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Let me show my work before presenting the problem itself.

Let $M=\{(x,y,z) \in \mathbb{R}^3 : x+y=5 x+z=cos^2y\}$.

We can easily see that $M$ is a submanifold of $\mathbb{R}^3$ of dimension $1$.

We can also see that we can construct a global atlas for $M$, say $\{(M,\varphi)\}$, where $\varphi:M\rightarrow\mathbb{R}$ is given by $(x,y,z) \mapsto y$.

If $P=(p_1,p_2,p_3) \in M$, then $T_PM=\langle (-1,1,1-2sin(p_2)cos(p_2)) \rangle$. In particular, for $P=(5,0,-4) \in M$, $T_{(5,0,-4)}M = \langle (-1,1,1) \rangle$.

Consider the morphism (smooth map) $F:M \rightarrow S^1$ given by $(x,y,z) \mapsto (\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}})$.

Remember that $S^1=\{(x,y) \in \mathbb{R}^2 : x^2+y^2=1\}$, so the given morphism is well defined.

Expressing $F$ in the atlas of $M$, we get $$F(x,y,z)=F(\varphi^{-1}(y))=(\frac{5-y}{\sqrt{2y^2-10y+25}},\frac{y}{\sqrt{2y^2-10y+25}})$$

We can see that $F(5,0,-4)=(1,0)$.

Now we choose a coordinate chart $\psi$ of $S^1$ in a neighborhood of $(1,0)$. For example, $\psi:U=\{(x,y) \in S^1 : x > 0\} \longrightarrow ]-1,1[$ given by $(x,y) \mapsto y$.

I now want to compute in the chosen charts of $M$ and $S^1$ the expression of $DF_{(5,0,-4)}$.

How can I proceed? Some help would be appreciated. Thanks in advance.

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At this point you have expressed $F : \mathbb{R} \to S^1$ as $$F(\varphi^{-1}(y))=(\frac{5-y}{\sqrt{2y^2-10y+25}},\frac{y}{\sqrt{2y^2-10y+25}})$$ You also have the mapping $\psi : U \to ]-1, 1[$ given by $$\psi(x, y) = y$$ and have demonstrated you would like the differential at the point $0$ in the domain of $F \circ \varphi^{-1}$.

So we can compose $\psi$ with $F$ to obtain: $$\psi(F(\varphi^{-1}(y))) = \frac{y}{\sqrt{2y^2-10y+25}}$$

Now $$D(\psi(F(\varphi^{-1}(y))))\big|_{y = 0} = \frac{d}{dy} \left(\frac{y}{\sqrt{2y^2-10y+25}}\right)\bigg|_{y = 0}$$

Evaluate the RHS of that and you will be done.

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  • $\begingroup$ Thanks for answering. Please correct if I'm saying something wrong. $DF_{(5,0,-4)}$ is a map $\mathbb{R} \longrightarrow \mathbb{R}$. What is the map? I also computed the RHS and I got $\frac{25-5y}{(2y^2-10y+25)^{3/2}}|_{y=0}$ and substituting I got $\frac{1}{5}$ $\endgroup$ – Leafar Jun 28 '15 at 1:08
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    $\begingroup$ @Leafar I'll be really explicit about the coordinates to make it clear. $DF_{(5, 0 -4)}$ is a map on the tangent space of your original curve at the point $(5, 0, -4)$ to the tangent space at the image $(1, 0)$ in $S_1$. If we let $G = \psi \circ F \circ \phi^{-1}$ (which is $F$ expressed in the charts), then $DG_0$ maps $\mathbb{R} \to \mathbb{R}$ which is the tangent space of $0$ to the tangents space of $0$. The map $DG_0$ is given by $y \mapsto y/5$ using the result of your computation. $\endgroup$ – muaddib Jun 28 '15 at 1:16
  • $\begingroup$ Just one more thing, why do you say ''and have demonstrated you would like the differential at the point $0$ in the domain of $F$''? Where did the zero comes from? $\endgroup$ – Leafar Jun 29 '15 at 21:16
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    $\begingroup$ @Leafar - Changed it to $F \circ \varphi^{-1}$. Where did the $0$ come from? $\varphi(x, y, z) = y$ so $\varphi^{-1}(0) = (-5, 0, 4)$. $\endgroup$ – muaddib Jun 29 '15 at 21:43

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