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Write a in the form a =aT+ avN at (he given value of without finding and N. r(t) = (3t + 5)i + ( ~ ZU)j + (?)k, (-3...

Question

Write a in the form a =aT+ avN at (he given value of without finding and N. r(t) = (3t + 5)i + ( ~ ZU)j + (?)k, (-3

Write a in the form a =aT+ avN at (he given value of without finding and N. r(t) = (3t + 5)i + ( ~ ZU)j + (?)k, (-3



Answers

Write a in the form $\mathbf{a}=a_{\mathrm{T}} \mathbf{T}+a_{\mathrm{N}} \mathbf{N}$ at the given value of $t$ without finding $\mathbf{T}$ and $\mathbf{N} .$

$\mathbf{r}(t)=t^{2} \mathbf{i}+\left(t+(1 / 3) t^{3}\right) \mathbf{j}+\left(t-(1 / 3) t^{3}\right) \mathbf{k}, \quad t=0$

Okay, We're going to start with a victor. Value function of our of tea is equal to t squared. I plus T plus 1/3 T cube, J plus T minus 1 30 cube. Okay, um, and what we want to do is eventually right. The acceleration in the form of the tangential component times that unit tangent vector plus the normal component times that unit normal vector. Okay. And we know that that tangential component of acceleration is given by the derivative of the magnitude of the velocity. Okay. And so philosophy is given by the derivative of that, um, position, vector value function. And so this is going to be equal to two t I plus one plus T square J plus one minus t squared. Okay. And the magnitude is given by the square root of to t squared. So basically were in Take the each of the components so to t squared plus one plus t squared, squared, plus one minus t squared, squared. And so let's go ahead and distribute all that out. So and then combine like terms to simplify. So fourty square plus one plus two t squared plus t to the fourth plus one minus two t squared plus t to the fourth. And so this is going to simplify down to, um Let's see, we've got the two t squares. Add it to zero. Ah, we have the square root of two t to the fourth plus four t squared plus two We're gonna factor out a to and we have t to the fourth plus two t squared plus one, which you can recognize that that is fact herbal. And so we get, um, the square root of two times t squared plus one squared. So this is gonna be a t squared, plus one times the square root of two. And so there is my magnitude of my velocity, and so that tangential component is equal to the derivative of that function. And so when we take the derivative of that, we get, um, to time discredited, too. Times t. So now what we need to do is find that normal component. And so, no, now that normal component of acceleration is given by the square root of the magnitude of the acceleration squared minus that 10 gentle components squared. Okay. And so what do we know? We know that acceleration is given by the derivative of velocity. So we're gonna go back up, take the derivative of the velocity function, which is a to I class A to T J minus A to t. Okay, so now I need to find the magnitude of the acceleration. So the magnitude of acceleration and I'm gonna go ahead and keep it squared is equal to to square toe say, four plus a four t squared, plus a fourty squared. So this is gonna be a four plus in eight t squared, So my normal component is equal to the square root of four plus 80 squared plus. And now we're square. Um, we're gonna square the, um, tangential component. And so they square that tension till component. I get an a another eight t squared. And so this will be equal to, um, four plus 16 t squared. We're going take out a four. So that would be two times the square root of one plus four t squared. Okay, um and so now my whips hate when that does that. So now my acceleration is equal to that tangential component which ended up being to times the square root of two t times that unit Tangent Vector plus two times the square root of one plus fourty squared, which is my normal component times that unit normal vector. And now we want to evaluate that at zero. So we notice that that will go to zero, and, um so this will be zero times that unit Tangent Vector plus two times that unit Normal vector. And so there is my acceleration at zero.

This question asked us to write the given information in the form of a Eagles. What we know is the first thing we can do is we can differentiate and get Viv T. To get velocity, not finding the magnitude we know we have the square root of nine plus one plus nine, which simply gives us a square root of 19 which means now different shooting velocity. To get acceleration, we have zero I plus zero j plus zero k, which means magnitude. We know just right off the bat by looking at this is simply zero cause there's no numerical numbers here and then we know that and same thing is zero, which means our final A is simply zero T plus zero on.

Okay, We're going to start with a victor Value function are of T is equal to one plus three t I plus T minus two J minus three TK And this is actually going to be a position that vector value function. And we eventually want to find, um, the acceleration, um, function, um, which is equal to that. Accelerate the tangential component of the acceleration times that tangential or unit tangent Victor plus the normal component of the acceleration times that unit normal vector. Um, and we don't want defying, um, the normal, the tangent unit, Tangent Vector or normal unit Vector. We just want to write this so we want to find a safety and a seven. And so we know that the velocity function is given by the derivative. So first of all, um, to do that on DSO, that would be equal to three. I, um plus J minus three. Okay. And the magnitude of that velocity is given by the square root of three squared plus one squared, plus a negative three squared. So basically the to find the magnitude you just take, um, those coefficients or those x component Why component in Z components um, and square them and then take the square root. Okay, so this will be equal to this square root of 19. And the reason we did that is because we know the tangential component of the acceleration is equal to the derivative of that magnitude of the velocity, so that will be equal to zero. We also know that that normal component is equal to the square root of the magnitude of the acceleration squared. Ah, minus that tangential put component squared. And so we also know that, um, acceleration is the derivative of the velocity. So this is going to be equal to zero. Um, and so my normal component is zero a swell. So this is equal to zero T plus zero in, which is just zero.

Okay, So we most like component wise here not for and three plus two and zero plus native. One, five, twelve minus five seven.


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