Question

Exercise 16 Find the tangent plane to f(x, y) = (x + y) ln(x2 + y2) at the point (1,1) in both standard form and linearized form.

Answer 1

Scalar form:
`(\ln 2+2)(x-1)+( \ln 2+2 )(y-1) - (z-2\ln 2)= 0`

Linear form:
`(\ln 2+2) x +(\ln 2+2) y-z=4`

Step-by-step explanation:

We need the partial derivatives of the function f(x,y):

Notice we need to use product rule: (uv)’ u’v+uv’ where here:

`u=x+y, v=\ln(x^2+y^2)`

Therefore:

`\displaystyle f_x(x,y)=(1)\cdot\ln(x^2+y^2)+(x+y)\cdot(2x)/(x^2+y^2)`

`\displaystyle f_y(x,y)=(1)\cdot\ln(x^2+y^2)+(x+y)\cdot(2y)/(x^2+y^2)`

We evaluate them in the given point (1,1):

`\displaystyle f_x(1,1)=(1)\cdot\ln(1^2+1^2)+(1+1)\cdot(2(1))/(1^2+1^2)=\ln 2+2`

`\displaystyle f_y(1,1)=(1)\cdot\ln(1^2+1^2)+(1+1)\cdot(2(1))/(1^2+1^2)=\ln 2+2`

We also need to evaluate the function f(x,y) at (1,1):

`f(1,1)=(1+1)\ln(1^1+1^1)=2\ln 2`

Then we plug the pieces into the formula of the tangent plane:

`z-f(x_o,y_o) =f_x(x_o,y_o)\cdot(x-x_o)+f_y(x_o,y_o)\cdot(y-y_o)`

Here:
`(x_o,y_o)=(1,1)`, so:

`z-f(1,1) =f_x(1,1)\cdot(x-1)+f_y(1,1)\cdot(y-1)`

`z-2\ln 2=(\ln 2+2 )(x-1)+( \ln 2+2 )(y-1)`

We collect everything on the left side to get:

`(\ln 2+2)(x-1)+( \ln 2+2 )(y-1) - (z-2\ln 2)= 0`

Which is the scalar form of the equation of the plane

Then we distribute to get:

`(\ln 2+2) x-(\ln 2+2) +(\ln 2+2) y-(\ln 2+2) -z+2\ln 2=0`

Then simplify by combining like terms:

`(\ln 2+2) x +(\ln 2+2) y-z-4=0`

Finally move the constant to the right side to get the linear form:

`(\ln 2+2) x +(\ln 2+2) y-z=4`