21. Find an equation of the tangent plane to the surface x^2...

21. Find an equation of the tangent plane to the surface x^2 + 2y^2 + 3z^2 = 6 at the point (1,1,-1). A. -x + 2y + 3z = 2 B. 2x + 4y - 6z = 6 C. x - 2y + 3z = -4 D. 2x + 4y - 6z = 0 E. x + 2y - 3z = 6 22. Find an equation of the plane tangent to the graph of f(x,y) = ? + sin(?x^2 + 2y) when (x,y) = (2,?). A. 4?x + 2y - z = 9? B. 4x + 2?y - z = 10? C. 4?x + 2?y + z = 10? D. 4x + 2?y - z = 9? E. 4?x + 2y + z = 9? 23. The differential df of the function f(x,y,z) = x e^{y^2 - z^2} is A. df = x e^{y^2 - z^2} dx + 2xye^{y^2 - z^2} dy + 2xze^{y^2 - z^2} dz B. df = e^{y^2 - z^2} dx + 2xye^{y^2 - z^2} dy + 2xze^{y^2 - z^2} dz C. df = e^{y^2 - z^2} dx - 2xye^{y^2 - z^2} dy - 2xze^{y^2 - z^2} dz D. df = e^{y^2 - z^2} dx + 2xye^{y^2 - z^2} dy - 2xze^{y^2 - z^2} dz E. df = e^{y^2 - z^2} (1 + 2xy - 2xz)

Transcript

00:01 Let us find the equation of tangent plane of this surface at the point 1 comma minus 1.

00:11 So the equation of the triangle of degree is given by z minus minus 1 is equal to f x at this point p into x minus 1 plus f y into y minus 1.

00:27 Fx and fy are partial derivatives with respect to x and fy's partial derivative with respect to y.

00:37 All right.

00:39 So now how to find that? so let's quickly find that.

00:44 Let's partially differentiate this particular equation with respect to x.

00:48 So i'll be getting 2x plus 6 z into dz by, do z by though x, which is my fx is equal to 0.

01:00 Now plug in x is 1 and z is negative 1.

01:03 X is 1, z is negative 1, so 6, 2 minus 6 times fx is 0, so fx is 1 of 3.

01:12 Now similarly let's partially differentiate with respect to y, you get 4 y plus 6 z, though z by the y which is f y, 0.

01:21 Now plug in y is 1 and z is minus 1, you get 4 minus 6 f5 0, f y is 0, f y is 2 by 3 so simply plug in fx 1 by 3 and f5 2 by 3 in this so we get the z plus 1 is equal to 1 third of x minus 1 plus 2 third of you know like y minus 1 the lcm is 3 cross multiply you'll get 3 z plus 3 is equal to x minus 1 plus 2 y minus 2 so finally you get x plus 2 y minus 3 z is equal to so this is the equation of the tangent plane to the surface given, all right? now we need to find the equation to tangent plane of this surface at the point given in the question as 2 comma pi alright, so this is my z now first letter is valued the z value at 2 comma pi so z is equal to pi plus sine of pi 2 square and 4 pi plus 2 times of pi sign 6 pi that sign 6 by 0 plus pi is 0 plus pi is pi so the equation of the tangent plane looks typically as z minus pi is equal to fx at the point if you do x minus 2 plus f y into y minus pi.

02:44 Now let's quickly find out the partial derivatives of f x and f y.

02:50 So we have f x is equal to the sign of pi x squared plus 2 y...

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