Prove that vector $[3,2]$ is tangent to the set $\{(x,y): x^2 - y^3 = 0\}$ at point $(1,1)$. As in title.
Prove that vector $[3,2]$ is tangent to the set $\{(x,y): x^2 - y^3 = 0\}$ at point $(1,1)$.
I start from 'differentiating' condition defining the set: $2x - 3y^2 = 1$.
Then $2(3-1) - 3(2-1)^2 = 1$.
Is it right justification?
 A: The kernel of a function $f$ from $\mathbb{R}^n \to \mathbb{R}$ (called a "scalar function"), can be visualized as an embedded submanifold of dimension $n$ inside $(n+1)$-dimensionsal space. This $(n+1)$-dimensionsal space, here, is the direct product (combination, addition) of the input space (here $\mathbb{R}^n$) and the output space (here $\mathbb{R^1}$).
Another representation of a scalar function is as colored sets inside $\mathbb{R}^n$; where negative output values are mapped to blue (darker as you approach negative infinity), the kernel is mapped to white, and positive values to red (darker as you approach positive infinity). A certain shade of color corresponds precisely to what is called a level set, the solution to an equation of the form $f(u) = C$. Your kernel is specifically the level set for $C=0$. A level set is a manifold of dimension $n-1$ in this context.
Inside this representation in $\mathbb{R}^n$, the gradient at a given point, geometrically, is always perpendicular to the level set on which it starts. Proving that a certain vector is tangent to a level set is then simply a question of proving that its dot product with the gradient is null. That is the justification you are looking for. Why ? Because the tangent space at a point on a manifold is always just the orthogonal complementary of the $1$-dimensional subspace generated by the gradient (in this configuration of a $(n-1)$-dimensional manifold in $\mathbb{R}^n$).
So, now for some calculation, your gradient is: $\nabla f(x, y) = [2x \space \space -3y^2]$. This means $ \nabla f(1, 1) = [2 \space \space -3]$. And $<[2 \space \space -3], [3 \space \space 2]> = 0$, so indeed the vector $[3 \space \space 2]$ is tangent to the level set defined by the kernel.
Finally, some visuals:
Fig 1: General point of view of your manifold, the $(n+1)$-dimensional view. Your kernel is the curved black line. The grey plane is the input space; the vertical blue axis is the output space. The gradient at $[1 \space 1]$ is the black vector showing the direction of greatest increase on the blue manifold. The other vector is clearly tangent to the kernel, and perpendicular to the gradient.

Fig 2: Top-down view, projecting the output space to zero, to give you an idea of the $n$-dimensional visualization. Imagine parts below the plane are in blue, and parts above a red, the black curve is white, and the further away from the plan points are, the darker they get.

