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Qs uakate () fe Cos" '('Jsin (')dt(II) f(anx +cotx) & (IIl) i2 ~&r+ed_ xit2(iv) | sin Ocos 0 d0(v) fL Ke Vr c:...

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Qs uakate () fe Cos" '('Jsin (')dt(II) f(anx +cotx) & (IIl) i2 ~&r+ed_ xit2(iv) | sin Ocos 0 d0(v) fL Ke Vr c:

Qs uakate () fe Cos" '('Jsin (')dt (II) f(anx +cotx) & (IIl) i2 ~&r+ed_ xit2 (iv) | sin Ocos 0 d0 (v) fL Ke Vr c:



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Find $\frac{d}{d t}[\mathbf{u}(t) \cdot \mathbf{v}(t)]$ $$ \mathbf{u}(t)=\sin ^{2} t \mathbf{i}-\cos ^{2} t \mathbf{j} \quad \text { and } \quad \mathbf{v}(t)=\mathbf{i}-\mathbf{j} $$

Okay, What we're gonna do is step through a process. It's a kind of a long process. And so what we want to do is we're gonna start with a vector value function are of t is equal to co sign t I plus sign T j plus t. Okay. And so everything we do is based off of T equal to zero. And so the first thing is, we want to know what the vector value function is at zero. And so that is just going to be equal to I. So that's the first thing. The second thing is, we want to find the unit tangent vector. And so we know that the unit Tangent vector is given by V over the magnitude of the and V is velocity. Um and so that is equal to the first derivative of that vector value function over the magnitude of that. And so let's go over here and find that derivative. And so that is gonna be equal to negative sign of tea. I plus co sign of T J plus K. And so the magnitude of that function is equal to the square root of negative sign of T and we're going to square that plus a co sign t. And we're going to square that plus a one squared. And so this is just gonna be the square root of two. And so, um, my unit vector my unit tension Victor then becomes, um, negative. Negative sign of tea over the square root of two. I plus co sign of t over this creative too. J plus one over this quota to Okay, okay. And we want to evaluate this again at zero. And so when we do, we get one over this. Grow to two J plus one over the square root of two K. Yeah, which is square to 2/2 j plus this grow to 2/2 k So there is that unit Tangent Vector evaluated at zero. Now we want to find now the unit Normal vector. And we know that that unit normal vector is given by the derivative of that unit Tangent Vector over that magnitude of the unit Tangent Vector. So let's go over here and take the derivative of that unit Tangent Vector. And so that is going to give me a negative co sign t over the square root of to I, um plus a native sign of tea over this grow did to J and then plus zero. So the magnitude of that is going to be the square root of the I component square. So this is a negative co sign t over the screw to two, and we're square that plus that negative sign of tea over the square root of two. And we're going to square that. And so what we get is the square root of 1/2 which is one over the square root of two. Okay, so now we're gonna go ahead and substitute that all in to that, um, unit normal vector. And so we get, um, and have to scroll back up a little bit. So we get, um, negative co sign of tea over the square root of two over one. Over this. Grow to to I plus a negative sign of t over this. Grow to two over one over this. Grow to two J. And so that simplifies down to negative co sign of tea. I plus a negative sign of tea J. And so when I evaluate this at zero, we get negative. I So there is that unit Normal vector evaluated at zero. Now what we want to dio another thing is we wouldn't know what that by normal vector is. And so that by normal vector is the cross product of that unit Tangent Vector and that unit Normal vector. And we're actually going to do it evaluated at zero itself. So it's gonna be that Unit 10 Victor evaluated at zero Cross product with that unit Normal Vector evaluated at zero. So let's set up our matrix until we have I, j and K uh, the unit Tangent Vector had zero for I the square root of two over to for Jay and the square root of 2/2 for K. Um, the unit normal vector had negative one for I zero and zero for K and so that by normal vector is equal to, um, the I component is going to be, um, zero. I plus for Jay is going to be zero plus a square e two or two j plus. Then for the K component, we get zero plus a square root of 2/2. Okay, so this is simply going to be the square root of two over to Jay, plus the square ready to over tu que So we have that now the by normal vector evaluated it. Zero. Um And so now what we want to do is we want to find equations for the escalating plane Normal plane in the rectifying claim. So that's step this through for the Oscar relating plane we know that is perpendicular to perpendicular. Oh, that's how you spell perpendicular perpendicular to, um, the by normal vector. Okay, um, and of course, we want this all at zero. So the by normal vector had zero for the eggs. Screw to over two for why, and the square to two over to for Jay. And they were going toe multiply. Each of those two are, um, the vector value function. And so that was just an eye. And so that means X minus one for the X position. Why? For the y position and just Z for the Z position. And so we're in Do zero times x minus one plus this grow to 2/2 times at why, plus the square root of 2/2 times at sea, and that is equal to zero. So for the Oscar relating equation for the escalating plane we have. Why Plessy? Equal to zero now for the normal plane? No, For that normal plane, we know the normal plane is perpendicular or thought journal. Whichever word you want to say to that, um unit Tangent Vector. Okay, so the unit tangent Vector happened have the same zero for the I screwed to over two and describe it to over two. And we're going to multiply each of those or their respective with X minus one. Why? NZ. So we have zero times x minus one plus screwed to over two times. Why? Plus square with a 2/2 times See equal to zero. Which leaves me toe. Why plus z equal to zero for that normal claim as well. So now let's do the rectifying playing So the rectifying playing. So if I do the rectifying plane, the rectifying plane is perpendicular or or thaw Gunnell Toothy unit Normal vector. So it's gonna be perpendicular orthogonal to that unit. Normal vector and the unit normal vector had, um negative one zero and zero. And so this is going to be eggs minus one. Why N. C and So we have a negative times and X minus one plus zero. Why? Plus zero z equal to zero. And so we have negative X plus one equal to zero. So x equal toe one. And so there is the equation for the rectifying plane and we're done.

Okay, we are going to find the 1st and 2nd derivative of the vector given. So with our different components, we're going to take their derivatives separately. So we take the derivative of sine to T. We go to co sign to T. And then we multiply by the derivative. The insides were going to get an extra two in front for RJ component notice. There's a negative sign here. Well, co sign will go to a negative sign so that will become a plus. And then we also multiplied by the inside and get that extra two in front. Okay, so now our next derivative, just like last time we take the drove a coastline, we get a negative sign of two t, multiplied by that extra two And take the driver's sign. We will get the co sign of two T and then again get that extra two.

Okay. We are going to find the dot product. It's derivative and the cross products derivative of R. Two vectors that are shown. So, remember there's two ways we're going to focus on dot product first, there's two ways to do our dot product first. We could do the dot product, then take the derivative. Or we can use the right side of the formula. Where we're looking at taking dot products between derivatives and the original of the other one. So looking at these, if I do multiply these together, I will have to um go ahead and use um um product rule. So instead I'm gonna go ahead and I'm gonna take my derivative of each of my vectors separately. So remember as you're taking a derivative of like a co sign to T. You have to multiply by the derivative of the inside. So we have those extra twos in, are you prime? And then um for our vector V, it's derivative is a little bit simpler because there's only a tea inside and not a to T. Okay so to find our actual derivative of our full cross product will now take our um derivative of are you? And we'll do the cross product with the V. So now we're going to multiply my eyes. So I'll have a negative to sign to T. All multiplied by Kassian T. And then I'll add that to my to co sign T. Times the sine of T. Okay so that's just my first base. No, I have to add the dot product between my you and my my derivative. So here you see that I have a negative to negative co sign of two tee times sine T. And then I have a sign of two tee times. Co sign T. Well notice we have some like terms here um We have the negative two and then we have a plus one of the same thing. So we end up with just negative one of that sign to T. Co sign T. I wrote a second because I like to have my positive one first. Um and then we also had like terms to combine. We had 2 -1 of the co sign of two teas um times the sine of T. Okay so that is our final answer for the derivative of our dot product. So to do are cross product um Cross product. You notice I'm just changing my equation. The equation is the same acceptance that adopt product. We're doing cross product. So in order to do my cross product of U. V. I again I'm using my prime and doing the cross product with Vot. So notice I'm writing in my matrix. See here my components of you prime and then under it the original V. So this is going to allow me to write out those values so you can see that I have that to co sign of two T. Fortunately it's multiplied by one and then it's minus the one, multiplied by zero. So my I and J are pretty quick because I get to multiply kind of that second piece by zero and that's helpful. Now my K. Is going to be more complex because I will have to um have kind of the two different pieces there as I multiply across. So I have my negative to sign of two tee times sine of T. And then I'll subtract my to co sign of two tee times cosine T. And that is my K. Component. Okay, so that was just the cross product um of the first two pieces, I have a whole second cross product that I need to do. So here Um I'm now writing and remember it is U. Times V prime. So the, you has to go on top and the V prime goes on the bottom so you can see that for my eye, I will have sign of two. T. Is time zero and then minus cosine of T times one. So I just have that negative cosign A T. Then for my J. Again it's going to be a zero minus a negative scientist. So that ends up being a positive sign T in the J. And then again my K. Is more complex because I have both the co signer to tee times co sign and then I have minus the negative of sign of two tee times sine. Okay, so to clean this up, I don't have a lot of space here, but what I am going to do is I'm going to consider my my final I component really is that um to co sign of to t minus cosine T. My final J component is that to sign of to t minus sine of T. Now my K component. Again, I have some things in common so I can clean this up. What I'm going to do is if I have a negative two of something and then add one of it, I'm going to have a negative one of it, so I'm going to factor that negative out and then you can see that I'll be left with that sign of two tee times sine T plus the co sign of two tee times cosine T. And so here is my final piece um of my K component. So the red markings show our final answer.

So we are going to take the 1st and 2nd derivative of the vector given. Remember when they say that sign with the two T. That really means sign of T. Squared. I'm going to write it that way because it's a little bit easier to use my chain rule in that form. So for my eye component my two comes down. I keep my scientist and I went to the first power and then I multiply by the derivative of the inside. So for my next derivative um it is a co sign T squared. So the two will come down. I get my co sign T. Now to the first power. And then when I multiply by the derivative that's gonna negative so it changes the sign and then sign T. And that's in the J. So notice both my I and J. Component have like a two with a scientist times co sign T. Well that's actually are um equivalent to the sign of two T. And so I'm gonna rewrite it because the next derivative is gonna be a lot easier as signed to the to T. Instead of having to use product roll with sign and co sign. So any time you can use your identities to clean things up, go ahead and do that. So the derivative of sine of two T. Is going to be a co signer to T. And then we multiply by the driver of the inside and so you get that extra two in front and that's going to be both in the eye and the J. Direction.


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