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

Answer:

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:

$\displaystyle f_x(x,y)=(1)\cdot\ln(x^2+y^2)+(x+y)\cdot\frac{2x}{x^2+y^2}$

$\displaystyle f_y(x,y)=(1)\cdot\ln(x^2+y^2)+(x+y)\cdot\frac{2y}{x^2+y^2}$

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

$\displaystyle f_x(1,1)=(1)\cdot\ln(1^2+1^2)+(1+1)\cdot\frac{2(1)}{1^2+1^2}=\ln 2+2$

$\displaystyle f_y(1,1)=(1)\cdot\ln(1^2+1^2)+(1+1)\cdot\frac{2(1)}{1^2+1^2}=\ln 2+2$

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:

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