How to compute the slope for a 3 or multi-dimensional equation.

Question

If I have an equation $Z=X^2+Y^2+3X+6Y+5$ and want to find the slope at the point $x=2$, $y=1$. How do we compute it? I know for a two dimensional equation we can compute it by differentiation of $Y$ with respect to $X$. But In case of multi-dimension. How do we solve it?

Answer 1

First of all, there are the partial derivatives. If you haven't learned anything about differentiating and multivariable functions, then I understand if you think they feel a bit like cheating, but please understand: They are important, and used a lot. If you have the function $f(x, y) = x^2 + y^2 + 3x + 6y + 5$, then the partial derivative of $f$ with respect to $x$ is defined as $$ \frac{\partial f}{\partial x} = 2x + 3 $$ where we simply pretend that $y$ isn't a variable at all, but a constant, and we differentiate with respect to $x$ as you normally would a single-variable function. Similarily, the partial derivative of $f$ with respect to $y$ is $$ \frac{\partial f}{\partial y} = 2y + 6 $$

Then there is the gradient, which tells you which direction the graph is increasing fastest, and what slope it has in that direction. Depending on how you think about the usual derivative, this might be the notion that feels closest. It's a vector field (i.e. each point in the $xy$-plane has a corresponding vector attached), and if your function is $f(x, y) = x^2 + y^2 + 3x + 6y + 5$, then the gradient of $f$ is $$ \nabla f = \begin{pmatrix}\frac{\partial f}{\partial x}\\\frac{\partial f}{\partial y}\end{pmatrix} = \begin{pmatrix}2x + 3\\2y + 6\end{pmatrix} $$ If you want to find the tangent plane at say $(x, y) = (3, 4)$, then you find the gradient at that point: $$\nabla f(3, 4) = (2\cdot 3 + 3, 2\cdot 4 + 6) = (9,14)$$ That means that you want a plane of the form $z = ax + by + c$ that has the same gradient, which is easy: $a = 9, b = 14$. Finally, we need to find $c$, and that's so that it actually touches the graph, so we need to know the function value: $f(3, 4) = 63$. Putting $(x, y, z) = (3, 4, 63)$ into the equation for the plane, we get $$ 63 = 9\cdot 3 + 14\cdot 4 + c\\ 63 = 27 + 56 + c\\ -20 = c $$ so the tangent plane to $f$ at $(3, 4)$ is finally given by $z = 9x + 14y - 20$, or $9x + 14y - z = 20$ if you prefer it that way.