Is t(s) perpendicular to the radius of the surface sphere at point γ(s)?

[ParisSpart]

whether γ=γ(s):I->R^3 curve parameterized as to arc length (single speed). Assume that γ is the surface sphere centered on the origin (0,0). Prove that the vector t(s) is perpendicular to the radius of the sphere at point γ(s), for each s.


i know that t(s)=γ΄(s) but i don't know how to continue to prove it , maybe i don't have undrstand the problem on how to show that t(s) is vertical on the radius of sphere

 

[micromass]

Since $\gamma(s)$ is on the surface of the sphere, you know
<\gamma(s),\gamma(s)> = 1
Now differentiate both sides.

 

[ParisSpart]

i do not have any other information... for γ(s)

 

[Quesadilla]

As was mentioned before, if $\gamma$ is on the surface of a sphere that means that
\begin{equation*}
\gamma(t) \cdot \gamma(t) = |\gamma(t)|^2 = \text{ constant}.
\end{equation*}
Then you should differentiate both sides with respect to $t$.

 

[ParisSpart]

with this i show that t(s) is perpendicular on the radius of the sphere? we do not use any radius r

 

[Quesadilla]

with this i show that t(s) is perpendicular on the radius of the sphere? we do not use any radius r

What does the vector $\gamma(s)$ look like?

 

[ParisSpart]

i think γ(s) is our curve and vector is t(s)=γ΄(s)

 

[Quesadilla]

i think γ(s) is our curve and vector is t(s)=γ΄(s)

Well, as you wrote yourself, $\gamma : [0,1] \rightarrow \mathbb{R}^3$. So $\gamma(s) \in \mathbb{R}^3$. A point in $\mathbb{R}^3$ is also a vector. Don't you agree?

 

[ParisSpart]

yes its my fault , γ(s) is a vector..

 

[ParisSpart]

if i differentiate i will find γ'(s)=0

 

[Quesadilla]

if i differentiate i will find γ'(s)=0

That is not correct. Differentiate $\gamma(s) \cdot \gamma(s) = c$ again. Remember the product rule.

 

[ParisSpart]

2γ(s)γ'(s)=0 but what i will do next?

 

[micromass]

2γ(s)γ'(s)=0 but what i will do next?

So the dot product is zero, what does that mean?

 

[ParisSpart]

γ(s) and γ'(s) are perpendicular , but we want t(s) to be perpendicular with radius.

 

[Quesadilla]

What does the vector $\gamma(s)$ look like?

Think about this.

 

[ParisSpart]

i am thinking it but i can't find any answer , its a radious? if yes why?

 

[Quesadilla]

i am thinking it but i can't find any answer , its a radious? if yes why?

Well, $\gamma(s)$ is a point on the surface of the sphere, but it is also a vector, right? What would that vector be? Try drawing it, if you are still unsure.

 

[ParisSpart]

and because t(s)=γ'(s) and γ(s) ιs radious t(s) is perpendicular on γ(s)?

 

[Quesadilla]

and because t(s)=γ'(s) and γ(s) ιs radious t(s) is perpendicular on γ(s)?

Yes.