Question

Find an equation of the tangent plane to the surface at the given point. f(x, y) = x2 − 2xy + y2, (1, 5, 16) with maple lab please

Answer 1

Solution:

The equation of the tangent plane to the surface at the point $(x_o,y_o,z_o)$ on the surface, is:

$$z-z_o=f_x(x_o,y_o)(x-x_o)+f_y(x_o,y_o)(y-y_o)$$

In the equation, the coefficients are the partial derivatives of the function.

By using the two-dimensional surface $z=f(x,y)$, we can evaluate the partial derivatives of the function with respect to x and y at the given point.

Given surface is $z=f(x,y)=x^2-2xy+y^2$ at the given point (1, 5, 16).

First, we find the partial derivatives of the surface at the given point:

$\frac{∂f}{∂x}=\frac{∂}{∂x}(x^2-2xy+y^2)$

$\frac{∂}{∂x}(x^2)-\frac{∂}{∂x}(2x)+\frac{∂}{∂x}(y^2)$

$f_x(x,y)=2x-2y$

$f_x(1,5)=2(1)-2(5)=2-10$

$f_x(1,5)=-8$

$\frac{∂f}{∂y}=\frac{∂}{∂y}(x^2-2xy+y^2)=\frac{∂}{∂y}(x^2)-\frac{∂}{∂y}(2xy)+\frac{∂}{∂y}(y^2)$

$f_y(x,y)=2y-2x$

$f_y(1,5)=2(5)-2(1)=10-2$

$f_y(1,5)=8$

Let us find an equation of the tangent plane to the surface at the given point:

$$z-z_o=f_x(x_o,y_o)(x-x_o)+f_y(x_o,y_o)(y-y_o)$$

$$z-16=(-8)(x-1)+(8)(y-5)$$

$$z-16=-8x+8y-32$$

$$z=-8x+8y-32+16$$

$$z=-8x+8y-16$$

An equation of the tangent plane to the surface at the given point is 

$$z=-8x+8y-16$$

Maple code:

TangentPlane(x^2 − 2xy + y^2,x=1,y=5,z=16)

output:

$8x-8y+z=-16$