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(42=ErAcontinuous ad Lat lis typc nccmalntsFindie polnta (i Jm whichithe unctions discontinuily tne n wtitc noneRcmeablcmnaloOsCllatu gUumno(y = 2'Infote%-2tW...

Question

(42=ErAcontinuous ad Lat lis typc nccmalntsFindie polnta (i Jm whichithe unctions discontinuily tne n wtitc noneRcmeablcmnaloOsCllatu gUumno(y = 2'Infote%-2tW 11. y = I_ItG12. Y = tanx

(42= ErA continuous ad Lat lis typc ncc malnts Findie polnta (i Jm whichithe unctions discontinuily tne n wtitc none Rcmeablc mnalo OsCllatu g Uumno ( y = 2' Infote %-2tW 11. y = I_ItG 12. Y = tanx



Answers

Write an equation that is of the form $y=\tan b x, y=\cot b x, y=\sec b x,$ or $y=\csc b x$ and satisfies the given conditions. Tangent, period: $\frac{\pi}{3}$

All right. So the first thing that we need to do for this question is meeting to sell for the ingredient of F s. So we know that half some acts, the great development simply f sub x and f sub y. Right. So the grating of his F sub X is equal to y divided by X squared plus y squared on F sub y is equal to negative acts divided by X squared plus y squared. Now evaluating the directional derivative off at F off one camel one, we end up with these A view of F one comma one is equal to zero and this a view off f off the square. 23 Come on, one. Once you plug that in, this is equal to one over four times the square root of two multiplied by the square root of three, divided by four times the square of two, which is equal to outsiders, minus which is equal to one minus the square to three, divided by four times square

Okay, we're going to write an equation for tangent, who has a period of .5. So remember tangent and co tangent have a normal period of pie, as long as there's a B value in there, it changes it. So we need to figure out what that be value is. So period is pi divided by B. So B equals pi divided by period. So if we have a period of 0.5 we're taking pie and we're dividing by a half, which means we will really be multiplying by two. And so we get to pie So we can write our equation as y equals the tangent of two pi x.

So the first thing they want you to do in this problem is show that negative 11 is a point on this curve and we have co sign of pi Y. And you can show that this is a point by plugging in -1 for X. And one. And for why. And then co sign of High Times one. And I would just expect my students to know that negative one Cuban is negative one and one squared is one. So negative one times one is negative one. And think about where co sign is a pie or pie radiance is over here and co sign is the X coordinate. So yeah checks out. So what you would have to do next is finally derivative. So yeah, I'll circle this. This work right here in red is part A. But then you have to do the product rule which is take the directive. Leave alone plus leave alone take the derivative. You y dx there's your implicit and then on the right side the derivative of co sign. His negative sign. You leave the inside alone and then the derivative of the inside is pie Dy dx. That's how I would write. It is not the only correct answer. What I would do to simplify this. I'm not sure if I would even simplify this because on this side now you're doing sine of pi. Um so what you're gonna get there will be zero and zero times anything is just zero. So that's what you have on the right side. Um And as far as everything else goes, we're still plugging in negative one for X and one for Y. So as I go through a negative one when I square this positive one, one square is still one negative one to the third, powers negative one times two times one times dy dx is people to zero. So to solve this, I'm going to subtract the three over And then divide by -2 And then not just reduces the three has.

Okay, So this problem, once you calculate the directional derivative of the function our tended of y over X plus rat three times the arc sine of X y over tune. And to capitalise Europe, the directional derivative, we're gonna need it. Actually, the Grady in of the function and the Yoon effect er of the direction vector that we're measuring in and take the dot product between them. So, looking at our direction factor you This is not is obviously not a unit vector, because the magnitude is not equal to one. So if we were to calculate the magnitude of this factor, it would be the square root of its components square. It will be nine plus four, and this is just squared off 13. And using the magnitude we can calculate, we could find the unit vector. You call it be in vector, which is just the original vector you divided by its magnitude. So for this factory would be one over the square root of 13. I'm the factor three I had minus two j hat. All right, so now that we found our unit vector V, let's go ahead. And frankly, the Grady in off the vector at point, Pete. Now the country, the grading of the vector we're gonna have to take its partial derivatives with respect of X and with respect to y And these would be the components off the green vector. All right, so let's first take the partial to relive with respect to X. So this involves an arc stand in function and in arc sine function. So we got to use chain rule here. So we took a derivative all the Arc Tangent function. It will be why over X squared plus one in the in the denominator. And now we have taken the red off by over X. Because, look, general, this would give us negative. Why over X squared. So that's a first term about the second term. Second term would be a bus Mother Fraction with one minus one minus x y over two squared in the denominator and squared of three. Forgot that it's gonna be a constant Now we figured the root of X y over two, which is because of Cheryl. And that's just why over two. Okay, so that's that's the partial derivative with respect to X. Now will take the partial derivative with respect to why And this again we're dealing with in our tent and functions. So we will have y over X square, plus one in the denominator and and just one over X in the numerator. Since we're just taking the derivative of y or X with respect to y now, the second term with the arson fine again, we're gonna have the square root of one minus. Next. Why over two squared in the in the denominator and in the numerator, we're just gonna have a squared off three times Heard of three times X over two. So now that we have our partial derivative partial derivatives with respect to X and with respect to why let's plug in the point P, which is one comma one into these derivatives to find value over if I the components of the Grady in at that point. So if you were to plug in the values for actually would end up with 1/2 every plugged in the value into parts iterated, it was stepped up with respect to y sorry would get three house so are radiant at the point 11 is just the vector 1/2 I had plus three house J hats. All right, so now that we've had calculated a unit vector and the great in that point P, let's go ahead and take the dark hearted between these two factors. To find the directional derivative point peak, find a directional derivative. The direction of you Angel age Oh, at point P, which is one comma one would be the top wanted. Between these two vectors through the top prize, you have won over screwed of 13. That's a fraction that we would have. Then we would have 1/2 times three, which is three house plus three halves times I go to, which is minus three. And this would end up this would give us ah fraction of negative three over to Rod 13 and this is your final answer.


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