Let $f(x,y) = \frac {-1}4 (3xy^2 - 5x^3y + 2x^4)$. Find the equation of the tangent plane to $f$ at the point $(2,4)$.

Question

Let $f(x,y) = \frac {-1}4 (3xy^2 - 5x^3y + 2x^4)$. Find the equation of the tangent plane to $f$ at the point $(2,4)$.

Using vector dot product with:

$a = 2$
$b = 4$
$f(a,b) = -8$
$\frac {\partial f} {\partial x}(a,b) = -56$
$\frac {\partial f} {\partial y}(a,b) = -22$

I worked out the answer:

$56x + 22y - 2 = 192$

but according to the solution sheet this is wrong. Would really appreciate help on where I am going wrong.

Thanks!

Answer 1

@samuel With the information above and the equation of a tangent plane to be $z-z_o=F_x(x-x_o)+F_y(y-y_o)$ you can now plug in all the info we have: $z + 8 = 32 (x-2)-2(y-4)$ Work this out and that should be it.

Answer 2

So, we have

$$f(x,y) = -\frac{1}{4}(3xy^2 - 5x^3y + 2x^4).$$

Then, $$f_x(x,y) = -\frac{1}{4}(3y^2 - 15x^2y + 8x^3),$$

and

$$f_y(x,y) = -\frac{1}{4}(6xy - 5x^3).$$

From this I get $f_x(2,4) = 212$ and $f_y(2,4) = -2$, which doesn't agree with what you got. Maybe that's why things aren't working out?