Tangent space for the differentiable manifold $S^1$

Since $$S^1$$ is a compact 1-dimensional regular submanifold in $$\mathbb{R}^2$$ (it's $$S^1 = f^{-1}(1)$$ for $$f : \mathbb{R}^2 \to \mathbb{R}$$ given as $$f(x,y) = x^2+y^2$$), we can find the tangent space for $$S^1$$ in (1,0) as $$T_{(1,0)}(S^1) = Ker(d(f)_{(1,0)}).$$ Intuitively $$T_{(1,0)}(S^1)$$ is the plane $$\{(1,y): y \in \mathbb{R}\}.$$

But we get $$Ker(d(f)_{(1,0)}) = Ker((2x,2y)_{(1,0)}) = Ker((2,0)) = \{(0,y) : y \in \mathbb{R}\}.$$

What it is wrong? Thanks in advance!

• You are calculating the tangent space as a subspace of $\mathbb{R}^2$. So, it passes through the origin. You can then translate it to the point $(1, 0)$. – Joe Johnson 126 Nov 18 '18 at 19:29
• Your result is a subspace of tangent space $T_{(1,0)}R^2$. Nothing wrong with your calculation. The result represent the components of tangent vectors. That is the elements of the kernel should be in the form $0 \partial_x + s\partial_y$ for $s\in R$. Which is exactly what you wanted (up to the identification). – kelvinn aja Nov 18 '18 at 19:29
• @KelvinLois: This looks like essentially an answer. Would you mind writing it up so that we can get this question off the Unanswered queue? – aleph_two Dec 29 '18 at 4:12