I'm doing a Calc III homework problem, and I cannot seem to figure out what the correct solution is.
$$ \text{Find the equation of the tangent plane to the surface }z = 9 y^{2} - 9 x^{2}\text{ at the point }\left( -1, 4, 135 \right). \\ z = \text{______ Note: Your answer should be an expression of }x\text{ and }y\text{; e.g. "}3x - 4y + 6\text{"} $$
Seems simple enough. First I move $z$ from the left side of the equation and then take partial derivatives.
$$ f(x,y,z) = -9^2+9y^2-z = 0 \\ $$ Calculate partial derivatives and plug in their respective points $$ f_x = -18x\\ f_x(x-x_0) = -18(x+1)\\ f_y = 18y\\ f_y(y-y_0) = 18(y-16)\\ f_z = -1 \\ f_z(z-z_0) = -1 (z - 135) $$ Which should then mean that: $$ -18(x+1)+18(y-16)-1(z-135) = 0\\ -18x-18 + 18y - 288 - z+135 = 0\\ -18x + 18y-z-171 = 0 \\ $$ We can then move $z$ back across the axis to get the answer $$ z = -18x+18y-171 $$
This seems like it should be correct, but webwork will not accept my result. Do I have something amis?
Hint.
Your partial derivatives are correct, now the tangent plane at $(x_0,y_0,f(x_0,y_0))$ is:
$$ z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) $$
and you have $x_0=-1$ , $y_0=4$, so $f_x(x_0,y_0)= 18$ and $f_y(x_0,y_0)=72$