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### Differentiation of Vectors

Find the derivative of x(t) = ( cost,t2,[1/t] )
• To find the derivative of x(t), we differentiate each of its components, that is x′(t) = ( f1′(t),f2′(t),f3′(t) ).
Hence x′(t) = ( − sint,2t, − [1/(t2)] ).
For x(t) = ( cos2t,5et )
i) Find x(0)
• Letting t = 0 yields x(0) = ( cos20,5e0 ).
So x(0) = (12,5(1)) = (1,5).
For x(t) = ( cos2t,5et )
ii) Find x′(0)
• We take the derivative of x(t) to obtain x′(t) = ( − 2costsint,5et). Recall that f(t) = cos2t = [ cost ]2.
• Letting t = 0 yields x′(0) = ( − 2cos0sin0,5e0 ).
So x′(0) = ( − 2(1)(0),5(1)) = (0,5).
For x(t) = ( cos2t,5et )
iii) Find x"(0)
• We take the derivative of x′(t) to obtain x"(t) = ( − 2cos2t + 2sin2t,5et). Recall that f′(t) = [d/dt][ − 2costsint ] = − 2cost(cost) + (2sint)(sint).
• Letting t = 0 yields x"(0) = ( − 2cos20 + 2sin20,5e0 ).
So x"(0) = ( − 2(1)2 + 2(0)2,5(1)) = ( − 2,5).
Let x(t) = (t,t2), y(t) = (cost,sint) and z(t) = (t2 − 4t,e − t).
Find x(t) + x′(t)
• We first compute x′(t) = ( [d/dt]t,[d/dt]t2 ) = (1,2t)
So that the sum x(t) + x′(t) = (t,t2) + (1,2t) = (t + 1,t2 + 2t). Note that this is vector addition.
Let x(t) = (t,t2), y(t) = (cost,sint) and z(t) = (t2 − 4t,e − t).
Find [ x(t) ×y(t) ]′
• We first find the scalar product x(t) ×y(t) = (t,t2) ×(cost,sint) = tcost + t2sint.
• Now we find the derivative of the scalar product [ x(t) ×y(t) ]′ = [d/dt](tcost + t2sint).
Computing yields [ x(t) ×y(t) ]′ = tsint + cost + t2cost.
Let x(t) = (t,t2), y(t) = (cost,sint) and z(t) = (t2 − 4t,e − t).
Find [ 3x(t) + 4z(t) ]′
• We first find the sum 3x(t) + 4z(t) = 3(t,t2) + 4(t2 − 4t,e − t) = (4t2 − 13t,3t2 + 4e − t). Note that 4(t2 − 4t,e − t) = (4(t2 − 4t),4e − t).
• Now we find the derivative of the sum product [ 3x(t) + 4z(t) ]′ = ( [d/dt]( 4t2 − 13t ),[d/dt]( 3t2 + 4e − t ) ).
Computing yields [ 3x(t) + 4z(t) ]′ = (8t − 13,6t − 4e − t).
Let u(t) = (etsint,t,2,cost). Find u′(0).
• First we take the derivative of u(t), so u′(t) = ( [d/dt]( etsint ),[d/dt]t,[d/dt]2,[d/dt]cost ).
• We obtain u′(t) = (etcost + etsint,1,0, − sint). Setting t = 0 we find u′(0).
Hence u′(0) = (e0cos0 + e0sin0,1,0, − sin0) = ((1)(1) + (1)(0),1,0, − 0) = (1,1,0,0).
Find the line tangent to the curve x(t) = ( t3,√t ) at t = 1.
• The line tangent to a curve is the line passing through x(t) in the direction x′(t). Recall that a line can be represented parametrically with a point and direction by L(t0) = x(t) + t0x′(t).
• We compute the derivative of x(t), so that x′(t) = ( 3t2,[1/(2√t )] ). We can now find our line tangent to the curve L(t0).
• Substituting yields L(t0) = ( t3,√t ) + t0( 3t2,[1/(2√t )] ), at t = 1 we obtain our desired L(t0).
Hence L(t0) = ( 13,√1 ) + t0( 3(1)2,[1/(2√1 )] ) = (1,1) + t0( 3,[1/2] ).
Find the line tangent to the curve y(t) = ( [sint/2],t,[cost/2] ) , 0 ≥ t ≥ 2p, at t = [p/2].
• The line tangent to a curve is the line passing through y(t) in the direction y′(t). Recall that a line can be represented parametrically with a point and direction by L(t0) = y(t) + t0y′(t).
• We compute the derivative of y(t), so that y′(t) = ( [cost/2],1,[( − sint)/2] ). We can now find our line tangent to the curve L(t0).
• Substituting yields L(t0) = ( [sint/2],t,[cost/2] ) + t0( [cost/2],1,[( − sint)/2] ), at t = [p/2] we obtain our desired L(t0).
Hence L(t0) = ( [(sin[p/2])/2],[p/2],[(cos[p/2])/2] ) + t0( [(cos[p/2])/2],1,[( − sin[p/2])/2] ) = ( [1/2],[p/2],0 ) + t0( 0,1, − [1/2] ).
The trajectory of a point is described by the parametric equation x(t) = ( t, − [1/2]t2 + 1 ) at the time interval of 0 ≥ t ≥ 2 seconds.
i) What is the velocity of the point at t = 1?
• The velocity is obtained by taking the derivative of the position, that is computing x′(1).
We have v(t) = x′(t) = ( 1, − t ) so that x′(1) = (1, − 1). Note that velocity is a vector, defined by both a quantity and direction.
The trajectory of a point is described by the parametric equation x(t) = ( t, − [1/2]t2 + 1 ) at the time interval of 0 ≥ t ≥ 2 seconds.
ii) What is the speed of the point at t = 1?
• The speed is represented by the norm of the velocity, that is || v(1) ||.
Hence the velocity at t = 1 is || v(1) || = || (1, − 1) || = √2 . Note that speed is a scalar.
The trajectory of a point is described by the parametric equation x(t) = ( t, − [1/2]t2 + 1 ) at the time interval of 0 ≥ t ≥ 2 seconds.
iii) What is the acceleration of the point at t = [1/4]?
• The acceeration is obtained by taking the second derivative of the position, that is, the derivative of the velocity.
• We have v(t) = ( 1, − t ) so that a(t) = v′(t) = x"(t) = (0, − 1). Note that this acceleration is constant no matter what time t.
Hence the acceleration of the point at t = [1/4] is (0, − 1). Note that acceleration is a vector.
Find the angle that separates the position and velocity of an object represented by the parametric equation x(t) = (cost,sint) for 0 ≥ t ≥ 2p at t = [p/2].
• Note that t = [p/2] the position x( [p/2] ) = ( cos[p/2],sin[p/2] ) = (0,1).
• To find the velocity at t = [p/2], we take the derivative of the position, so x′(t) = ( − sint,cost).
• We have x′( [p/2] ) = ( − sin[p/2],cos[p/2] ) = ( − 1,0). Note that the vectors (0,1) and ( − 1,0) are orthagonal (the scalar product is equal to zero).
Hence the angle that separates the position and velocity of x at t = [p/2] is 90o.

*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.

### Differentiation of Vectors

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
• Differentiation of Vectors 0:18
• Example 1
• Definition 1: Velocity of a Curve
• Line Tangent to a Curve
• Example 2
• Definition 2: Speed of a Curve
• Example 3
• Definition 3: Acceleration Vector
• Two Definitions for the Scalar Part of Acceleration
• Rules for Differentiating Vectors: 1
• Rules for Differentiating Vectors: 2
• Rules for Differentiating Vectors: 3
• Rules for Differentiating Vectors: 4
• Example 4

### Transcription: Differentiation of Vectors

Hello and welcome back to educator.com, welcome back to multivariable calculus.0000

Last lesson we introduced the notion of a map, of a parameterization, a way of representing a curve in any dimensional space.0004

Today we are going to start doing calculus on those curves, so let us just jump right on in.0013

Hopefully, before we actually give any definitions or things like that, so example 1.0024

We will let x(t) = a curve in 3-space, defined by the following: (cos(t),sin(t),t).0031

If you recall from last lesson, this happened to be the curve of a spiral in 3-space that goes around the z-axis.0044

Okay, how do we differentiate a vector?0053

Well, it is exactly what you might think, you differentiate every component of that vector, and those become the components of a new vector.0057

That is exactly what we are going to do. So x', we will do x'(t), or we will also sometimes notate it as dx/dt.0065

In keeping with what we know from single variable calculus as far as notation. That is going to be equal.0078

Well, the derivative of cos(t) is -sin(t), the derivative of sin(t) is cos(t), and the derivative of t is 1.0085

There you go, that is it. The derivative of this vector is this vector.0094

We just differentiate the component functions one at a time, it is that simple.0101

Now, let us start with a definition, and we will draw some pictures.0108

Definition: the velocity of a curve x(t) is x'(t), in other words the velocity vector is equal to dx/dt at t = some value.0113

Let us stop and think about what this means, let me go ahead and draw some things here.0147

Just take some random curve in n-space, okay, so this is my curve in 2-space.0154

In other words, this is my x.0163

I get these vectors here for different values of t, I get different vectors, then when I connect the arrows of these vectors, I get this curve in 2-space.0167

Let us just say that this is one of the, let us just call this one x(t), this happens to be a particular value of t, so that happens to be the vector, so it is at that point.0178

Let me draw this a little darker here. There we go, that is our curve so we will start moving in that direction.0194

The velocity vector is equal to the derivative of this and as it turns out, this happens to also be a vector like here.0204

We had the cos(t),sin(t),t, the derivative was -sin(t),cos(t),1, this was also a vector.0215

This vector as it turns out is going to be in that direction.0221

This ends up being x'(t), let me write that a little bit better, x'(t).0228

We define that to be the velocity vector.0243

As it turns out, a vector in any space is a vector that originates from the origin, but of course I can take that vector and I can move it anywhere that I want to, it is still the same vector.0250

They are parallel. The direction actually is... you know, where a vector starts does not matter as long as it is the same vector translated somewhere.0260

So when we take a vector and we move it up here, and place the beginning of it here... something like that... this is the velocity vector.0272

It is a tangent vector. It is the vector tangent to the point x(t). In some sense, we are... this is sort of a pictorial way of showing you what the velocity of a vector really is.0285

Yes, it is the derivative. If you remember from single variable calculus, that is what a derivative is.0301

It is a tangent line, so, this is our way of staying analogous to single variable calculus.0305

The vector -- the velocity vector -- is really the vector that originates from here, but it does not matter, it is the same as this vector over here, so this is how we think about it.0313

What the velocity vector does is it is telling you the direction that that point on the curve is moving at that point.0322

So, the curve is going this way.0330

If I stop it at that point, that point, the point is actually moving in that direction, that is the whole idea.0335

The point is here, that is the position vector, the velocity vector is that way, it is going in that direction.0341

In a minute we will define the speed. As it turns out, the length of the vector, the norm of that vector is going to be the speed at which that point is travelling along that curve.0347

Again, this is all just normal stuff. We are just moving from actual functions, f(x) and f(t) now to vectors.0355

But we are treating it the same way, we are just working component wise. That is all that is happening here.0365

So, now, we will go ahead and talk about a line tangent to a curve in space.0371

The line... phew, I have got to learn how to write... so the line tangent to a curve in RN, or n-space, is the line passing through x(t), that is the position vector, right?0383

This is going to be the same as if we did the parameterization of a line, in the direction of x'(t).0414

Let me redraw this, the thing that I just drew. We had a curve, we had the position vector, then we said we had the velocity vector which we can think of as over here.0429

The line tangent to the curve at p, so that point, that is going to be the tangent line to the curve.0444

In RN, it is the line passing through x(t), well x(t) is right here in the direction of x'(t).0450

x'(t) is here so it is going to be passing in that direction. That is all this is.0456

Let us do an example. Example 2.0463

We want to find a parametric representation of the line tangent to the curve -- let us call it c(t) -- = (sin(t),cos(t),t) at t = pi/4.0470

So our curve is sin(t),cos(t),t and when t=pi/4, we want to find the equation of the line, the parametric equation of the line that is tangent to the curve, wherever that point happens to be.0513

Well, let us go ahead and just use our little definition right here.0529

The line, l(t) is going to equal x(t) the point that passes through that in the direction of t × x'(t). That is our generic equation.0534

Now we just need to take this, and take this, and put in t and see what we get.0550

So, x(pi/4), so we put pi/4 here, here, and here.0557

The sin(pi/4) is 1/sqrt(2), the cos(pi/4) is 1/sqrt(2), this is of course pi/4, so that is x(pi/4).0567

So let us do x'(t).0580

x'(t), which is the velocity vector right? so I will just go ahead and write that also.0583

Which is equal to v(t). So, we take the derivative.0590

The derivative of sin is cos(t), the derivative of cos is -sin(t), and the derivative of the t is 1.0595

That means, so now over here, let us go ahead and do x'(pi/4), well the cosine, it is still 1/sqrt(2).0604

This time it is -sin(t), so it is -1/sqrt(2), this one is just plain old 1.0621

Let me make a little more room here.0630

So, our l(t) becomes, I am going to write this in vertical notation, in other words I am going to take this and I am going to write it vertically rather than horizontally.0636

It becomes (1/sqrt(2), 1/sqrt(2), pi/4), that is that vector, that is x(t), + t × (1/sqrt(2),-1/sqrt(2),1).0649

That is it. That is what is going on here. This is the line, that is tangent to this curve.0685

I will use red. This is our curve. This equation that we came up with is the line that is tangent to the curve when t = pi/4.0697

It passes through this point, and it is moving in this direction. That is it.0708

Just a couple of basic definitions from some previous lessons, and this, everything is exactly the same.0718

Everything that you have learned from single variable calculus still applies, except that now we are just applying it to components.0724

Now instead of differentiating one function, in 5-space we are going to be differentiating 5 functions.0729

We just have to keep track of that. That is it, that is all that is going on here.0734

Now let us go ahead and define what we mean by the speed.0742

The definition, the speed of the curve, x(t) is the norm of the velocity vector.0747

You just did exactly what you expected, it is exactly what you know from physics. Basically you have a vector, a vector has direction and it has magnitude, the magnitude is the actual speed.0776

It is a number, it is a scalar, it is the scalar quantity that is associated with that vector. The vector itself gives you the direction, the norm gives you the scalar property.0786

I am going to write this one as speed of... you know what, that is fine, I will just call that s, it does not really matter what we call it.0797

So, s(t) = let us take x'(t) at a given value of t, and go ahead and take the norm of it. That is it, that is the speed of the curve.0808

There is so point that is travelling... so at any given point it is going to be moving at a certain speed, that is the algebraic definition of speed.0823

So, let us do another example.0834

We will let c be a map from r to R3, defined by c(t), again we will use our (cos(t),sin(t),2t), so it is a parameterization except... it is still a spiral, but it is a spiral that is moving up faster. That is the 2t part.0841

Our question is, how fast is a point travelling along the curve.0870

Well, we said that speed = the norm of the velocity vector, so the velocity vector is we are going to take the derivative of that vector.0894

Let us go ahead and take c', let us find what that is. c'(t), the derivative of cos is - sin(t), and the derivative of sin is cos(t) and the derivative of 2t is 2.0907

Now, the norm, c'(t)... I know this gets a little notationally tedious writing everything out, but it is actually worth writing everything out, do not do anything in your head, it is not worth it, you want to be correct.0923

You do not want to impress anybody with what you can do in your head.0936

c'(t), so we said that a norm of a vector is the vector dotted with itself, and then you take the square root.0943

So -sin(t), -sin(t) is sin2(t), so that is going to be sin2(t) + cos2(t) + 4, all under the radical.0950

Well sin2 + cos2 = 1, so that would be sqrt(1+4), so it is going to be sqrt(5). That is it.0971

So, this particular curve in 3-space, a point travelling along that curve is moving at a speed of sqrt(5) units/sec, units/minute, whatever it is the units have to be if we are dealing with a physics problem.0980

Now, let us define the next thing after speed. We will define the acceleration vector.0998

The acceleration vector is exactly what you think it is, it is d(x'(t))/dt, in other words it is the derivative of the velocity vector, which is, if x is the original curve, it is x''.1006

It is the second derivative, just like it is for normal single variable calculus.1030

That is it, nothing going on. It is still a vector, because you are differentiating vectors component wise.1035

Now, slight thing going on with the acceleration, I will just mention it, we will not particularly spend too much time discussing it. I just want you to be aware of it.1042

There are 2 definitions for the scalar part of acceleration.1060

One, there is the rate of change of the speed.1082

The rate of change of the speed, so you know that the rate of change of the speed is the acceleration.1088

Well the rate of change of the speed is d(s)/dt.1099

That speed is a scalar quantity, right? We said the speed was the norm of the velocity vector. A norm is a number.1108

So d(s)/dt is a number, it is a scalar. It is the rate at which the speed is actually changing.1117

Another way of defining acceleration, the scalar part, is by actually taking the acceleration vector, which is x'', and taking the norm of that.1123

That is analogous to defining what we did with speed in the first place.1136

We have a velocity vector, if we take the norm of that it gives us the magnitude, it give us the speed.1140

Here if we have an acceleration vector, which we do have, we can just differentiate the vector twice, the curve twice.1145

We can actually take the norm of it, however these two numbers are not actually the same.1153

More often than not they will not be the same. This has to do with how acceleration is defined based on the curve1158

I will not say anymore than that, but just know that we have 2 definitions for the scalar.1167

We can talk about the norm of the vector, or we can talk about the rate of change of the speed.1172

They are not the same. Again we will not deal with it if we have to deal with it, we will deal with it then. It will make more sense in a certain context.1177

It is not worth doing an example simply for the sake of doing an example. There may be one in your book, by all means I encourage you to take a look at it.1184

Now let us go through some rules for differentiating vectors. They are going to be pretty much exactly the same, but I just want to list them.1193

They are going to be the same as what you know from single variable calculus.1200

Let us go through them. Rules for differentiating vectors.1204

Okay, let us start with let x(t) and y(t) be 2 differentiable curves defined over the same interval.1218

In other words, for the same values of t. The values of t that work for x work for y, and differentiable just means that we can take the derivative of it, it has a derivative.1250

The components of the functions are differentiable, that is what that means.1261

Then, if I add the two curves, and then differentiate, in other words the sum of the curves, the derivative of the sum = the sum of the derivatives. That is what we are saying here.1265

= x' and y'. Noticed I left off the t just to save some notational tedium.1282

This is the same as first variable, if you have the sum of 2 functions f + g, well the derivative of f + g is the derivative of f + the derivative of g, that is it.1290

If I have some curve x and I multiply it by a constant and then I take the derivative, it is the same as just taking the constant and multiplying by the derivative of the function c(x)'.1300

Again, if you have t2, or 2t2, you can take 2 × derivative of t2 instead of taking the derivative of 2t2. It is the same thing.1313

Three, we have, okay, if we have x · y, if I have 2 curves and I take the dot product of those, and if I want to take the derivative of that, that is... this is analogous to the product rule.1324

Remember, 1 × the derivative of the other + that one × the derivative of the first. It is exactly what you think it is.1346

Here when you multiply the vectors, we are talking about the dot product, the scalar product.1353

We end up with the following, we end up with x · y' + x'... you know what let me do this in terms of t, this one I want to write out because it is important to see.1360

So, let us do x(t) · y(t)... now that we are getting stray lines down at the bottom, I think I am going to move to another page here, so we are doing this up here, so x(t) · y(t), if I want to take the derivative of that, that is equal to x(t) · y'(t) + x'(t) · y(t).1380

That is it, it is the normal product rule.1420

Notice, the dot product is a scalar, so these are going to be coordinate functions.1424

When I end up doing this, I am going to end up getting actual functions of t, right?1430

Because when you dot two vectors, you get a number, so when you dot two curves, which are numbers, you are going to get functions of t, just normal functions of t.1435

Again, just trust the definition and work it out component wise. There is nothing strange here.1445

Let us go ahead and do rule number 4. Then we will go ahead and finish off with a slightly lengthy example, but it is important to see the example.1454

The algebra, what it is that we actually do.1468

Let x(t) and y(t) be as before, in other words, both differentiable curves and both defined over the same interval, the same boundaries of t, and let f(t) be a function, just a normal differentiable function.1473

That is defined over the same interval.1508

So I am saying I have a curve x, a curve y, and I also have this curve function which is f(t), which could be anything t2, t3 , whatever.1516

Then if I happen to take that function and multiply it, scalar multiplication, by the vector x(t), and then if I try to take the derivative of this product, the product of a normal function × a vector function, that equals exactly what you think it is.1527

This times the derivative of that and that times the derivative of this.1551

It would be f(t) × x'(t) + f'(t) × x(t).1556

Do not let the notation throw you off. Everything is exactly still the same.1570

Multivariable calculus, a lot of the... you remember from your calculus courses how the problems started to get very long. Just in terms of algebra.1576

Just making sure that everything was written out. Multi-variable calculus will be no different.1585

You have the notation issue, that now, because you are working with components, the problems are going to be... they are going to be kind of long, simply because you have to work everything out in components, when you are working with a specific case.1590

When you are doing proofs of course, you just work with these things, you do not really have to work with components, sometimes you do.1605

Do not let the length of a problem discourage you. It is not that you are on the wrong path, this is the nature of sophisticated mathematics.1610

Now, let us do this length example that I mentioned. Example 4.1620

Let c b a mapping from R to R2, so a curve in 2-space, defined by c(t) = (e(t)cos(t),e(t)sin(t)).1631

You can see that this component is a product of two functions, and this component is a product of 2 functions.1650

Our goal in this problem is to show that the tangent vector to the curve, and the position vector, which is the curve itself, c(t), they form a constant angle of pi/4.1658

Okay, let us make sure that we understand what it is that this thing is saying.1708

So there is some curve in 2-space, (e(t)cos(t),e(t)sin(t)). Whatever it looks like.1713

Show that the tangent vector to the curve and the position vector form a constant angle.1718

I am just going to take... we just want to make sure that we know what this looks like.1723

Let us just take some piece of the curve... I do not have to know what the curve looks like, I do not want to know what the curve looks like.1728

That is the power of this, this is algebraic, I am just using geometry to help me get a grip on the problem.1732

The position vector is going to be some vector, let us actually go ahead and draw some, so the position vector is going to be this thing there.1742

That is c(t), well tangent vector is going to be this thing right here. That is going to be the velocity vector.1753

As this curve moves, they say that the position vector and the tangent vector, as the curve moves, it always keeps an angle of pi/4 at the angle between those 2 vectors always stays π/4, that is what we have to demonstrate.1760

So, what we are going to end up using of course, the thing that we know, is that the angle between the 2 vectors, the dot product between the two vectors divided by the norm... we are going to run through that entire algebraic property and hopefully at the end we end up with what we want which is this pi/4 business.1778

This is just what it means physically, now we can start doing the algebra based on our rules of differentiation.1795

This is where the problem gets a little long, but it is nothing strange going on here. It is just long, that is all it is.1801

Let us take c(t), so c(t) = e(t)cos(t) and e(t)sin(t).1810

I am going to rewrite that. That is the same as some function e(t), notice that e(t) is in both and I am going to pull it out of that vector and I am going to write it as cos(t) and sin(t).1822

This is akin to rule number 4, I have some normal function of t × some vector, that is all this is here, I've just rewritten it.1834

I did not have to, but I just thought I might as well use rule 4 since I wrote it down.1841

Now, let us go ahead and take the c'(t), let us go ahead and take the derivative of this.1848

c'(t), well we said that it is going to equal this × the derivative of that + that × the derivative of this.1853

That was rule 4, if I have a function times a vector, when I take the derivative of it, it is just the normal standard derivative they always take, 1 × the derivative of the other + 1 × the derivative of the other.1870

It is going to be this × the derivative of that, so e(t) × -sin(t)cos(t) + that × derivative of that.1881

Well derivative of e(t) is e(t) and that is just cos(t)sin(t).1900

There we go. Now, when I add these, I can go ahead and actually add them because e(t) is in both, it actually becomes e(t) and we are going to write -sin(t)+cos(t), I am going to add this component and this component.1909

Now I am going to add this component, and this component, cos(t)+sin(t). That is it.1931

So that is c'(t). Now let us go ahead and find the norm of c and t.1940

Again, we need the norms because... let me just write this down... remember we said the cos(θ) is going to equal c · c'/norm(c) × norm(c').1949

Now that we have c, we have c', now we need to calculate the norms.1966

The norm of ct, and what was the norm? The norm is the vector dotted with itself and then you take the square root.1972

So, c(t), that is going to be this thing, this dotted with itself... okay, so we are going to end up with e(2t) right?1985

e(t) × e(t) is e(2t).1998

This dotted with itself, you are going to end up with cos2(t) + e(2t) sin2(t), all under the radical.2006

This is going to equal e(2t), I am going to... e(2t) factors out cos2 + sin2 = 1 so... that... so it actually just comes out being e(t).2029

That is the norm of the vector, the position vector.2044

Now let us go ahead and do the norm of this thing.2047

This is going to be a little involved, we just have to keep up with the algebra. It is actually not that bad.2051

Now, c'(t), let us go ahead and calculate the norm of this, we are going to be calculating the norm of this vector right here.2058

That is going to equal e(2t), I will put a little parentheses here. This dotted with itself.2066

You are going to end up with sin2(t), this × itself is going to be -2sin(t)cos(t) + cos2(t) + e(2t) × cos2(t) + 2sin(t)cos(t) + sin2(t).2078

All of that is going to be under the radical.2114

Again, I take this vector right here... I will do it in red... I have taken this vector, and I have dotted it with itself.2117

So, e(t) × e(t), that is e(2t).2126

When you dot something with itself it is going to be this × this, so it is a binomial. When you do this × itself, you end up with this term.2132

When you do this × itself, you end up with this term. This e(t) distributes over both, that is why it shows up in both places.2143

Now, let us go back to black. We end up with sin2 + cos2 is 1, cos2 + sin2 is 1, - 2sin(t)cos(t) + 2sin(t)cos(t), those cancel.2154

You end up with the following. You end up with e(2t) × 2 under the radical, and you end up with e(t) sqrt(2).2171

We are almost there. We have all of the components that we need except just one more.2183

Now let us go ahead and move to the next page.2190

Let us write the definition of cos(θ).2192

cos(θ) = c(t) · c'(t)/norm(c(t)) × norm(c'(t)).2197

That will equal... I will come down here so I have a whole lot of room.2215

When I take c(t) · c'(t), here is what I end up getting.2220

I get e(2t) × - sin(t)cos(t) + cos2(t) + e(2t)sin(t)cos(t) + sin2(t), all over e(t) × e(t) sqrt(2).2227

Now, this and this factor out, I have to combine these, this - sin(t)cos(t) + sin(t)cos(t) they cancel, cos2 + sin2 is 1, you end up with the following.2261

You end up with e(2t) × 1/e(2t) × sqrt(2), that goes away and I am left with 1/sqrt(2).2277

cos(θ) = 1/sqrt(2), therefore θ = pi/4.2290

So, we got what we wanted, we showed that for this particular curve, the vector itself, the position vector and the tangent vector, the velocity vector always maintain an angle of pi/4.2300

We did this by just utilizing the formula that we know of, regarding the dot product, regarding the angle θ between those 2 vectors.2313

We just worked with differentiation and we worked with.. that is it... we worked with differentiation.2324

We made sure to be very slow and easy and to write everything out.2331

Do not let yourself get intimidated by the length of these things. Do not feel like you are getting lost in the algebra.2335

Trust yourself and trust your definition and just run through the process.2342

You do not have to know where you are going. These problems are too long to know what is exactly happening along the way.2346

But you have to know where you are going to start, you have to know what it is you are doing.2353

If you are sort of wondering where is the problem going, I am not quite sure, do not stop yourself, keep going.2357

If you end up with something that is wrong, it is not a big deal, you just go back and you start again.2363

Mistakes... that is the nature of mistakes, especially when you have this kind of algebra going on.2368

Thank you for joining us here at educator.com and we will see you next time for some more multivariable calculus. Bye-bye.2374