Hello. In the last session of curvature factors account We have seen ways. With a clear representation of the curvature function y is equal to that fx x in the function of As we can show we can show as a parametric 've seen. So we found two different ways. The curves in this plane but for in the plane I repeat again, the best curves we will be able to understand the concepts in space. It is possible to understand, but also directly in space E, I thoroughly to internalize this useful reminder. Now we'll see a third way and in space this third the road will be valid generalization. Now we have a tangent to the curve the truth When we draw it fx y is equal to the slope of the was given. y is the base for the tangent. We also know that since middle school. Derived slope's. Here we can obtain a vector. When we get a unit horizontal vertical the tangent As for when we go here u feature vector. But this is not one length of the vector. Thus from the vector e, units vector to go to this length, We need to partition. As you can see, E, root comes below the length. If we look at it as a geometric e, and y The following base year equal to the tangent of the base. BI in the denominator is the length. One plus tangent frame fi. Because the base year equal to the tangent of the plug. When we look at it this denominator Let us remember again. One plus frames for the tangent there. This is a split cosine square fi. So up, share a cosine fi income. For when we hit the cosine of the cosine fi fi here and sinus occurs. This geometrically we can find bi Şiyar. Because we move the x direction cosine fi time we will proceed in the y direction for sinus you saying this. As long as it still Pythagorean theorem sine cosine square frame. Where a is the length of a we see. These two vectors, it is also a unit vector u T We show through. This geometric and algebraic representations like this. These two vector, the vector perpendicular to the second There vector. We call this the orthogonal vectors. This is a vector perpendicular to T in the perpendicular vector. We define it like this:This is where As shown in this way the normal vector You can also choose direction. E, we can choose it in the opposite direction. Among them the adoption of this decision got to give. Both could not but indecision. This content is bowing in the direction of N were selected. Tion of describing it, the way k T get vector multiplication. This is a vector perpendicular to T as we know it will give. N as we want. We can do this kind of account. Since T is for sine cosine for this vector old when he calculated by multiplying the sine fu cosine fi income. Or when we look in terms of algebraic With this one year minus the base year income base Get multiplication of vectors. We are now able to provide bi. We claim that N together T'yl steep. Really be multiplied by the cosine of minus fi times sinus to the first two, the first component the product comes from. The second component of the product of the same size but with a plus sign, this time really zero giving. Again, this is similar algebraic representation to me, when we get N T times base year plus to minus y, y, again as a base gives zero. You can ask the question:Why ourselves Besides yoral at bi N I T Let? This has two meanings. How is it with unit vectors i and j point We tanımlıyabiliyo position wherein T and N acts are the same. And moreover, this curve better ta, They included the possibility to define. Because if you know the point T and N T curve E, E, N also what gives the tangential direction toward the side that gives curl. Because such a right tangent of this toward the side opposite to a bendable can be curled. N is also for us which side curls shows. This bi them from the application of a along the trajectory to calculate acceleration. Momentum in the direction of a tangent to the tangent at b that the centrifugal acceleration in the direction perpendicular to it we say. E, there are components. E, they are also in real life things encountered. For example, in amusement park death train called rails b out of the train when suddenly comes down. Meanwhile, our house would be a little bad. But that's going a little fun in the have no interest in them. This is what makes our inner bad, modeled either car in a curve Since the time to enter, itilme Us As the centrifugal force, gives impetus. They say that what works in practice things. Now another way the relationship between this tangent and perpendicular vector We can examine. Sine cosine fi fi t we said this. Let s based on this derivative. Look, when you get the derivative with respect to s N is happening. We can not directly. Because T as a function of s not given as a function of f given. Thus, indirectly, ie chained derivatives We take the derivative rules. After the first plug derivative of f at According to s derivative. As the first term but is kalıyor fi We know that split off from the DS. By definition, we know. Here, too, we find that close. Therefore derivative by T s We can find just taking off. Two encores, the second bi thing you know, we find. The following derivative with respect to t we get See here mA, minus sine cosine fi fi here is coming. E, is the unit vector that we, We know that the perpendicular vector. I once long one. A sine squared plus cosine squared is the. In addition to this, because they see that other T. both When multiplied by e, we find zero. So it's one, easily perpendicular vector we find. T is the fie derivatives. This geometric. With a more algebraic way it functions As we arrived there a way second. Let t multiplied with itself. This is due to the inner product of T's neck be square. T is a length. A frame. Thus T is T is a point. T and bury it, e, h, according to the derivative of T dT Taking the first one E derivative, dt ds split times T plus e first, the second of the first times derivative. But this may change as the inner product ds dt T will divide twice. One derivative of the derivative at the right side is zero. Therefore, the result is very interesting bi we find. Two of course does not matter. Here is simplified. DS with T dt divided inner product is zero. E, I've learned a lot already. If the inner product of two vectors is zero is the angle between ninety degrees. Because the length of the inner product means that it intermediate length cosine fi. Means that the cosine fi is zero. F is zero, the cosine of the two vectors be other shows. So here we divided dt T DS We see that it is perpendicular. But of these units is sized to E b, There's no guarantee. For this reason, there will be general, dt divided ds in the direction of N but in order to do unit divided by dt we divide the length of the DS. When we divide dt ds inasmuch as it splits The size of N is that it closes. Because here we saw that divide dt ds N, these dT I have seen for what it is divided by d. Once we find immediately to cover that. Here we see that the curvature to our unit We find the tangent vector. Also, the unit vector perpendicular to the tangent We find from the vector. That's us in space, space curves in will provide the opportunity to expand. Now here's an example I'm doing. This example perfectly clear in detail functionally y is equal to the circle when given as f x We give here in this circle. This circle also parametric representation We know. Locate the unit tangent vector of this circle here, the derivatives we want to take the curvature we can find. Open because it will pass quickly function, la fine detail We did and the curvature of the circle and a slash a and the radius of curvature of the circle We have found that the radius. This is consistent with our intuition as a result stated. DS already, so the curvature of the arc divided by df That length coming from derivative results of our own Find intuitively consistent in what we are is chosen to be. Parametric t of this circle a point on the center combining When we get the job right angle We found all of these formulas. The denominator of the second derivative of the curvature of the first with derivatives wherein it is obtained from the supplier We know. That means that all the work to be done first derivatives account to the second derivative to account for this from the x and y. They also put in place. Here, as the gene should be divided by the radius of curvature so that the radius of curvature the now is the radius of the circle we find. The third way and K, E, vector equation of circle as a function identify the unit tangent to tangent here find vectors. This unit tangent vector, the re- come to find that taking the derivative of curvature we see. Results are the same. Because the derivative of x where the x now we find. Derivatives of the components. This ds, dx was the length. So, divided by the length of the derivative dT we see that length. Here is just a involved. dt just as easy as it slashes DS 's. Here's the t ds dx divide We define, here's the t geometric as We find the results that we found and DT shut at times n reaffirm that, We verify the cover means and here simple DS dt divided by the length of the 're getting. E divided by dt dt found here. We find dt divided the DS. See also firmly than the length thereof wherein the length of the vector part In a split because of a curvature of a we find. This means very simple geometry with vector operations able to achieve without thinking. I'm here to give an example homework basis. We had found the curvature of a circle. Now here's an ellipse, and let ellipse different point would be different curvatures easily see. See here for a circle oturtsak on the long side of the ellipse this is a small splitter to the point, radius A circle will appear. But when we sat down to point b great will be a circle of radius. Them to account, the path is as follows: We know the equation of the ellipse. y squared divided by x squared plus b squared divided by the square represents a. Here to show the open function We can now. Parametric representation in this way. There is a difference from the circle as follows:in the circle a cosine t y de Gene koi, was a sine t. However, where a and b are different. We really provide it immediately. x divided by a, b sine cosine theta divided by y theta. That's when we take the square of the ellipse obtain more generally known functions We are. This parametric representation of the x vector with By defining these components tangent t by vector been achieved in three ways I'm waiting to compare. To check your account, this After guiding the answers promise. divide does that square off at point b, on the x axis. Therefore, the show rode it a'yl b of radius is divided by the square on the y axis at a closes divided by b squared is quite symmetric, not quite fully symmetric and b one. Although such a two though b See here for a while the radius of öbürkü two would be. Four would Öbürkü. That's according to this account, making We expect to find the results. Now, to reach the curve in space We're ready. Be it out of this issue After you give a little time to learn them After waiting at least to understand and also to do homework Looking forward, we will move on to the space curve.
