speed of particle

[logistic_guy]

here is the question

Find the speed of a particle tracing the curve $\displaystyle \boldsymbol{\varphi}(t) = (t,t^2,t^3)$ at time, $\displaystyle t = 1$.


my attemb
$\displaystyle \boldsymbol{\varphi}'(t) = (1,2t,3t^2)$
$\displaystyle \boldsymbol{\varphi}'(1) = (1,2,3)$
this look like the tangent vector i find in my previous post
if $\displaystyle \bold{T} = \boldsymbol{\varphi}'$
it mean tangent vector and speed is the same thing
first time to know this
so the answer to the original question
the speed of the particle in the $\displaystyle x$-direction is $\displaystyle 1$
the speed of the particle in the $\displaystyle y$-direction is $\displaystyle 2$
the speed of the particle in the $\displaystyle z$-direction is $\displaystyle 3$

 

[topsquark]

here is the question

Find the speed of a particle tracing the curve $\displaystyle \boldsymbol{\varphi}(t) = (t,t^2,t^3)$ at time, $\displaystyle t = 1$.


my attemb
$\displaystyle \boldsymbol{\varphi}'(t) = (1,2t,3t^2)$
$\displaystyle \boldsymbol{\varphi}'(1) = (1,2,3)$
this look like the tangent vector i find in my previous post
if $\displaystyle \bold{T} = \boldsymbol{\varphi}'$
it mean tangent vector and speed is the same thing
first time to know this
so the answer to the original question
the speed of the particle in the $\displaystyle x$-direction is $\displaystyle 1$
the speed of the particle in the $\displaystyle y$-direction is $\displaystyle 2$
the speed of the particle in the $\displaystyle z$-direction is $\displaystyle 3$

Good so far. So what is the speed of the particle?

-Dan

 

[eipi45]

here is the question

Find the speed of a particle tracing the curve $\displaystyle \boldsymbol{\varphi}(t) = (t,t^2,t^3)$ at time, $\displaystyle t = 1$.


my attemb
$\displaystyle \boldsymbol{\varphi}'(t) = (1,2t,3t^2)$
$\displaystyle \boldsymbol{\varphi}'(1) = (1,2,3)$
this look like the tangent vector i find in my previous post
if $\displaystyle \bold{T} = \boldsymbol{\varphi}'$
it mean tangent vector and speed is the same thing
first time to know this
so the answer to the original question
the speed of the particle in the $\displaystyle x$-direction is $\displaystyle 1$
the speed of the particle in the $\displaystyle y$-direction is $\displaystyle 2$
the speed of the particle in the $\displaystyle z$-direction is $\displaystyle 3$

tangent vector and speed is the same thing - not quite, velocity is the vector, speed is the magnitude of the velocity (irrespective of direction)

 

[logistic_guy]

Good so far. So what is the speed of the particle?

-Dan

thank

$\displaystyle 1,2,3$ depend on the direction the particle go

tangent vector and speed is the same thing - not quite, velocity is the vector, speed is the magnitude of the velocity (irrespective of direction)

thank

but my calculation show they're the same

 

[eipi45]

thank

$\displaystyle 1,2,3$ depend on the direction the particle go


thank

but my calculation show they're the same

hi,
your calculation found (1, 2, 3) is the velocity vector
the speed is found by calculating |(1,2,3)|
do you know how to find |(a,b,c)| ? (hint - pythagoras)

 

[topsquark]

thank

$\displaystyle 1,2,3$ depend on the direction the particle go


thank

but my calculation show they're the same

Yes, 1, 2, and 3 are the components of the velocity in the x, y, and z directions. How do you find the speed, the magnitude of the velocity vector?

-Dan

 

[logistic_guy]

hi,
your calculation found (1, 2, 3) is the velocity vector
the speed is found by calculating |(1,2,3)|
do you know how to find |(a,b,c)| ? (hint - pythagoras)

yes i know to find magnitude of vector $\displaystyle \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$
but i thought the speed along arc is $\displaystyle \frac{ds}{dt}$

Yes, 1, 2, and 3 are the components of the velocity in the x, y, and z directions. How do you find the speed, the magnitude of the velocity vector?

-Dan

i think you mean speed = $\displaystyle \sqrt{14}$ but i don't understand why

 

[topsquark]

yes i know to find magnitude of vector $\displaystyle \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$
but i thought the speed along arc is $\displaystyle \frac{ds}{dt}$


i think you mean speed = $\displaystyle \sqrt{14}$ but i don't understand why

Yes, the speed is $\sqrt{14}$. You found it in your other thread on this topic. It is the same as $|d\textbf{v}/dt|$.

You have mastered Differential Geometry, but you don't understand why a vector has a given length in terms of its components?

Again, we expect you to put some work into this, too. Look it up.

-Dan

 

[logistic_guy]

You found it in your other thread on this topic. It is the same as $|d\textbf{v}/dt|$.

where which thread?

$\displaystyle \frac{ds}{dt}$ is the speed and it's derivative and $\displaystyle |d\bold{v}/dt|$ is magnitude
how can derivative be magnitude of vector? i'm confused

You have mastered Differential Geometry

yes, not only mastered but also i'm very good in it

but you don't understand why a vector has a given length in terms of its components?

it's not i don't understand. it's just confuseing because i know $\displaystyle \frac{ds}{dt}$ is the speed
it mean i take the derivative

Again, we expect you to put some work into this, too. Look it up.

what work do you expect? i do all the work i know

 

[topsquark]

where which thread?

The one where you found the tangent vector. The tangent vector is the velocity.

$\displaystyle \frac{ds}{dt}$ is the speed and it's derivative and $\displaystyle |d\bold{v}/dt|$ is magnitude
how can derivative be magnitude of vector? i'm confused

yes, not only mastered but also i'm very good in it

If you are very good in it then you already know how to take the derivative of a vector. My God, do you even know where Differential Geometry comes from??

it's not i don't understand. it's just confuseing because i know $\displaystyle \frac{ds}{dt}$ is the speed
it mean i take the derivative

The derivative gives you a velocity.

what work do you expect? i do all the work i know

I expect you to look up things that you are already expected to know from previous coursework. What would you say to someone who claims to know all about Plane Geometry, but doesn't know the equation of a line?

-Dan

 

[logistic_guy]

What would you say to someone who claims to know all about Plane Geometry, but doesn't know the equation of a line?

i'm not claim i know all about plane geometry
i just know approximateltly $\displaystyle 99\%$

i look up the definition of speed in calculus, i get this

if $\displaystyle s$ is the arc length, $\displaystyle \frac{ds}{dt} = \sqrt{(\frac{dx}{dt})^2 + (\frac{dx}{dt})^2 + (\frac{dx}{dt})^2}$ is the speed

the definition don't make sense because the left side say take the derivatiove and the right side say take the magnitude

 

[topsquark]

i'm not claim i know all about plane geometry
i just know approximateltly $\displaystyle 99\%$

i look up the definition of speed in calculus, i get this

if $\displaystyle s$ is the arc length, $\displaystyle \frac{ds}{dt} = \sqrt{(\frac{dx}{dt})^2 + (\frac{dx}{dt})^2 + (\frac{dx}{dt})^2}$ is the speed

the definition don't make sense because the left side say take the derivatiove and the right side say take the magnitude

Must be part that 1% of plane Geometry that you don't understand. It's called the Pythagorean Theorem.

-Dan