In this video, I go over the True-False Quiz at the end of the chapter on vector functions from my calculus book. There are 14 questions, and I cover vector equations of lines, parabolas, different...
Transcript
the true, false quiz, and I'm going to read all the questions out, and then go one by one, just in case you want to solve the quiz yourself. All right, so question one, vector equation of a line, and this one states the curve, r of t equals to t cubed i plus two t cubed j vector, plus three t cubed k vector is a line, and then question two parabola, the curve, this right here, r of t, then it has the 0, t squared 4T is a parabola. Question 3, and again, is this true or false? Let me know how you do, and then I'll go over the solutions after I read all these. So, question three, line through the origin, the curve r of t is, yeah, which equals to 2T, 3 minus T, 0 is a line that passes through the origin. Question 4, differentiation by components. The derivative of a vector function is obtained by differentiating each component function. and then question 5, derivative of cross product. If you and V are differentiable vector functions, then the derivative of the cross product is just the derivatives of them, basically derivatives of each vector function, and then do the cross product of the derivatives. Question 6, derivative of the magnitude of a vector. So if R of T is a differentiable vector function, then the derivative of its magnitude is just equal to the magnitude of its derivative. Interesting. Question 7, curvature. So if T is the unit tangent vector of a smooth curve, then the curvature is Kappa equals to DT over DT, basically the derivative of the unit tangent vector in terms of the parameter T. Alright, question 8, binormal vector. So the binormal vector is b of t is equal to n, n of t cross t if this is the normal vector, and this is the tangent vector. All right, question 9, curvature and inflection point. Suppose f is twice continuously differentiable. At an inflection point of the curve y equals effects, the curvature is zero. Question 10, curvature of a straight line. If Kappa of t is equal to 0, the curvature, yeah, for all T, so the curvature is 0 for all T, the curve is a straight line. Question 11, constant magnitude of a vector. If R of T is equal to 1 for all T, then the magnitude of its derivative is a constant. Alright, question 12, space curve on a sphere, so if... If the magnitude of a vector is equal to 1 for all T, then derivative R, r prime of T is orthogonal or perpendicular to R of T for all T. Question 13, oscillating circle. The oscillating circle of a curve C at a point has the same tangent vector, normal vector, and curvature as C at that point. Alright, and now question 14, parameterizations of a curve Different parameterizations of the same curve result in identical tangent vectors at a given point on the curve. Interesting setup there. So again, you can pause video and then try to solve it yourself, and then I'll go over the solutions, and you can see how that goes. All right, so solution to question one, vector equation of a line. So the curve r of t equals to t cubed i vector plus 2t cubed. J vector plus 3T cubed k vector is aligned. And solution, well, this is true. If we re-parameterize the curve by replacing U equals to T-cubed, we have, you know, we get, so now we're going to replace, well, T with U, so this is going to be R of U is equal to, so forever, for all the T-cubes, you just replace it with U, so that's going to be, well, U and then I vector, plus, well, now we're going to have a 2U, and then J-vector, and then we're going to have a plus 3UK vector, like that. All right, and now this vector, this is just a line through the origin with direction vector is going to be, well, direction vector is just going to be, well, this is going to be, well, 1, 2, and then 3, like that. And, yeah, for completeness, I'm going to move this over and write direction vector. Basically, when you just set that equals to 1. R of 1 equals to this, and that's just equals to I, yeah, just in standard basis vector, set up I plus 2J, and then plus 3K, and again, through the origin, it goes through the origin because, well, yeah, because let's go R of 0, this just equals to, well, everything goes to zero, so that's just going to be the zero vector. Yeah, that's just the origin. All right, and now let's take a look at the solution to question 2, and it's on a parabola. So the curve right here, R of T, this should not be bold, which equals to 0, t squared, and then 4T is a parabola. And solution, well, this is actually true. So parametric equations of the curve are, so for the x component, that's just x equals to zero. For the Y, that is going to be y equals to t squared, and then the Z component, that's just going to be the other component for, yeah, 4T. So if we rearrange this, so this is going to be, well, let's put an arrow like this, this is going to be t is equal to, move it over to the other side, Z over 4, and then move this, plug this inside, what we end up getting as y is equal to Yeah, so we get Y's equal to, yeah, just get rid of the T, it's going to be Z over 4 squared, like that, which equals to Z squared over 16. Yeah, so thus what we have is, thus, we have, we have X equals 0, and then the other equation, just Y equals 2, Z over squared, yes, Z squared over 16. This is just a problem. This is a parabola on this, so basically it's going to be a vertical parabola. Actually, it's going to be, yeah, that'll be the Y component. And, yeah, this is actually going to be a vertical parabola. So if I draw the X, Y, components like this, this is going to be Y, this is Z, and then this is going to be our X. So X is going to be on the Y, Z plane, and then, so Y, is the... component there, so Z is going to increase. No, actually, no, Y is going to be the one increasing. So actually, it's a horizontal parabola. It's going to be like this. So Y is going to be increasing, basically through this parabola like setup there. So that's going to be this Y there. This is going to be a problem. And this is going to be, again, along this whole setup there. This is going to be a Y, yeah, this is the order. Yeah, a parabola there. Yes, that was pretty cool stuff. All right, yeah, so this is an equation of a problem in the YZ plane. So again, this whole plane like that. So, yes, fascinating setup there. All right, and now let's take a look at the solution to question three, and this is the line through the origin. So the curve R of T equals to 2T3 minus T zero is a line that passes through the origin. A solution, this is actually false, and again, you can see that because, well, This one is a 0, but then this setup, these are not similar, that probably is going to be false. So anyways, the vector function RFT represents the line, but the line does not pass through the origin, and you could see the parametric equations. If you go x equals to 2T, and then we have, yeah, so at x equals 2T, what we'll do is at, at x equals 0, what we get is, well, T is 0. All right, so that's fine, and dandy. Alright, yeah, so now the Y component, that's just going to be 3 minus T. 3, I'll fix this up, 3 minus T, and now if we plug in T equals 0, because that's when X is equal to 0, we get, well, Y is equal to 3 minus 0, in other words, we get a y equals to 3. Like that, I'm going to circle this, and then so we have X equals 0, Y equals 0, and then Z is always equal to 0. So we can box this out. So we get the point. So this means we have point, yeah, we have a point, this is going to be, actually it's a point out of vector, so just put 0, now Y is 3, and then we have 0. This is not, this is a point here is not the origin. It's not 0, so it's false. So this is not through the origin. Alright, now let's move on to the next solution to Question 4, and this is differentiation by components. So the derivative of vector function is obtained by differentiating each component function. And this is actually true, and you can see my earlier video again, there's a link to the video notes, and then scroll down, you'll see a bunch of thumbnails. on the individual sections, videos, and plans, and so on. All right, so see my early video on the derivative of vector functions by components. So derivative of vector functions by components, you have, if the other function, RFT, H of T, then basically the derivative is just derivative of each component F prime of T, G prime of T, T, and H prime of T. And again, here's the proof using the definition. of derivative for vector functions and so on. So make sure to watch that to learn more. Alright, so let's take a look at the solution to question 5, and this is the derivative of a cross product of vector functions. So if the vector U and F are differentiable vector functions, I mean, U and V, not F, then we have D, D, D, D, T of the cross product of these vectors. equals to the cross product of their derivatives. And solution is actually false, so you can see my earlier video, on the derivative of a cross product of vector functions, and here's what you actually get is a product rule type set up. So the derivative of u cross v is actually u prime, the derivative, cross with v plus u cross with v prime. So yeah, that's basically a cross part of and there's the derivation. Basically you split it up into components and do that and yeah, you get this giant thing. All right, so that's that. And now, now let's take a look at the solution to question 6, and this is derivative of the magnitude of a vector. So if r of t is a differentiable vector function, then d over d t of r, yeah, this is the magnitude of the r vector is equal to, well, the magnitude of the derivative. And this is actually false. And for example, we can do a simple example, just to illustrate this is not the case. So if we let, let R of T equals to the vector, just a 2D vector, cosine of T, and then sine of T for the Y component. So now if we take the magnitude of this, so the magnitude of this one, using the distance in 3D equation. This just equals to, well, square root, and then we have to basically square each component and add them up, so cosine square t plus sine squared t. Now this equals to, again, this is trigonomic identity, this just equals to 1. And that's, I mean, that's the Pythagorean identity, and that equals to 1 there, like that. Like this, and then squared of all that just equals to 1. Alright, and then, and, yeah, and now, and, so we have the magnitude. Now, if we take the derivative of its magnitude, right here, D over DT of the magnitude, this is going to be R of T like that. This just equals to, well, this is a constant, this should be D1 over DT, I'll just put it like this, how should I do it better, D over DT, and then 1. This just equals to 0. So it's just a constant. So that's just equals to 0. All right. So we have that, but if we take the derivative first, what we get, and then take the magnitude, but the magnitude of the derivative, this equals to, well, we take the magnitude of the derivative, is going to be derivative cosine, due by components, is going to be negative sine t. and then sine t, derivative is cosine T, like that, and then the magnitude of this is just equal to, well, same thing as before, so we would do square root, and then we do minus sine squared, uh, t, yes, this is going to be, and then, yeah, minus sine squared and squared and square that, plus cosine, um, yeah, plus cosine, uh, square t. and I'll put it better, okay, cosine, square t, and now this is, again, this is just going to be positive, so this equals to 1, and then squared all that equals to 1. But this is not equal to 0, so it's false, then. Yeah, so it's false. This is, so thus false. Yeah, so the statement is false there. Yeah, so that statement is false there, yeah, this is not equal to, that's not equal to zero. And now quickly, just recall the equation for the length of a vector in 2D and 3D for my earlier video, and again, that's where I was using these. formulas right there. So from my earlier video, if you have a distance from this point to that point, basically you're going to have each component subtracted by each other, then squared and sum it all up. If you go from the origin, you just, all you do is just the point, the x2 there, or in this case, just these values there, because you're just starting from the origin, and then you just square them up. So the distance in 2D and 3D, and 2D, this just all goes to zero, the exact component. Alright, so that's distance from in 3D. And also recall the Pythagorean identity from my earlier video. It's a screenshot from, I found online. So basically, A squared plus B squared equals C squared. And this stems from, if you have a unit circle of radius 1, and then basically, this is going to be our A squared plus B squared equals to C squared, where C is equal to 1. And then the A is going to be cosine, this adjacent over our hypotenus. yeah, so that's going to be the x equals to cosine over 1, and then this is when the sign is y over 1 and so. Let's watch the derivation for that, and again, I use that over here. So, sine squared plus co squared equals to 1, like that. So I guess, fascinating set up there. All right, so now let's take a look at the solution to question 7, and this is on curvature. So if T, capital T of T, T, is a unit tangent vector of a smooth curve, then the curvature is Kappa equals 2, the magnitude of the DT, capital T over DT, basically the derivative of the unit tangent vector divided in terms of the parameter T. And solution, this is false. The curvature is the magnitude of the rate of change of the unit vector with respect to arc length, not with respect to T as described my earlier video. And I believe this was so that it doesn't matter what, how you reprameterize the function, it's always going to be the same thing. Something like that, so yeah, make sure watch my earlier video. So again, there is the actual definition. capital equals DT over Ds, where this is the arc length. The actual arc length, you calculate from the derivative of, yeah, the derivative of the unit tangent vector. No, I mean, not the unit tangent vector from the actual derivative, just from the tangent vector, as we went over earlier in the concept review. Yeah, so make sure to watch out. So basically it's in terms of arc length, not that. So that's basically just false. Yeah, so it's not DT, it's DS. Like that. And if you were to convert it into parameter T, yeah, there it is, you're going to get this, this is going to be the derivative, the unit tangent vector derivative in terms of T. That's this right here, this is, this is T prime of T, but then you're going to have to divide this by the magnitude of the tangent vector in terms of T. So that's the other one, so that's false. Yeah, so that is false. It's in terms of arc length, and if you convert it back into the parameter T, you get this, and not just this. See, it's divided by that setup there, so fast setting is set up there. Alright, so now let's take a look at the solution to question 8, and this is on the binormal vector. So, this is stating the binormal vector is b of t equals to n of t cross the unit tangent vector. So this is a normal vector cross with the unit tangent vector. Solution, this is actually false. So as shown in my earlier video, the bnormal vector is equal to t, the unit tangent vector cross. the normal uh yeah uh yeah the normal unit normal vector so n of t and it's it's the order matters so this is this order matters this one is on the wrong side this is this should be switched around all right so that's that's that's the wrong one series of my earlier video on that and here is the unit tangent uh vector is this and then you have the unit normal and then the unit bin normal is t of uh yeah t cross with n of t like that so it's basically unit tangent cross the unit normal and so recall from my earlier video of my also recall from my earlier video that switching the order of the cross product obtains a negative value so if you have a cross b this equals to negative b cross a so again the order order matters so this one's reversed it's actually going to be negative so thus so thus What we get is, so if we have B of T, this equals to, well, this is going to be the unit tangent vector of T, cross with the unit normal vector of T, and then this equals to, if you were to reverse the order, so this is the unit tangent, I mean, this is the unit binormal, if you reverse the order, you have to put a negative sign. So this equals two, negative, and then this is going to be N. of T cross with capital T of T like that. Yeah, so that's order matter. All right, now let's take a look at the solution to question nine, and this one looks at curvature and inflection points. So suppose F is twice continuously differentiable, in other words, you can take a derivative twice. At an inflection point of the curve y equals fx, the curvature is zero. Alright, so let's take a look at the solution, and this is actually true. So at an inflection point where F is twice continuously differentiable, we must have that the second derivative, F double prime, equals to zero. Thus, the curvature is zero there. I'll put it like this, yeah, is zero by there by my earlier video. Here's earlier video on curvature. So if you have curvature like this, right here, and there it is for the plain curve y equals f of x there, and this is a curve of y equals fx. So recall the equation, we have, well, F, uh, your curvature as a function of x equals to f double prime of x divided by one plus f prime of x squared and then all square, and all to the power of three over two. Like that, well, if this is, uh, if this is zero, the curvature is zero, then this equals zero. Yeah, if that goes zero, that goes zero. Yeah, so that's pretty straightforward. So, and now also recall from my earlier video way back, years ago I did this one, on inflection points and concavity, and I didn't have, didn't make a good... thumbnail, so I'm just took screenshots from the calculus book right here. So definition at point P on a curve y equals of X is called an inflection point. If F is continuous, there, and the curve changes from concave upwards. So basically it's concaving like this, up, and then goes back to down, or the other way around, to concave downwards, or from concave downward to concave upward. All right, at P. And also note a continuous, spell this properly, a continuous function. that changes concavity means that at the inflection point, the second derivative, aka acceleration, is zero. So basically if you have a function like this, it's concaving up, and now of a sudden it's slowing down, the acceleration's going backwards. Yeah, now it's accelerating the other way. So it's concaving downwards there. You could kind of think of it as it's kind of like putting the brakes on here, slowing down, but now it's accelerating downwards, and then here it starts to accelerate upwards, so the acceleration has to change, there's concaving, like that, and concaving, and concavity test, so if F double prime of x is gradient zero for all x in an interval, so in case this one here, it's concave upwards, and the graph is concave up. If it's negative for all in an interval, for example, here to here, this is going to be concave downwards, this is negative. at L prime is less than zero there. That's all down, and again. And these are inflection points. These spots where the concavity is changing, basically where the acceleration changes from positive negative or negative to positive. Yes, epic stuff. Alright, now let's take a look at the solution to question 10, and this is a curvature of a straight line. So, if the curvature kFT is equal 0 for all T, the curve is a straight line. So solution, this is actually true. And we could use the earlier equation for curvature in terms of the parameter T. So the curvature, k of t, this equals to, well, this equals to the derivative of the unit tangent vector. the magnitude of it, divide this by, right here, the tangent vector. Yeah, there are…