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Find equations for the tangent plane and the normal line at point Po (xo Yo Zo) (1,3,0) on the surface 8 cos (TX) 6x2y+5 e Xz + 5yz = 31. Using coefficient of 18 fo...

Question

Find equations for the tangent plane and the normal line at point Po (xo Yo Zo) (1,3,0) on the surface 8 cos (TX) 6x2y+5 e Xz + 5yz = 31. Using coefficient of 18 for X, the equation for the tangent plane is

Find equations for the tangent plane and the normal line at point Po (xo Yo Zo) (1,3,0) on the surface 8 cos (TX) 6x2y+5 e Xz + 5yz = 31. Using coefficient of 18 for X, the equation for the tangent plane is



Answers

Find equations for the
(a) tangent plane and
(b) normal line at the point $P_{0}$ on the given surface.
$$
x^{2}+y^{2}-z^{2}=18, \quad P_{0}(3,5,-4)
$$

In this problem. You're given these show information and asked to first for by the equation for the tangent plane. To do this, we're first going to rearrange this equation by moving eat of X y Z to the other side. So we're going to subtract it from both sides and get X plus. Why, plus Z minus e to the X Y Z is equal to zero. Now we're going to pretend that this is a function f and find the partials of F with respect to X. With respect of William, with respect, dizzy, we're only gonna do that the left side, because the derivative of a constant is always zero. And so we'll take the partial with respect to X of X y c, and that is equal to one Maya's E to the x y z times y Z in the person with respect to why is going to be equal to one minus e to the X y z times X c and then the partial with respect to Z going to be equal to one minus e to the x y z times X y. Now we're going to Valerie. All these at the point Given. And so when we do that, get the following one minus E to the zero to the time zero times one should just gonna be each of the zero times zero times one. So that whole term is zero. And so we just get one that's actually gonna happen in a person's for y and Z two because Z and why both equal to zero. And so we're just gonna get one for the person with respect to why and one for a partial. With respect, to see. And to get the equation of the tangent plane, we're going to take what we just found and write it in this little. In a way, we're going to write the partial with respect to X, which is just the one. So I'm not going to write that at all. And then we were both pile at times X minus the X value from up here, which is zero so x minus the room. Then add to that the personal with respect to why, which again, this one. So I'm not gonna write it times why, minus zero from up here. And add to that the partial with respect to see which again is once I won't write it. Clumsy purposes, A plus Z minus one. So the one is multiplying times two C minus one and all of that is equal to zero. And I'm going to simplify this. When I do, I get X plus y close Z minus one is equal to zero. And that is the equation of our attention plane. And now, to get the equation of the normal line, I'm going to do the following. I'm going to take X, subtract its value, appear so just zero. So I don't have to do anything. They're actually. So I guess I'll just write minus zero, just to show that is over the partial. So X minus here over one. And then it's going to be equal to why, minus its value up here, which is zero, although that is gonna be over the partial. And then it's equal to Z minus. Its value appeared over its partial, and so that would be X is equal toe. Why is equal to C minus one and now all of these air equal to T. And so I'm going to write them that way as well. And when I do that I get the following X is equal to T. Why is equal to T s? NC minus one is equal to t Z is equal. Go is going to be equal to t plus one and so you can write them either way. Usually you see them as this way. I chose to write them in this way in this case.

So we have the function X squared plus y squared plus Z squared equals three and that surface contains the point one one one. In the first part of the problem, we want to find the equation of the tangent plane. So to find the equation of the tangent plane, the first thing we have to do is find the Grady int of our function evaluated at the point that we have. So we have to do the partial derivative of our function with respect to each variable. So as we're doing the partial derivative with respect to X, anything that doesn't have an axe is treated as a constant so that partial derivative will be two x times the vector in the extraction I plus now the partial derivative with respect to lie. So that'll be, too. Why times the component J, which is in the direction of the Y axis and then the partial derivative with respect to Z, which will be a to z times its component in the direction of Z axis, which is K. And now we have to evaluate that had our opponent 111 So we plug those in for X Y and Z, and we end up with the Grady int being the vector to I plus to J plus two K. And that's the grading of s And the equation of the tangent plane will be that victor times are X, y and Z variables minus the coordinates of our point. So we would have to times a quantity X minus one plus two times the quantity y minus one plus two times a quality Z minus one equal zero. So there's the first part of the problem part A. We just have we have the equation for the tangent plane. Uh, we can also distribute these and combine our constant terms if we would like and have two X plus two Why plus a to Z and two times in minus 12 times the minus 12 times the minus one gives us a minus six equal zero or two x plus to I plus A to Z equal six. And then, if you wanted to divide out the common factor, we can simplify this to X plus why? Plus Izzy equals three. And there's the equation of the tinge of plane. Now the normal line is going to be using the same vector, giving us the direction of the line. And then we just have to multiply that by our parameter t based off of the starting point of 111 So for the line, we're going to have X equals are starting value of one plus two times t and then the Y coordinates of the mine will come from our starting. Why value one plus to t also because of the two and the why direction and then Z equals one plus to t using the

First the engine plane. Chris is unique. Technically. Why Haynes 90 X X plus. See time's into my square gate Clang. See in peace your 40 Palmer won re traded him first attendant complain that think that part A these partial derivatives the partial off we set up to that not personally video breeding the lettering green which is equal to eating extent. Why I close e to the X remains to take e to the wise where temps white Dempsey times j e to me Why squared? Plus on. Okay, so the app the greedy in F zero comma 01 coaches hero I plus j means cute. Okay, so then this one this part that tangential plane would be Why minus two tame See things one nickel zero Why q c equals minds to her party. Then way have trying to in the region it turned the whole the normal line for X He since 6000 it will be and the slope physical zero Here I'll be a zero While he's cooled is equal to T. C t. Called wise here which is one plus remains to cames t This is equal toe My two key

Wire times E to the X minus Z e to the wife square is equal dizzy And this initial point and we are asked to first and part a fines the equation of the tangent plane and then in part, b equation of the normal line. So first off, looking at this is this isn't in the normal form that we've seen the previous problems I have equal to variable. So first, I just wanna put it in the same form as everything else. I must attract the sea to the other side. I get why use X minus Z eats the Y squared minus Z Siegel zero and then out from here to find the equation of the change in plan. I want to get the grading it of us and evaluate at key. Not so when I do that when I take the radiant I am when take the partial derivative with respect to each variable and Adam together. So taking the partial derivative with respect to X of this newly formed equation, I see my only term that has an access his 1st 1 So when I take the druid of this, I just get my same thing back, since this is an exponential and then times my vector I next to take the derivative with respect to eye. So with respect, why so for its 1st 1 my ex is just a constant, my wise. Very well. So I just get you to the ex and then for this next one again, this is an exponential someone have Is the u to the y squared still there, following the rules of the exponents exponential. And then take the derivative of the exponents When I take the drift of why squared I get to Why, Yeah, that's it for my wise. So multiply this by a back to J. And I'm gonna take the derivative with respect. Izzie, this first term is going to go to zero the second term. I'm just going to get a minus eat y squared. And then for my last time, I get a minus one, then times a vector, Okay? And I'm gonna evaluate this at a given point 001 So putting in X equals zero y equals zero and C is equal to one for this first time. I'm just gonna get a zero for this next one eat zeros, just one. And for the rest of this since why's he was there? I get zero here, so I just get the plus one times J and then first next one eats of zeros again. Just once I get negative one minus one. So I get a minus two terms. Okay, so following equation one from your book, I get that my equation for the change in plane is going to be first time. It's zero because this coefficient, the next one's one times the doctor J which is just why mice? Why not? So why minus zero plus or minus two times the vector K, which is my C minus c not using mice one. So I called zero and then just putting this into a nicer form, I get that this is equivalent to why minus two Z is equal. It's a negative, too. And now, for the next part asks us to find the equation of the normal lines. So that's gonna be in the form of X equals like ALS. The equals are all these are the initial point, plus the partial derivative evaluated at our initial point. Cumpsty. So if the 1st 1 initial 10.0 When I evaluated the partial derivative respect a point, I also got zero. So x is just equal zero for why I have the initial point is zero plus. I evaluated the point at the partial derivative with respect to why I got a whole one. So that's just gonna equal t Remember Z following the same thing? I have one minus duty.


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