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J |vnd +"_ '2-f-lvmt' 4e') - (% 4 + < t' > N + |vrkl...

Question

J |vnd +"_ '2-f-lvmt' 4e') - (% 4 + < t' > N + |vrkl

j |vnd +"_ '2-f-lvmt' 4e') - (% 4 + < t' > N + |vrkl



Answers

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

Alright problem five. Doing ours we go to t squared Come a T plus 1. 30 Q com A T minus 1 30 q. Find acceleration. A T equals zero first fund V, which is three rd t giving us to T come Ah one plus T squared Coma one minus T square now confined. Oh well, that's fine. Magnitude to be first each component square something, then take square root. So that's route to T Square plus one A F. T is derivative of the magnitude, which is to rooty great post one and a of zero bag material is go to to the normal. It's squared of a square, so that's two squared form on a zero squared equals two. So a of zero is you go to zero t plus two and

Say we are adding fractions with Unlike common denominators in this problem we're adding for 3/4, N plus four plus six over and squared -12. No, in order to add any fractions we need at least common denominators when dealing with polynomial such as these, we want to factor to find our least common denominator. So if we factor for N plus four we can take out of four, we're left with N plus one and now factoring and squared minus and minus two. Yeah, I left with N and it's gonna be -2 Plus one. Yeah. So are at least common denominator will be everything you see in the factors written once which is thus four and plus one And an and -2. So that will be really is common denominator. Now, in order to add, we need to re evaluate these fractions into their equivalent at least common denominator forms the fastest way to do that is to show them as they are factored in this case. The first one is factored like so And you can see now clearly what it is missing from the least common denominator, it is missing and -2. So we will multiply The top and the bottom by end -2, Simplifying that by distributing the three gives us three and -6 over are less common denominator of four and plus one. And my institute. No We will do the other side which is six over. The factor form of end -2 And plus one. We can now clearly see that it is missing a four. So we'll multiply the top and the bottom by four that gives us 24 over or at least common denominator. Okay, now that they both share a denominator we can add the numerator, Adding the numerator is here, it gives us three and and then 24 -6 is 18, So of course 18. And that's going to be over our common denominator. Mhm. So your answer to your problem is going to be this whole fraction right here.

M plus three Divided by N -2 Equals 4/5 mm. So we're solving for N, which is in the denominator. So I'm gonna cross multiply. So I have four and -2 equals five times M plus three. So when I distribute I get four, N -8 equals five. M plus 15, add my eight, So I have four. N equals five. M plus 23 divide by four, And we have n equals five M-plus 23 over four.

So here again we have that A vector is equal to the vector 235 B vector is equal to the vector negative four. I plus three. J. Um is equal to the vector negative for 30 And we're going to find c vector. Let's suppose that c vector is equal to Director X, Y. Z. And then we have that a vector plus B vector is equal to see vector. So um just equating our X components, we have a sub X plus B. Sub X is equal to see setbacks. That implies um Here that to minus four is equal to X. So this implies that X is equal to negative two. And then equating our Y components, we have a sub Y plus B. Sub Y is equal to see supply. So therefore we have three plus three is equal to Y. So therefore we have that Y is equal to six. And equating our Z components. We then have that five plus zero would be equal to Z. So um that implies that well um Z is equal to 2, 5. So therefore we have that our vector here see vector is equal to the vector X, Y. Z. So see vector is going to be equal to the vector negative, too comma sex comma five.


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