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Givenimhakukhakkyare Rhketkauns erukhe HaakmekUdlcated raxe ofuehameekld Find Hwhenlmalu Idi @MetwenyhakanyM dtWRIRHHHMANHO e ecimnalplereskalskmeeded...

Question

Givenimhakukhakkyare Rhketkauns erukhe HaakmekUdlcated raxe ofuehameekld Find Hwhenlmalu Idi @MetwenyhakanyM dtWRIRHHHMANHO e ecimnalplereskalskmeeded

Givenimhakukhakkyare Rhketkauns erukhe HaakmekUdlcated raxe ofuehameek ld Find Hwhenlmalu Idi @ Metwenyhakany M dt WRIRHHHMANHO e ecimnalplereskalskmeeded



Answers

Find: a. $t(18,0.90)$ b. $t(9,0.99)$ c. $t(35,0.975)$ d. $t(14,0.98)$

So we're asked to look at different critical values using T. So the 1st 1 we're gonna do is look at T and then it's going to be 12 comma 0.1 So what you could do is look at typically, it's in the back of the book in appendix, or you can even find any of the resource is this is just typically going to be something that is given to you a table or chart teacher might be able to give to you. So what you need to do is just look at that table and match the critical value. So you're gonna look at 12 and point a one, and you should see that the value on your table should equal 2.68 Okay, so that should be the first critical value and then were asked also find T for 22 and then 0.25 So again, look at that chart. Look at that table. Find 22 1st and the 220.25 and you should get the solution or the answer 0.2 point 07 And then we're also gonna look on the same table for tea when it's 50 and 0.10 and you should get 1.30 for that critical value. And then finally, we're gonna look at T when it's eight and then 0.5 and you should see that it would be a critical value of 3.36 So again, when we're given that, we would look at a table or chart that is given to you and find the first number that is given to us and then the second number and align those to find the critical value.

Okay, we are given a vector and we're going to find the 1st and 2nd derivative. So unfortunately um for both our I. Component and RJ we do have to use the product rule, notice that E to the negative T. Times the trig function is a product. So for my first derivative I'm going to do a keep. Then the derivative of cosine is a negative sign. And then plus derivative. So when you take the derivative of E to the negative T, you do get an extra negative out front and then keep. Now that's all my I. Component. My J. Component. Again, I'm doing a keep derivative was derivative, brings in a negative and then a keep and that's in the J. And then from my K. I just have the one K. So I can write K. K. To clean this up a little bit. Notice my first two terms both have a negative E. To the negative T. And so I'll factor that out and I'll be left with a sign T. Plus co sign T. In my other case I'm just going to factor out a positive E. To the negative T. And I'll be left with a co sign T minus sign T. Now by doing that. Um I I again in the second derivative I will have to still use product rule. Um But not so many as if I went to a factor those ease out. Okay so now we're doing a keep and then the derivative will be cosign T minus sign T. And then plus the derivative. And so now that goes to a positive and then the keep. So that is all my I. Component. Before I go too far. Let's consider if we distribute that negative that's in front into the coastline T minus 70. I now have a negative cosine T. Plus sign T. Well if I factored out E. To the negative T. For that piece what I'd be left with is that negative co sign plus sign plus sign plus co sign. So the negative co signing the plus co sign actually cancel out. And I just have two of the sign teas. So now I have to do the J. Piece. So look at my J. I'm going to do keep derivative. So I get negative sign T. Minus co sign T. Plus derivative. So a negative comes in and then my keep. So again I can see how if I distribute that negative into the second piece. Then I have a negative sign and a positive sign that cancel out. And I just end up with the two of the co sign T. In the J. Direction. And then my K. Was just one. So when I took its derivative it's now zero. So I don't have to write it.

Hello. So here we have our F. T. Is equal to E. To the negative tee times I had minus the natural log of T square times jihad. And we want to find here while the derivative of R F. T times the second derivative of R F. T. That's going to be equal to the R F T times the triple derivative here. So the third derivative of r f T plus the first derivative. Our prime ft times are double prime. So we're going to define our prime are double prime and our triple prime. So the first derivative here, our prime of teeth. Well that is going to be equal to we get a negative E. To the negative T. So a negative E. To the negative tee times I had. And then we get um the derivative of the next log of T squared is going to be again by the chain rule is going to be a minus to over tea time's jay hat. Okay, there's the first derivative. Then we find the second derivative. So it's our double prime. Um That is going to be equal to one of the derivative of the first derivative. Um is going to give us an E. To the negative tee times I had. And then um a minus a negative two over t squared times J hat. Okay. And then the third derivative um is going to give us well. Again here we get a negative E. To the negative tee times I had. And then we get the derivative of two over T squared. Again by the train rule, that's gonna be a minus four over t cubed time's jay hat. And then putting the values of R. T R prime are double prime are triple prime into our formula. Here we get well let's see. We get this is going to be equal, chew A um a negative E to the negative two T. Plus four over T cube times the natural log of T squared. And then we take this and then we're adding um a negative E. To the negative to T. And then minus four over T cubed. So um that is going to give us A negative two times. Eat to the negative to tea and then a -4 over tea cube plus four over T. Times The Natural Log of T squared.

Okay, we're asked to find the derivative of the dot product. So I'm going to bring in my formula here. However, I am just going to do the simpler one um kind of the left side where I first take the dot product and then I take its derivative. So let's go ahead and write out our dot product. So we have an eat of the tee times T plus and eat the negative tee times a negative T squared. So now we'll be going to take the derivative. So in order to take the derivative notice, I am going to need product rule, I need to keep derivative plus derivative keep. And of course the derivative of E to the T. Is E to the T. Now I'm going to do that for the second part. I have to use product rule with having those two functions. So I'm going to do a keep derivative. So I get a negative to T. Plus derivative, which makes that negative times keep, which also then had a negative in it. So we're able to get rid of that. So to clean this up, I am going to be factoring out and eat of the tea and I'll be left with a one plus T. I'm also going to factor out an E. To the negative T. And you can see that I have that T squared minus two T. So I'm just writing my answer a little clearer and cleaned up.


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