For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

### Differentiation of Vectors

^{2},[1/t] )

- To find the derivative of →x(t), we differentiate each of its components, that is →x′(t) = ( f
_{1}′(t),f_{2}′(t),f_{3}′(t) ).

^{2})] ).

^{2}t,5e

^{t})

i) Find →x(0)

- Letting t = 0 yields →x(0) = ( cos
^{2}0,5e^{0}).

^{2},5(1)) = (1,5).

^{2}t,5e

^{t})

ii) Find →x′(0)

- We take the derivative of →x(t) to obtain →x′(t) = ( − 2costsint,5e
^{t}). Recall that f(t) = cos^{2}t = [ cost ]^{2}. - Letting t = 0 yields →x′(0) = ( − 2cos0sin0,5e
^{0}).

^{2}t,5e

^{t})

iii) Find →x"(0)

- We take the derivative of →x′(t) to obtain →x"(t) = ( − 2cos
^{2}t + 2sin^{2}t,5e^{t}). Recall that f′(t) = [d/dt][ − 2costsint ] = − 2cost(cost) + (2sint)(sint). - Letting t = 0 yields →x"(0) = ( − 2cos
^{2}0 + 2sin^{2}0,5e^{0}).

^{2}+ 2(0)

^{2},5(1)) = ( − 2,5).

^{2}), →y(t) = (cost,sint) and →z(t) = (t

^{2}− 4t,e

^{ − t}).

Find →x(t) + →x′(t)

- We first compute →x′(t) = ( [d/dt]t,[d/dt]t
^{2}) = (1,2t)

^{2}) + (1,2t) = (t + 1,t

^{2}+ 2t). Note that this is vector addition.

^{2}), →y(t) = (cost,sint) and →z(t) = (t

^{2}− 4t,e

^{ − t}).

Find [ →x(t) ×→y(t) ]′

- We first find the scalar product →x(t) ×→y(t) = (t,t
^{2}) ×(cost,sint) = tcost + t^{2}sint. - Now we find the derivative of the scalar product [ →x(t) ×→y(t) ]′ = [d/dt](tcost + t
^{2}sint).

^{2}cost.

^{2}), →y(t) = (cost,sint) and →z(t) = (t

^{2}− 4t,e

^{ − t}).

Find [ 3→x(t) + 4→z(t) ]′

- We first find the sum 3→x(t) + 4z(t) = 3(t,t
^{2}) + 4(t^{2}− 4t,e^{ − t}) = (4t^{2}− 13t,3t^{2}+ 4e^{ − t}). Note that 4(t^{2}− 4t,e^{ − t}) = (4(t^{2}− 4t),4e^{ − t}). - Now we find the derivative of the sum product [ 3→x(t) + 4→z(t) ]′ = ( [d/dt]( 4t
^{2}− 13t ),[d/dt]( 3t^{2}+ 4e^{ − t}) ).

^{ − t}).

^{t}sint,t,2,cost). Find →u′(0).

- First we take the derivative of →u(t), so →u′(t) = ( [d/dt]( e
^{t}sint ),[d/dt]t,[d/dt]2,[d/dt]cost ). - We obtain →u′(t) = (e
^{t}cost + e^{t}sint,1,0, − sint). Setting t = 0 we find →u′(0).

^{0}cos0 + e

^{0}sin0,1,0, − sin0) = ((1)(1) + (1)(0),1,0, − 0) = (1,1,0,0).

^{3},√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(t
_{0}) = →x(t) + t_{0}→x′(t). - We compute the derivative of →x(t), so that →x′(t) = ( 3t
^{2},[1/(2√t )] ). We can now find our line tangent to the curve L(t_{0}). - Substituting yields L(t
_{0}) = ( t^{3},√t ) + t_{0}( 3t^{2},[1/(2√t )] ), at t = 1 we obtain our desired L(t_{0}).

_{0}) = ( 1

^{3},√1 ) + t

_{0}( 3(1)

^{2},[1/(2√1 )] ) = (1,1) + t

_{0}( 3,[1/2] ).

- 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(t
_{0}) = →y(t) + t_{0}→y′(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(t
_{0}). - Substituting yields L(t
_{0}) = ( [sint/2],t,[cost/2] ) + t_{0}( [cost/2],1,[( − sint)/2] ), at t = [p/2] we obtain our desired L(t_{0}).

_{0}) = ( [(sin[p/2])/2],[p/2],[(cos[p/2])/2] ) + t

_{0}( [(cos[p/2])/2],1,[( − sin[p/2])/2] ) = ( [1/2],[p/2],0 ) + t

_{0}( 0,1, − [1/2] ).

^{2}+ 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).

^{2}+ 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) ||.

^{2}+ 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.

- 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).

^{o}.

*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

### 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
- Differentiation of Vectors
- 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

- 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

### Multivariable Calculus

### 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

*We are actually going to start with an example.*0019

*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 sin ^{2}(t), so that is going to be sin^{2}(t) + cos^{2}(t) + 4, all under the radical.*0950

*Well sin ^{2} + cos^{2} = 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 curve*1158

*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 t ^{2}, or 2t^{2}, you can take 2 × derivative of t^{2} instead of taking the derivative of 2t^{2}. 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 t ^{2}, t^{3} , 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 cos ^{2}(t) + e(2t) sin^{2}(t), all under the radical.*2006

*This is going to equal e(2t), I am going to... e(2t) factors out cos ^{2} + sin^{2} = 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 sin ^{2}(t), this × itself is going to be -2sin(t)cos(t) + cos^{2}(t) + e(2t) × cos^{2}(t) + 2sin(t)cos(t) + sin^{2}(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 sin ^{2} + cos^{2} is 1, cos^{2} + sin^{2} 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) + cos ^{2}(t) + e(2t)sin(t)cos(t) + sin^{2}(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, cos ^{2} + sin^{2} 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

1 answer

Last reply by: Professor Hovasapian

Wed Dec 30, 2015 12:24 AM

Post by Shih-Kuan Chen on December 24, 2015

In example 2, don't you have to plug pi/4 as the coefficient for (xprime of t) in and get pi/4*(1/root2, -1/root2, 1)?

2 answers

Last reply by: Jamal Tischler

Mon Dec 21, 2015 7:05 AM

Post by Hen McGibbons on August 28, 2015

@11:10, why did you write t (followed by that vector)? Don't we have to use a different variable, such as c, since t was used in the vectors x(t) and x'(t)? or are these the exact same t?

2 answers

Last reply by: Denny Yang â™• [Moderator]

Tue Aug 19, 2014 4:45 PM

Post by Denny Yang on August 12, 2014

When you did example 4, how do you know e^t and e^t gives you e^t? I understand adding the two vectors but not so much about the two functions. I know they are the same. Can you please clarify?

1 answer

Last reply by: Professor Hovasapian

Tue Feb 4, 2014 1:18 AM

Post by Alex Emil on February 3, 2014

Thank you, great lecture!!!

2 answers

Last reply by: Professor Hovasapian

Tue Sep 24, 2013 1:05 AM

Post by yaqub ali on September 22, 2013

correct me if i'm wrong but is e^2t suppose to be on the top at all??

1 answer

Last reply by: Professor Hovasapian

Wed Mar 27, 2013 7:48 PM

Post by Jawad Hassan on March 27, 2013

thank you for this lessons, i think i will pass my class thanks to this, by far the best teacher online on tis topic!

1 answer

Last reply by: Professor Hovasapian

Fri Dec 21, 2012 5:48 PM

Post by SOUFIANE LAMOUNI on December 21, 2012

Great lecture .. straight to the point! thank you

1 answer

Last reply by: Professor Hovasapian

Sun Aug 5, 2012 4:13 AM

Post by Mohammed Alhumaidi on August 4, 2012

Isn't 1/sqrt(2) = Pi/6 = 30 degrees? ( Pi/4 = 45 degrees )

you almost kept writing that mistake in the whole lecture !!