] ) ) = (1,0,0)
Find the normal vector N for a sphere of radius 1 (centered at the origin) at ( [(π)/3],[(π)/6] ).
  • The normal vector N for a sphere of radius R is N(ϕ,θ) = (R2sin2ϕcosθ,R2sin2ϕsinθ,R2sinϕcosϕ) for 0 ≤ ϕ ≤ πand 0 ≤ θ ≤ 2π.
At N( [(π)/3],[(π)/6] ) = ( 12sin2( [(π)/3] )cos( [(π)/6] ),12sin2( [(π)/3] )sin( [(π)/6] ),12sin( [(π)/3] )cos( [(π)/3] ) ) = ( [(3√3 )/8],[3/8],[(√3 )/4] )
i) Find the normal vector N to the surface described by P(t,u) = (2t + eu,e − t,t3u2)
  • To find the normal vector N we compute [dP/du] ×[dP/dt].
  • Now, [dP/du] = (eu,0,2t3u) and [dP/dt] = (2, − e − t,3t2u2).
  • So [dP/du] ×[dP/dt] = (eu,0,2t3u) ×(2, − e − t,3t2u2) = det(
    i
    j
    k
    eu
    0
    2t3u
    2
    − e − t
    3t2u2
    ) = (0 − ( − e − t2t3u))i − (3t2u2eu − 4t3u)j + ( − eue − t − 0)k
  • Simplifying yields [dP/du] ×[dP/dt] = e − t2t3ui − (3t2u2eu − 4t3u)j − eue − tk
N(t,u) = (e − t2t3u, − 3t2u2eu + 4t3u, − eue − t).
ii) Evaluate N(0,0), where N is the normal vector to the surface described by P(t,u) = (2t + eu,e − t,t3u2)
  • We have N(t,u) = (e − t2t3u, − 3t2u2eu + 4t3u, − eue − t) and so N(0,0) = (e − 02(0)3(0), − 3(0)2(0)2e0 + 4(0)3(0), − e0e − 0).
Simplifying yields N(0,0) = (0,0, − 1).
iii) Evaluate N(0,1), where N is the normal vector to the surface described by P(t,u) = (2t + eu,e − t,t3u2)
  • We have N(t,u) = (e − t2t3u, − 3t2u2eu + 4t3u, − eue − t) and so N(0,1) = (e − 02(0)3(1), − 3(0)2(1)2e1 + 4(0)3(1), − e1e − 0).
Simplifying yields N(0,0) = (0,0, − e).
i) Find the normal vector N to the surface described by P(t,u) = ( √{t2 + u2} , − √{t2 + u2} ,u )
  • To find the normal vector N we compute [dP/du] ×[dP/dt].
  • Now, [dP/du] = ( [u/(√{t2 + u2} )], − [u/(√{t2 + u2} )],1 ) and [dP/dt] = ( [t/(√{t2 + u2} )], − [t/(√{t2 + u2} )],0 ).
  • So [dP/du] ×[dP/dt] = ( [u/(√{t2 + u2} )], − [u/(√{t2 + u2} )],1 ) ×( [t/(√{t2 + u2} )], − [t/(√{t2 + u2} )],0 ) = det(
    i
    j
    k
    [u/(√{t2 + u2} )]
    − [u/(√{t2 + u2} )]
    1
    [t/(√{t2 + u2} )]
    − [t/(√{t2 + u2} )]
    0
    ) = ( 0 − ( − [t/(√{t2 + u2} )] ) )i − ( 0 − ( [t/(√{t2 + u2} )] ) )j + ( − [tu/(t2 + u2)] − ( − [tu/(t2 + u2)] ) )k
Simplifying yields [dP/du] ×[dP/dt] = [t/(√{t2 + u2} )]i + [t/(√{t2 + u2} )]j + 0k, and so N(t,u) = ( [t/(√{t2 + u2} )],[t/(√{t2 + u2} )],0 ).
ii) Evaluate N( √2 ,√3 ), where N is the normal vector to the surface described by P(t,u) = (2t + eu,e − t,t3u2)
  • We have N(t,u) = ( [t/(√{t2 + u2} )],[t/(√{t2 + u2} )],0 ) and so N( √2 ,√3 ) = ( [(√2 )/(√{( √2 )2 + ( √3 )2} )],[(√2 )/(√{( √2 )2 + ( √3 )2} )],0 ).
Simplifying yields N( √2 ,√3 ) = ( √{[2/5]} ,√{[2/5]} ,0 ).
iii) Find the length of N( √2 ,√3 ), where N is the normal vector to the surface described by P(t,u) = (2t + eu,e − t,t3u2)
  • Recall that the lenght of a vector is equal to its norm, so the length of N( √2 ,√3 ) is || N( √2 ,√3 ) ||
Hence || N( √2 ,√3 ) || = || ( √{[2/5]} ,√{[2/5]} ,0 ) || = √{[2/5] + [2/5] + 0} = [2/(√5 )].
i) Find the normal vector N to the surface described by P(ϕ,θ) = (cos2θ,sin2ϕ,1)
  • To find the normal vector N we compute [dP/(dθ)] ×[dP/(dϕ)].
  • Now, [dP/(dθ)] = ( − 2sin2θ,0,0) and [dP/(dϕ)] = (0,2cos2ϕ,0).
  • So [dP/(dϕ)] ×[dP/(dθ)] = ( − 2sin2θ,0,0) ×(0,2cos2ϕ,0) = det(
    i
    j
    k
    − 2sin2θ
    0
    0
    0
    2cos2ϕ
    0
    ) = (0 − 0)i − (0 − 0)j + ( − 4sin2θcos2ϕ− 0)k
Simplifying yields [dP/(dϕ)] ×[dP/(dθ)] = 0i − 0j − 4sin2θcos2ϕk and so N(ϕ,θ) = (0,0, − 4sin2θcos2ϕ).
ii) Evaluate N( [(π)/2],[(π)/4] ), where N is the normal vector to the surface described by P(ϕ,θ) = (cos2θ,sin2ϕ,1)
  • We have N(ϕ,θ) = (0,0, − 4sin2θcos2ϕ) and so N( [(π)/2],[(π)/4] ) = ( 0,0, − 4sin( 2 ×[(π)/4] )cos( 2 ×[(π)/2] ) ).
Simplifying yields N( [(π)/2],[(π)/4] ) = (0,0,4).

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Tangent Plane & Normal Vector to a Surface

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Tangent Plane and Normal Vector to a Surface 0:35
    • Tangent Plane and Normal Vector to a Surface Part 1
    • Tangent Plane and Normal Vector to a Surface Part 2
    • Tangent Plane and Normal Vector to a Surface Part 3
    • Example 1: Question & Solution
    • Example 1: Illustrative Explanation of the Solution
    • Example 2: Question & Solution
    • Example 2: Illustrative Explanation of the Solution

Transcription: Tangent Plane & Normal Vector to a Surface

Hello and welcome back to educator.com and multi variable calculus.0000

In today's lesson, we are going to discuss the tangent plane and the normal vector to a surface.0004

Remember we described how to parameterize a surface with two parameters, now we are going to talk about the plane that is going to be tangent to it, the same way if you have a curve, you have a tangent line, if you have a surface you have a tangent plane.0010

We are going to define the normal vector, the vector that sticks away from the surface and we are going to set the stage for the integration of functions and vector fields over surfaces.0022

So, let us get started. The first thing I am going to do is draw a little picture here, so that we have something that we are working with.0035

I am going to draw your normal xyz coordinate plane, actually you know what, I think I am going to draw the surface first, and then I am going to put the axes on top of that.0040

So, let me go something like this, and this, and this, and that.0052

So, let me draw one line that goes like that and goes along the surface that way, and let us say that we have another line that goes along the surface that way.0062

Our axes, let us go ahead and put them here... this is going to go like that, and this is going to go over like that, so this is the y, this is the x, and this is the z, so we have this surface in 3-space.0071

okay. So, let us go ahead and start talking about what is going on. We have been using the letter p for parameterization of a surface, so let p(t,u) = well, we have f1, which is a function of (t,u), these are the three coordinate functions.0089

We have f2 which is a function of (t,u), and we have f3 which is a function of t and u.0108

So, Let this be a parameterization for a surface.0115

Okay. Excuse me... now, if we pick a specific value of t, if we hold it constant and if we hold it constant and we just vary the variable u, what you end up with is a curve along the surface.0132

If we end up holding u constant, pick a specific value of u and then just vary t, again we get another curve. Essentially you just get a function of 1 variable.0153

Really what we are doing when we parameterize this surface is we are making a curve this way, and then we are parameterizing, we are taking that curve and we are sweeping it out to create the rest of the surface. That is what is happening.0165

So, imagine this curve, and then imagine sweeping it back and forth in this direction and this direction, and that gives you the parameterized surface.0178

So, if we hold one of those constant, we actually get curves along the surface. So, let me write that down.0189

If we hold... if we pick a value for t, hold it constant and vary u, we get a curve... so let us call that one c1.0196

It is actually just a function of u, but it is actually the parameterization when we are holding t constant. That is what it is.0236

We can do the same for the other ones. So, this is c1(u), let us just call this curve over here c2(t), so, if we hold u constant and vary t we get a second curve.0247

This is actually a really, really good way of thinking about surfaces. Think about them as a curve in this direction, and a curve in this direction and the curves cross and it will give you a sense of what is happening on that curve near that point.0276

So, this point right here, that is the p(t,u) for some value of t and u.0290

Again, if you are there and if you happen to hold t constant but you are going to vary u, you are going to move in this direction and in this direction. 0296

If you are going to hold u constant, you are going to move that way and that way. If you vary both of them, you are going to vary all over the surface. That is what is happening.0304

So, you get a second curve. We will call this one c2 and it is a function of t, but it is still just the actual parameterization t and u, where one of them is held constant.0312

Okay. Now, here is what is nice, we know that we can differentiate curves. Well, differentiating these curves, taking c1' and c2' is the same as just taking the partial derivative of the parameterization with respect to the particular variable that you are varying.0324

What you end up with when you take the derivative is you get the tangent vector to the curve.0342

So, let me go ahead and write this down and then I am going to redraw the picture and I am going to redraw some tangent vectors on top of it.0348

We can differentiate these curves because we have been doing so all along.0355

Well, c1'(u) is nothing more than the partial of the parameterization with respect to u, and c2'(t) is nothing more than the partial of the parameterization with respect to t.0369

We know how to do partial derivatives, we have been doing them the entire course. So, that is it. That is all that is going on. Hold one variable constant, vary the other one, and you can take partials.0387

What you get are the following. Let me redraw this surface here, i am just going to draw the surface without the axes.0399

Let me draw it over here and give me a little bit more room to work. We have something like this, like that, like that, like that and we said we had a curve here and we have this curve here... like this.0405

So, when we take... so we pick a particular point, this point is our p(t,u), if we hold one of the variables constant and we take the derivative of it, it is like taking the derivative of the curve.0417

What you get is this vector, the tangent vector. Let us just call this one dp du.0432

If you move along this curve, where you are varying t holding u constant, you get a vector this way. You get the tangent to the curve at that point.0441

This is dp dt. I hope this makes sense. You have got these 2 curves and at a particular point, you can take the partial of the parameterization you are going to get two tangent vectors that are the tangents to the curves the way we have done before.0451

Now, these are... let me actually write all of this down. So these are just the tangent vectors to the curves.0469

Well, these two vectors, they span a plane, that is what it is. Any time you have two vectors, what you have is the plane that they cover.0495

These two vectors span a plane. This plane is tangent to the surface, right? it is tangent to the surface at pt(u). 0508

So, if I have a particular point, or a particular t and a particular u, if I take the partial with respect to u I get a tangent vector.0536

If I take the partial with respect to t at that point, I get another tangent vector. Well these two tangent vectors span a plane, and that plane is actually the plane that is tangent to the surface at that point. That is what is going on.0545

I have got a surface, I have got a vector this way, I have got a vector this way, this vector spans a plane, this plane is what is tangent to this surface at that point. Pretty fantastic.0557

Okay. Now, here is where it gets great. So, if dp du -- oops, not cross yet -- and dp dt are non 0 vectors, which for all of our problems they will be, then the cross product dp dt cross dp du, or perhaps in the other direction depending on the orientation that we want for the surface is a vector perpendicular to orthogonal to both dp du and dp dt.0572

So, given two vectors, if I take their cross product, I get another vector. This particular vector happens to be perpendicular to this one and this one. It is perpendicular to each. 0630

That is very, very convenient because now we have a plane that touches a surface, and at that point, we are going to create a normal vector like this.0639

Well, this normal vector is going to be at that point the normal vector to the surface, the same way we had a normal vector to a curve. 0650

Now, it just happens to be two curves that cross each other, those 2 curves that cross each other define a surface at that point... you have a normal that is... you have a vector that is normal to both curves. It is a vector that is normal to the surface.0658

We are constructing a tangent plane and we are constructing the normal vector at that point. That is what we are doing.0671

So, let me draw this time, let me draw it on top of this. So, we have dp du, actually you know what, let me do it one more time... so, let me go this way, make it a little bigger, so we have this is 1, this is 2, so we have... that is 1 tangent vector, that is another tangent vector.0678

Let us call this one dp du, and this is going to be dp dt. Now let me go to blue. Now when I take the cross product of dp du cross dp dt, and remember the right hand rule? I am going to get a vector that points this way.0707

It is going to be perpendicular to that, and it is going to be perpendicular to that.0726

So, this vector right here, it is dp du cross dp dt.0732

Now it is specifically dp du cross dp dt, if I went dp dt cross dp du, I would get the vector that is going in the other direction, that is actually going into the surface.0742

We want the one that is going away from the surface. That is a question of orientation.0754

We make that choice when we are doing these problems. We always want to pick the vector that is pointing away from the surface. You do so just by choosing which comes first, the dp du or the dp dt.0758

Okay. SO that is it, this is what we have created. Nice.0772

Let us go ahead and let us see, so let me just make sure we have got everything, dp du, dp dt. Let us go ahead and label this point, this is the point p of t, u, okay.0776

Now, this vector right here, we actually call it the vector n, for normal, capital N, and if you want to put a vector sign on top of it you can. I am going to do both. Sometimes I am going to have it on top, sometimes I am going to have the vector arrow now on top of a capital letter. 0790

Generally, when I do lower case letters and it is a vector, I will always put the vector mark on top, but normally for capital letters, I will not have the vector notation unless there is a point of confusion.0810

So, let us go ahead and define N. N is the normal vector, that is also a function of t and u, it is dp du cross dp dt.0824

You have a parameterization p, you take the partial with respect to t, you take the partial with respect to u, you are going to get 2 vectors, you take the cross product of that vector and what that gives you is the normal vector n. So this is N.0843

The normal vector to the surface and again let me write, making sure we choose the orientation such that N points away from the surface.0859

Now, again, for the most part this is not going to be too much of a problem.0891

When you are doing these problems, when you are dealing with a particular surface you are going to be dealing mostly with familiar surfaces and you are going to be able to sort of look at, once you come up with N, you are going to be able to see which direction the vector is pointing in.0896

If it is pointing away from the surface, or if it is pointing into the surface, and you can adjust accordingly.0910

By adjusting accordingly, all you are really doing is taking a negative sign, so dp du cross dp dt is just negative dp dt cross dp du.0917

When you are actually doing these problems, the only thing that changes is the sign. If you end up with a negative and you want the positive, you just change the sign.0926

What that means is you have taken the opposite orientation. The magnitudes and numbers do not actually change. There is nothing that you... ultimately there is nothing that you really have to do.0935

You are going to end up with the same number. It is just the sign might be different.0944

That is not really a problem for most applications. Okay. So, hopefully this is clear.0949

Now, we also define one other thing. Once we actually find N, so N is not in general a unit vector. It is not a vector of length 1, we can make it a unit vector by dividing by its norm, so we also define that thing.0958

So, let me go ahead and go to the next page and define that. So, we also define small n, that is just n divided by its norm.0972

Well, n is dp du cross dp dt, and it is divided by the norm of the vector which is dp du cross dp dt, because dp du cross dp dt is a vector.0990

This is the unit vector -- excuse me -- in the direction of n.1012

That is it. So, be very, very careful. Now that we have introduced the cross product and we have been dealing -- you know -- extensively with the dot product, you are going to get them confused. I get them confused.1030

In fact, yesterday when I was working on a problem, I ended up taking a dot product when I meant to take the cross product. In some of the things that we are going to be dealing with, like integrals of a vector field, integrals of a function over a surface, you are going to have a mix.1041

You are going to have dot products and you are going to have cross products, so just be very, very careful. There is going to be a lot of notation on a page. 1055

Just be very clear about what you are doing and -- you know -- go slowly and go carefully. When you want to take a dot product instead of a cross product, when you want to take a cross product instead of a dot product. 1063

Dot product is a number, it is a scalar. Cross product is a vector, you actually get a vector.1074

So, let us just do an example and hopefully everything will start to come together here. So, example number 1. We will take the sphere of radius r, a given radius.1080

Well we have a parameterization for a sphere of radius r, so let us go with this particular parameterization. φ and θ.1100

That is going to equal rsin(φ), cos(θ), rsin(φ) sin(θ).1108

Now remember, we have two parameters. r is not a parameter, r is a constant, we have chosen the radius that we want because we are not talking about the volume, the whole thing, we are talking just about the surface.1119

We have r -- oops, make it a little bit clearer -- this is, I know we know what it is but -- we have rcos(φ), and of course φ is going to run from 0 to pi, and θ is going to run from 0 to 2pi.1130

That is our parameterization. Let us go ahead and construct our particular tangent plane and our normal vector.1148

Let us take dp dφ, so if I take dp dφ, well, I just -- you know -- take the derivative of this with respect to φ derivative of this with respect to φ, derivative of that with respect to φ, I should end up with the following.1158

Again I hope that you are going to confirm this for me because I do tend to make a lot of little errors.1175

So, rcos(φ), cos(θ), and then we have rcos(φ)sin(θ), again we are doing it with respect to φ so the θ stays the same, and this is going to be -rsin(φ).1182

There you go, that is dp dφ, well now let us go ahead and do dp dθ, so again when we differentiate this, this is 3-coordinate functions. We have a vector.1203

Now dp dθ, this is going to be -rcos -- oops, not cos(φ), see again, there you go, this is the original right here... let me actually circle that one... this is our parameterization, so you have a lot of things floating around so let us make sure -- so rsin(φ)cos(θ).1216

This is going to be -rsin(φ)sin(θ), and if I take the derivative of this, this is going to be rsin(φ)cos(θ), and there is no θ here so the derivative with respect to that is just plain old 0.1237

Okay. So now I have the two partials of the parameterization and I am going to go ahead and take the cross product of these two partials... dp dφ, dp dθ.1263

Again, it is just a lot of notational tedium, not a problem. Of course you are dealing with something that is reasonably complex. It is a surface in 3-space, there is a lot going on.1273

So dp dφ cross dp dθ, and again we are going to go ahead and use our symbolic representation to solve this. We are going to have i here, and I will put j here, and I will put k here, and now I have got this vector... the dp dφ.1285

So this is going to be rcos(φ)cos(θ)... this is rcos(φ)sin(θ), this is -rsin(φ), and here we have -rsin(φ)sin(θ).1308

This is rsin(φ)cos(θ), and this is 0, thank god. I have one thing that is going well here.1331

So, we are going to end up taking the determinant of that. Oops, got our crazy little lines here towards the bottom of the page again... hopefully the will not interfere with what it is that we are doing.1341

That is that, and we are going to expand it of course along the first row, so when I do that I am going to... well hopefully you are ok with determinant expansions, so let me write down what it is that you actually get here, and I am going to write it on the next page.1350

So, just to show you, the first one is going to be this × that - this × that.1367

The second is going to be this × that - this × that, with a negative sign in front because this is +, -, +.1374

The third one is going to be the third component function... is going to be this × that - this × that... okay?1382

Let me write it all out so that we actually see what we are dealing with here.1390

We are going to have r2sin2φcos(θ), and this is going to be the first coordinate function i + r2sin(φ)sin(θ), when you actually work out all of the negatives and positive signs, all of the negative and positive coefficients, this is what you end up getting, + r2cos(φ)sin(φ)cos2(θ) + r2sin(φ)cos(φ)sin2(θ) × k. 1396

Well sin2(θ) and cos2(θ)... uhh... sin2 + cos2 = 1, so our final, and I am going to write this in normal notation without the i, j, k, okay? 1453

You have the following vector. You have r2sin2(φ)cos(θ), you have r2sin2(φ)sin(θ), and you have r2sin(φ)cos(φ).1468

That is our normal vector. So this is n. This equals n. This is n, the normal vector to the surface of the sphere.1497

Okay. Now let us go ahead and find the norm of n. Now we are going to end up taking the norm of this vector, now let me go to blue, so now that we have n, we are going to take the norm of this.1522

Well, the norm of the vector is this squared + this squared + this squared, the 3 components under the square root sign.1536

So, the norm of n is going to equal r4sin4cos2(θ)... actually you know what, I do not know if I want to go through this whole thing... I would much rather just, ahh it is okay, I will go ahead and go through it. It is not a problem, it is nice to sort of see everything.1543

Okay. r4sin4sin2(θ) + r4sin2(φ)cos2(φ), this is going to be sin4(φ), I have to put an angle in there... this is also going to be sin4(φ). Okay... and all of this is of course under the radical.1574

When we simplify some things we end up with r4sin4(φ) + r4sin2(φ)cos2(φ), and all of this is under the radical.1605

I can pull out an r4 and I can pull out a sin2, so this is going to be r4sin2(φ) × sin2(φ) + cos2(φ), which is equal to 1.1623

So, this is equal to r4sin2(φ), and the solution to this is r2, you can just take the square root. The absolute of sin(φ). 1645

When you take the square root of something that is squared underneath, it is the absolute... but since φ is > or = 0, and < or = pi, the sin(φ) is always going to be positive, so the absolute of sin(φ) is just plain old sin(φ)... we can get rid of the absolute value signs.1657

So, the norm of this vector is equal to r2sin(φ)... there we go. We have everything that we need to deal with the normal vector if we want to take this vector and divide it by this thing, we end up getting the unit vector in the direction of n, and that is it.1684

Given a parameterization, take dp dφ, take dp dθ, take the cross product, and then take the norm if you need the norm. You do not always need the norm, but we thought that we would go through the process. That is all that is going on here.1705

So let us go ahead and take a look at what this actually looks like. Let me draw out a circle.1718

Now, I am going to go ahead and give you a... I am going to pick a curve here, it is going to be one curve, and I am going to pick another one... so another curve.1732

So, where they meet, this is going to be some p(t) -- oh, p(φ,θ). This is going to be some point on the sphere... p(φ,θ).1744

Okay. Now, let me go ahead. So if I have... so this curve right here, if I pick a θ... if I hold θ fixed and I vary φ, I am going to get this curve right here. That is this one.1759

So, hold θ fixed and vary φ and of course this other curve is then the opposite.1778

If I hold φ fixed and vary θ, I get this curve. Well, if this is the one where φ is varying, I can get... and if I take that... this is dp that is dp dφ.1793

Well if this is the one where θ is varying, I get that vector. This is dp dθ.1817

When I take the cross product of those two, I end up with a vector that is perpendicular to both. It is perpendicular to this, and it is perpendicular to that.1825

This is my vector n, this is the cross product of dp dφ, dp dθ, this way and it points out of the sphere.1838

There is the center of the sphere. That is it. That is all that I have done. Very nice.1850

Okay, let us do another example. Just to sharpen up our skills here, so example 2 we will let the parameterization of t and θ be equal to tcos(θ), tsin(θ) and t2. This happens to be the parameterization for a paraboloid.1857

Of course t is going to be > or = 0, and θ is going to run greater than 0 and less than 2pi.1885

So, let us go ahead and calculate things. We do not need to -- actually, we do not even need to worry about what the surface is.1895

It is nice to know what it is, but really we are just working algebraically, so let us just go ahead and do the calculus, do the math.1902

So, if we take dp dθ), let us do θ first. It does not matter what you take first, you are going to end up with -tsin(θ), you are going to end up with tcos(θ), and you are going to end up with 0.1910

Let us go ahead and take dp dt, that is going to equal cos(θ)sin(θ) and 2t.1934

Now, when I go ahead and I take dp dθ cross dp dt, this vector cross that vector, I am going to end up with my normal vector.1948

Well, that is going to equal i,j,k, my symbolic representation - tsin(θ), this is going to be tcos(θ), this is going to be cos(θ), this is going to be sin(θ), and this is going to be 2t. I take the symbolic determinant of that expanded along the first row.1967

So, now I am going to move to the next page. When I do that I will go ahead and let you confirm what it is that I end up getting. I end up with the following.1994

I end up with 2t2cos(θ)i + 2t2sin(θ)j + t × k, and expressed in regular list notation, without the i,j,k, I end up with 2t2cos(θ), 2t2sin(θ, and t. This is my n.2004

That is my normal vector. Now, if I want to take the norm, well, not a problem. square of this, square of this, square of this, square of this... all under the square root sign.2045

You end up with 4t4cos2(θ) + 4t4sin2(θ), and we like it when we sin2 and cos2 because that is going to equal 1 + t2, all under the radical, and the cos2 and the sin2, I factor out the 4t4.2060

I end up with 4t4 + t2 all under the radical. This gives me the norm. The length of that vector at a particular value of t.2082

My normal vector is this one. Okay? It is a vector.2097

This is the norm of that vector. It is the magnitude of that vector. It is the length of that vector.2104

Now, let us see what this looks like. I am going to draw a paraboloid here, so we have something like this and of course this is a 3-dimensional paraboloid, but I will not put the axes on there.2111

Well, there is going to be some particular curve that goes this way as you hold one of them... if you fix θ and vary t, you are going to get this thing.2125

If you fix t and vary θ, you are going to get that thing.2135

Well, this point right here, this is a particular p(tθ). You are going to get a tangent vector that way, this is going to be dp dt.2140

You are going to get a tangent vector that way. That is the dp dθ, and when I take dp dθ, cross dp dt, I end up with my normal vector to the surface.2157

This is a vector as the surface changes, this normal vector is always going to be perpendicular to it... no matter which, that is the whole idea.2174

So, it is perpendicular to that one and it is perpendicular to that.2185

This happens to be our n, which again is dp dθ cross dp dt.2190

If you end up doing dp dt cross dp dθ, what you end up with is the vector going into the paraboloid.2199

It is still normal to the surface, but it is just pointing in. That is it. This way instead of this way.2206

It is just a question of orientation, positive, negative, that is all it is, the only thing that changes is the sign. 2213

Okay. So, that is tangent plane and normal vector to a surface.2219

Thank you for joining us here at educator.com, we will see you next time. Bye-bye.2223