space curve

[logistic_guy]

here is the question

For the space curve $\displaystyle \boldsymbol{\varphi}(t) = (t,t^2,t^3)$, find a tangent vector and the tangent line at the point $\displaystyle (1,1,1)$.


my attemb
i can find tangent vector by the derivative of $\displaystyle \boldsymbol{\varphi}(t)$
but how to find tangent line?

 

[Harry_the_cat]

You have a point and you have the tangent vector, so go from there.

 

[logistic_guy]

You have a point and you have the tangent vector, so go from there.

thank

my tangent vector $\displaystyle \bold{T} = (1,2,3)$
how to relate it to the tangent line?

 

[Harry_the_cat]

What does the equation of a line look like? What's the format?

 

[logistic_guy]

What does the equation of a line look like? What's the format?

i'm a little confused

i'll show how i understand differential geometry
if $\displaystyle \bold{p} = (1,1,1)$ and $\displaystyle \bold{T} = (1,2,3)$ is a tangent vector in $\displaystyle \bold{p}$
we call $\displaystyle \bold{T}$, $\displaystyle \bold{T}_{p}$
we call $\displaystyle \bold{p}$ in this case a point of application
it mean vector $\displaystyle \bold{T}$ start in $\displaystyle \bold{p}$ and end in $\displaystyle \bold{p} + (1,2,3) = (1,1,1) + (1,2,3) = (1 + 1, 1 + 2, 1 + 3) = (2,3,4) = \bold{q}$
so vector $\displaystyle \bold{T}$ start in $\displaystyle \bold{p}$ and end in $\displaystyle \bold{q}$

my think tell me a tangent line in $\displaystyle \bold{p}$ must have the same direction as $\displaystyle \bold{T}$, it only won't have start and end
but i can't link this to create an equation for the tangent line

 

[topsquark]

i'm a little confused

i'll show how i understand differential geometry
if $\displaystyle \bold{p} = (1,1,1)$ and $\displaystyle \bold{T} = (1,2,3)$ is a tangent vector in $\displaystyle \bold{p}$
we call $\displaystyle \bold{T}$, $\displaystyle \bold{T}_{p}$
we call $\displaystyle \bold{p}$ in this case a point of application
it mean vector $\displaystyle \bold{T}$ start in $\displaystyle \bold{p}$ and end in $\displaystyle \bold{p} + (1,2,3) = (1,1,1) + (1,2,3) = (1 + 1, 1 + 2, 1 + 3) = (2,3,4) = \bold{q}$
so vector $\displaystyle \bold{T}$ start in $\displaystyle \bold{p}$ and end in $\displaystyle \bold{q}$

my think tell me a tangent line in $\displaystyle \bold{p}$ must have the same direction as $\displaystyle \bold{T}$, it only won't have start and end
but i can't link this to create an equation for the tangent line

Differential Geometry?! Didn't you take Calc III?

See here.

-Dan

 

[fresh_42]

thank

my tangent vector $\displaystyle \bold{T} = (1,2,3)$
how to relate it to the tangent line?

A tangent is the straight defined by the location, here $\mathbf{p}=(1,1,1),$ and the tangent vector, here $\mathbf{T}=(1,2,3).$ A tangent is an affine linear space, which means we take the straight through the origin $(0,0,0)$ in direction $\mathbf{T}$ and shift it by the position vector $(1,1,1).$

You had it almost right, but you considered only one point on that straight. @Harry_the_cat's question was about the equation for the entire straight (through the origin). If you have it, then you can add $\mathbf{p}.$ The point $(1,2,3)$ is one point on that straight (through the origin) and $\mathbf{q}$ is one point on the tangent. But what are the others?

 

[logistic_guy]

thank

Differential Geometry?!

yes

Didn't you take Calc III?

i do long time ago. now i'm very bad in clculus as i forgot most of the ideas

See here.

the website use notation i'm not understand

A tangent is the straight defined by the location, here $\mathbf{p}=(1,1,1),$ and the tangent vector, here $\mathbf{T}=(1,2,3).$ A tangent is an affine linear space, which means we take the straight through the origin $(0,0,0)$ in direction $\mathbf{T}$ and shift it by the position vector $(1,1,1).$

You had it almost right, but you considered only one point on that straight. @Harry_the_cat's question was about the equation for the entire straight (through the origin). If you have it, then you can add $\mathbf{p}.$ The point $(1,2,3)$ is one point on that straight (through the origin) and $\mathbf{q}$ is one point on the tangent. But what are the others?

i'm not quite understand this but i think Harry want to tell me what happen if you have a point and tangent?
which lead me go back to my favorite subject Algebra

Algebra have nice equation for the tangent line $\displaystyle y = mx + b$
i think Harry is telling me in space $\displaystyle \bold{T} = m$ and $\displaystyle \bold{p}= b$
so the tangent line $\displaystyle \bold{S}(t) = \bold{T}t + \bold{p}$

i'm wrong?

 

[topsquark]

the website use notation i'm not understand

Then you need to go back and review Calc III before you go any further. This is a basic Calc III problem.

-Dan

 

[logistic_guy]

Then you need to go back and review Calc III before you go any further. This is a basic Calc III problem.

-Dan

why i need calculus when i'm very good in differential geometry?

 

[fresh_42]

Algebra have nice equation for the tangent line $\displaystyle y = mx + b$
i think Harry is telling me in space $\displaystyle \bold{T} = m$ and $\displaystyle \bold{p}= b$
so the tangent line $\displaystyle \bold{S}(t) = \bold{T}t + \bold{p}$

i'm wrong?

No, that is true. Your first attempt had only the point at $t=1.$

$\mathbf{T}\cdot t$ is a one-dimensional vector space. It contains the origin at $t=0.$ In this sense, this straight is the tangent space. But the tangent itself is shifted by $\mathbf{p},$ so it is $\mathbf{S}(t)= \mathbf{T}\cdot t +\mathbf{p}.$ This tangent is not a vector space because it contains no zero vector. It is an affine space, a vector space $V= \mathbf{T}\cdot t$ shifted by $\mathbf{p}.$ It is a bit confusing since we call the affine space the tangent, but the vector space $V,$ the tangent space since we want it to be a vector space. In your case, these are two parallel straights, one with a zero vector, and one that is affine linear because $p$ is its "origin".

 

[topsquark]

why i need calculus when i'm very good in differential geometry?

Evidently because you cannot solve this problem? And you can't be good at all in Differential Geometry if you aren't good at Calculus. You need Calculus to understand Differential Geometry in the first place. From what I've seen of your posts you need a serious review of a number of basic topics.

-Dan

 

[logistic_guy]

No, that is true. Your first attempt had only the point at $t=1.$

$\mathbf{T}\cdot t$ is a one-dimensional vector space. It contains the origin at $t=0.$ In this sense, this straight is the tangent space. But the tangent itself is shifted by $\mathbf{p},$ so it is $\mathbf{S}(t)= \mathbf{T}\cdot t +\mathbf{p}.$ This tangent is not a vector space because it contains no zero vector. It is an affine space, a vector space $V= \mathbf{T}\cdot t$ shifted by $\mathbf{p}.$ It is a bit confusing since we call the affine space the tangent, but the vector space $V,$ the tangent space since we want it to be a vector space. In your case, these are two parallel straights, one with a zero vector, and one that is affine linear because $p$ is its "origin".

once again thank very much for explaining it very nicely
i understand what you're saying

Evidently because you cannot solve this problem? And you can't be good at all in Differential Geometry if you aren't good at Calculus. You need Calculus to understand Differential Geometry in the first place. From what I've seen of your posts you need a serious review of a number of basic topics.

-Dan

i do solve this problem. i discvover the solution of the tangent line by my own in post 8
it's very bad habit to underestimate other people skiils
i'll let it go this time because i think you're a good teacher
next time don't do this behavior it's very wrong
not everyone in this website will accept like i accept
thank

 

[Harry_the_cat]

i'm not quite understand this but i think Harry want to tell me what happen if you have a point and tangent?
which lead me go back to my favorite subject Algebra

Algebra have nice equation for the tangent line $\displaystyle y = mx + b$
i think Harry is telling me in space $\displaystyle \bold{T} = m$ and $\displaystyle \bold{p}= b$
so the tangent line $\displaystyle \bold{S}(t) = \bold{T}t + \bold{p}$

i'm wrong?

That is correct.

 

[topsquark]

once again thank very much for explaining it very nicely
i understand what you're saying


i do solve this problem. i discvover the solution of the tangent line by my own in post 8
it's very bad habit to underestimate other people skiils
i'll let it go this time because i think you're a good teacher
next time don't do this behavior it's very wrong
not everyone in this website will accept like i accept
thank

Don't patronize me. I clearly upset you.

Good! You need a stiff review. This is a simple Calc III level problem. If you cannot solve it then you need to go back and look it up instead of coming back here and whining about how bad you are in Calculus. If you don't like the nomenclature in the link that I gave you, pull out your old Calc text and look it up.

Feel free to get mad. I really don't care. Just do the review.

-Dan