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(1 point) Find an equation of the tangent plane to the surface Z = 22 + y2 + xy at the point (_2,-2,12)....

Question

(1 point) Find an equation of the tangent plane to the surface Z = 22 + y2 + xy at the point (_2,-2,12).

(1 point) Find an equation of the tangent plane to the surface Z = 22 + y2 + xy at the point (_2,-2,12).



Answers

Find an equation of the tangent plane to the given surface at the specified point.
$z=\sqrt{x y}, \quad(1,1,1)$

Hi. It's clear. So wouldn't you, Right here here are gonna find the equation of the tangent plane. Here's the formula. Right here. No, for, um now, for Z, we're going to do C equals, uh, X comma. Why? It was just equal to our equation. Three x minus one square plus two. Why pull us three square plus seven and then P is going to be equal to too common negative to palma 12. So off of X is equal to six X minus one. And then this goes to, uh, X too common negative too equal to six for f of y x Comment. Why? Says equal to four times live plus three then if of why too common? Negative too is equal to four. Now we're gonna use substitution. Go to six X minus two plus y times. Why? Minus negative too. Lindsay is equal to six x plus for why? Plus eight. And they were gonna box or in sir

Were given a surface and a point on the surface and were asked to find an equation of the tangent plane to the surface. At this point, the surface has equations equals three times X minus one squared, plus two times why plus three squared plus seven. And the point is to negative 2 12 right away. We see the weekend rates C as a function of X and y se F, which is three times X minus one squared, plus two times why plus three squared plus seven. And therefore it follows that the partial derivative F with respect to X this is six times X minus one in the partial derivative f with respect to why is four times why plus three. Therefore the partial derivative of F with respect to X, evaluated at a point which has X Y coordinates to negative two. This is six the partial derivative of F with respect to y at too negative, too. This is four, therefore, by Equation two. Equation for the Tangent plane is Z minus. The Z value over point 12 equals partial derivative of F with respect to x Too negative, too. Times X minus the X value of her point, which is to plus the partial derivative of F with respect to y at too negative, too. Times y minus the Y value over point, which is negative. Two and substitution and simplifications give us Z minus 12 equals six times X minus two plus four times Why plus two and rearranging. We get that Z is equal to six x plus four y, and then the constant term is eight.

In discussion, we recall about the formula to compute the, you know the attention plan. It is a formula the equal to zero plus the first derivatives which the X times x minus x zero plus the first derivatives between the Y. In terms choir minus square zero. And in this question were given the functions E eco choose square root on the X squared plus y square. And the .345 here we can citizens equal to the FXY. And then the first time we need to compute those quantity here. So the first day referred to whisper to the X here it will blind the derivative which is equal to a fraction here and then we have the X give anyway square in the X squared plus y square. And now evaluate the quantity and the point we should get echo to executed three and then X squared plus y squared will be in the square under 25. So we'll be five here. Similarly in the first director was switching the Y by symmetry will have the wire on the top square on the X squared plus y square. Therefore, F Y evaluate and the point that we could get a far over five. Now we should be able to apply the formula to computer tension plan Z equals two. Z zero gravity five. And then F x will be the 305 times X minus three and plus five out of five times y minus far. And we simplify we will have Z ego to five plus 305 X minus nine out of five plus five out of five y minus 16 hour five. So we bring everything to one side. On the abortion, we will have minus it will be three, our five X. Blust four out of five y minus Z. And then it will echo to hear these two are sorry He's still here and we ended up we go to the -25. will be uh -5 -5 concerned I want to five so it doesn't equal to zero. And this will be the question we're looking for.

This question asks us to find the tension line given a plane and a point on the plane. So as a refresher, the equation for a tangent line is T. Is equal to f sub X at a comma B times x minus a plus F sub Y at a comma b times y minus B plus Z at a comma B. So to start, I am going to factor out rz equation. Ze equation is x plus two squared minus two Times. Why -1? All squared minus five. So to start, I'm going to factor this out. So from here I have, Z is equal two, X squared plus four, X plus four minus two times Y squared minus two, Y plus one minus five. And if I simplify this even further, I have that Z is equal to X squared plus four, X -2, y squared Plus four, Y -3. So from here I can solve for F sub X and F sub Y. So my f sub x is two, x plus four is two, X plus four, two X plus four. And my F sub y is negative four. Why plus four? If we then plug in our point? So we have ffx at a comma B. We get that ffx at a comma B is eight and our f of y at a comma B is negative eight. It's taking all of these values. We can plug these in to our equation for a tangent plane. So we get that T is equal to eight times X -2 minus eight Times Y -3 plus three. And that is for our X. So from here we can simplify, We get that T is equal to eight X -16 minus eight. Why plus 24 Plus three. And if we add up all the like terms you get That T is equal to eight X -8, Y plus 11.


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