Question

Find a tangent plane to a point ex: to z=x²+y² at (1,2)

a) The tangent plane does not exist at the given point
b) The tangent plane equation is ( z = 5x + 3y - 1 )
c) The tangent plane equation is ( z = x² + y² + 5 )
d) The tangent plane equation is ( z = 5x² + 3y² - 13 )

Answer 1

The tangent plane to the surface z = x² + y² at the point (1, 2) is z = 2x + 4y - 3. This is derived by calculating the partial derivatives at the point, which are 2 and 4 respectively, and using them to find the equation of the tangent plane.

Step-by-step explanation:

To find the tangent plane to the surface z = x² + y² at the point (1, 2), we first need to calculate the partial derivatives of the surface equation with respect to x and y. These partial derivatives will give us the slope of the tangent plane in the x and y directions, respectively.

The partial derivative with respect to x is 2x, and the partial derivative with respect to y is 2y. At the point (1, 2), these partial derivatives evaluate to 2(1) = 2 and 2(2) = 4 respectively.

The equation of the tangent plane can be constructed as: z = z0 + fx(x0, y0)(x - x0) + fy(x0, y0)(y - y0), where fx and fy are the partial derivatives evaluated at (x0, y0). Applying these to our point (1, 2), where z0 = 1² + 2² = 5, gives us: z = 5 + 2(x - 1) + 4(y - 2), which simplifies to z = 2x + 4y - 3.