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Deter mine arc lenqth interva L 31 Le;2)e (4t,- Cos(t), Sialtl)...

Question

Deter mine arc lenqth interva L 31 Le;2)e (4t,- Cos(t), Sialtl)

Deter mine arc lenqth interva L 31 Le; 2)e (4t,- Cos(t), Sialtl)



Answers

$$ \frac{2 t-30}{t^{2}+6 t-27}-\frac{2}{3-t} $$

We have our number 64 which is T minus two equal to one and willing to solve this. Since here too is subjected from T. So we need to add do at two in both sides. So we will be getting D minus duel plus two equal toe one place to the street. Cancel out de equal Do three. Something should be done.

Okay, so we need to use the chain rule to find the derivative of this function. And the whole premise of the change rule is to look for an inter function and an outer function. The one negative one hand power. All right, Eso What I'm talking about is the inner function is to look for the parentheses. Okay, So that that's my inner function. And then my outer function is going to be all the way to the negative one. Have function power eso When I'm tasked with finding the derivative d y DX, what I need to do is a derivative of the outer function first. So what I'm talking about is bringing this negative one half in front. And then when you subtract one from your exponents, you know that's the same thing as named one half minus two halves. Eso again negative three halves. That's your new exponents. And the change rule then says OK, leave the parentheses. Leave the inter function alone. But then you multiplied by the derivative of the inter function. So that's where we bring the 60 in front. Sorry, the two in front two temporary is six. Don't change the very well, like I just did. And then plus two. Now, as faras simplifying this answer, some teachers actually let you leave your answer like this. Oops. So I can circle this. Other teachers might ask you to go ahead and simplify. What does that? What does that mean? Well, you can actually distribute this negative one half into the numerator or into the red part. Eso maybe rewrite that as negative three t minus one. And then the negative exponents would stick that into the denominator to the three halfs power. And we're left with three t squared plus two T or other equivalent forms. This is not the only correct answers, is the one I came up with. So there you go. There's two options for you.

Hello, guys. So now we have this cost function here on. We need the first thing that we need to do for the letter. A Well, it's just blood dysfunction. So the letter A is already done because he is well, how dysfunction looks like in the intervals. 06 for the part V. We need to see if this limit here. Well, this limit here exist. So to doing this, we're going to use getting some help from a table. And we're going to see the behavior of dysfunction as we approach to the number 3.5 here on, uh, we're we're also getting some help from the graph on this part here. So let's see, How is the behavior of dysfunction? Well, at three. Discuss function have a value 3.7 on If we check here on this on this plot, three is in this point. So from here to here, we're going toe have the same value. And actually, this point here is 3.5. So all the values in the stable are going to have the same value. 2.72 point 7.7. Sorry. On the from here from this 0.3 point six. We're changing the behavior. We're going to be one step above in this function. So we're going to be in this part to here. So let's see, at this point, we're having values off 8.78 point seven and finally Well, a 0.7. So what is happening here? If we approach from the right off the function to 3.5, we're getting some, uh, limit to 8.7. But from the left, we're going a totally different values, which is, well, these seven points. So these two values here are not the same. So this implies that the limit not exist. So finally, what we can say about this limit? Well, this limit not exist. Finally, we need to check what happened with the limit off the cost function as we stand de 10 to 3. So we're going to do this same thing. We're going to use some table on getting some support from the graphic, so let us start. So where is to here in in this plot where two is in this this part here and we're going to be until 2.5 here, the same values from here to here on the value on that on that interval, for the cost function is going to be 6.7 at 2.5 is the last point where we're going to have this value in the cost function. That this 6.7 on from 2.9, which is really close to three there now in this site off the function. So here we have the behaviors before 7.7 at three. We already know that this value 7.7 on at 3.1, it's 7.7 on the last value where discussed function is 7.7 is 3.5. So here we have 7.7 and finally, at four. We're having this value that we check before that This a 0.7. So now we have a different behavior. A zoo we can see from the table and also from the graphics were approaching 23 from the left side. Onda from the right side here, we're getting to the same value. So in this case, if we're moving 23 from the right on from the left, we're getting the same value here. These two values coincide. So the limit off the Scots function Ste Ste MHM 10 to 3 well is just 7.7 on that.

This problem we're gonna be using lap has transformed tables to evaluate the lack class. Transfermarkt t sweat minus three. T minus to you to the negative t sign three t. This is easily done by linearity. It means we only need to calculate the lack. Last transport squared, the last last transom of tea and the last last transform of each of the negative D sign three teams. Now, the last ransom of T squared is quite easy. It's equal to two over s cubes. House transom of T is equal toe one of the square on the blacklist transform of each of the negative. T sign three t is gonna be equal. Tow three over s plus one squared plus nine and that's it.


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