Consider the paraboloid z = X? + 2y2 and the plane z=X+y+27, which intersects the paraboloid in curve C at (4,3,34) as shown in the figure to the right. Find the eq...

Line tangent to an intersection curve Consider the paraboloid $z=x^{2}+3 y^{2}$ and the plane $z=x+y+4,$ which intersects the paraboloid in a curve $C$ at (2,1,7) (see figure). Find the equation of the line tangent to $C$ at the point $(2,1,7) .$ Proceed as follows. a. Find a vector normal to the plane at (2,1,7) b. Find a vector normal to the plane tangent to the paraboloid at (2,1,7) c. Argue that the line tangent to $C$ at (2,1,7) is orthogonal to both normal vectors found in parts (a) and (b). Use this fact to find a direction vector for the tangent line.

We are given a surface s in our three defined by the equation X y squared plus two y z equals 16 in part A We're asked to find the normal back here to the surface for normal vector n We know from this section the formula for a normal vector to a surface to find implicitly by an equation in variables X, y and Z. Well, this is going to be equal to if we call f of x y z x y squared minus sorry plus two y Z minus 16 so that our surfaces defined by f of x y Z equals zero Then our normal vector is given by partial of F and the X direction I plus the partial derivative of F in the Y Direction Joe plus the partial derivative of F in the Z direction. Okay. And we find that the partial derivative of F with respect to X from our equation, this is why squared partially as an equity respect to why is two x y plus two z and the partial derivative of F with respect to Z is to y yeah, so we have that are normal vector and is equal to why Squared I plus two XY plus two z jay plus two y Okay, Now, in part B, pressed to find a tangent Plane H to the surface s at the point equals 123 Now, in order to find the tangent plane, we already know the normal to the surface at the point p and of P, this is end of 123 which from part A. You know, this is two squared or four I plus two times one times two or four plus two times three or six or four plus six or 10 jay plus two times two or four. Okay, so we have our normal vector At T. We know that this normal vector is also normal to our surface s at P. So an equation of our tangent plane h must be of the form two X plus five y plus two z. So I simply divided each of these coefficients by two. So divided by the greatest common divisor equals some constant C find See will substitute key into this equation. So we have it. C is equal to two times one or two plus five times two or 10 plus two times three or six, which is 18. Therefore follows that the Tangent plane H two s F p is defined implicitly by the equation two X plus five y plus two z equals 18.

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