Question

Find the equations of the tangent plane and normal to the surface z=xy at the point (2,3,6)

Answer 1

Answer:f(x,y,z)=2x  

2

+y  

2

+2z=3

∂x

∂f

​  

=4x,  

∂x

∂f

​  

=2y,  

∂x

∂f

​  

=2z

​  

 

At point (2,1,−3)

∂x

∂f

​  

=8,  

∂x

∂f

​  

=2,  

∂x

∂f

​  

=2

Equation of tangent plane

8(x−2)+2(y−1)+2(z+3)=0

4x+y+z−6=0

Equation of normal is  

8

x−2

​  

=  

2

y−1

​  

=  

2

z+3

​  

 

Then,

We get  

4

x−2

​  

=  

1

y−1

​  

=  

1

z+3

​ v

Answer 2

Answer:

The equation of the tangent plane is $3x+2y+z+6=0$

The equation of the normal line is $2x+3y+6z-49=0$

Step-by-step explanation:

Given: z-xy=0 at the point (2,3,6)

Derivative w.r.t (x):

$f'(x)=-y$

Differentiate $-xy+z$term by term:

The derivative of the constant $z$  is zero.

The derivative of a constant times a function is the constant times the derivative of the function.

The derivative of a constant times a function is the constant times the derivative of the function.

Apply the power rule: $x$ goes to $1$

So, the result is: $y$

So, the result is: $-y$

The result is: $-y$

The answer is: $-y$

Derivative w.r.t (y):

$f'(y)=-x$

Differentiate $-xy+z$term by term:

The derivative of the constant $z$  is zero.

The derivative of a constant times a function is the constant times the derivative of the function.

The derivative of a constant times a function is the constant times the derivative of the function.

Apply the power rule: $y$ goes to $1$

So, the result is: $x$

So, the result is: $-x$

The result is: $-x$

The answer is: $-x$

Derivative w.r.t (z):

$f'(z)=1$

Differentiate $-xy+z$term by term:

The derivative of the constant $-xy$  is zero.

Apply the power rule: $z$ goes to $1$

The result is: $1$

The answer is: $1$

Find a:

$f_x=-y\\f_x(2,3,6)=-y\\f_x(2,3,6)=-3$

Find b:

$f_y=-x\\f_y(2,3,6)=-x\\f_y(2,3,6)=-2$

Find c:

$f_z=1\\f_z(2,3,6)=1\\f_z(2,3,6)=1$

$x_0=2,y_0=3,z_0=6\\a=-3,b=-2,c=1$

The equation of the tangent plane is

$a(x-x_0)+b(y-y_0)+c(z-z_0)=0\\-3(x-2)+(-2)(y-3)+1(z-6)=0\\-3x+6-2y+6+z-6=0\\3x+2y+z+6=0$

The equation of normal line is

$\frac{x-x_0}{a}+\frac{y-y_0}{b}+\frac{z-z_0}{c}=0\\ \\\frac{x-2}{-3}+\frac{y-3}{-2}+\frac{z-6}{1}=0\\\\2(x-2)+3(y-3)+6(z-6)=0\\2x-4+3y-9+6z-36=0\\2x+3y+6z-49=0$

Reference Link

• https://brainly.in/question/12425311