Let $f:\Bbb{R}^2\to\Bbb{R}$, with $f(x,y)=\frac{1}{6}(3xy^2-4x^2y+y^3+10)$. Find the equation at the tangent plan of f at the point $(3,2)$.

Question

I've answered this question and I got:

$f_x(a,b)=-6$

$f_y(a,b)=2$

$f(a,b)=-3$

and my answer is: $-6x+2y+9$

Just wanted to know whether I am on the right tracks here?

The second part to this question then asks us to find the directional derivative of $f$ at point $(3,2)$ in the direction of vector $u=\binom{-4}3$

I got $6$ for this answer.

Thanks for your help,

Struggling Statistics Student.

Answer 1

Your equation is almost right but I don't know where the 9 came from.

The tangent plane at point $(x_0 = 3, y_0 = 2)$ at $(x_0,y_0)$ will be the map $$ \bar{f}(x,y) = f_x (x - x_0) + f_y (y- y_0) + f(x_0,y_0) $$ which comes out to $$ \bar{f}(x,y) = -6 (x -3) + 2(y- 2) -3 = -6x + 2y + 11 $$

The directional derivative in direction $\hat{u}$ is $\hat{u}\cdot \nabla f$. The direction given is not unit length so you have to divide by its length of $5$ to get: $$ \frac{1}{5} (-4, 3) \cdot \left( \begin{array}{c} -6 \\ 2 \end{array} \right) = 6$$