Find Normal and Tangential components of accelaration

[RyanH42]

I found the answer The question is not about $\vec{T}$ or $\vec{N}$.We are in wrong way the equation is $\vec{a}=d^2s/dt^2(\vec{T}+k\vec{N})$ here the components is $d^2s/dt^2$ for $\vec{T}$ and $d^2s/dt^2k$ for $\vec{N}$.

$ds/dt=√(1+(dy/dx)^2)$
$y=(t^2+1)\vec{j}$ → $dy =2t\vec{j}$,
$x=t\vec{i}$ →$dx=t\vec{j}$
$ds/dt=√(1+(2t/1)^2)$
$d^2s/dt^2=8t/2√(1+4t^2)=4t/√(1+4t^2)$
Which I checked I found the right answer.I will do the same thing for $\vec{N}$ but there's k. k=magnitude of $d\vec{T}/ds$. $d\vec{T}/ds=d\vec{T}/dt/(ds/dt)$ $d\vec{T}/dt=2j$ and $ds/dt=√(1+(2t/1)^2)$ so $d\vec{T}/ds=2j/√(1+(4t^2)$ and the magnitude is $2/√(1+(4t^2)$ but the answer is 2/(1+(4t^2)^3/2
What I am missing ?

 

[RyanH42]

I guess I mixed everything

 

[ChrisVer]

Well I found some mistakes...
First of all avoid writting x,y as vectors when they are not...
It would be correct to write x=t \Rightarrow dx = dt and y=t^2+1 \Rightarrow dy= 2 t dt

Also the reason you don't get the right answer is because in \frac{d \vec{T}}{ds} you didn't take the derivatives of the unit-tangent vector, just a tangent vector...you used \vec{T} = \hat{i} + 2t \hat{j} \Rightarrow |\vec{T}|= \sqrt{1+4t^2} \ne 1...
If you do the derivative with the unit-tangent vector correctly (I did it) you will get the result you ask for.

 

[RyanH42]

Well I found some mistakes...
First of all avoid writting x,yx,y as vectors when they are not...
It would be correct to write x=t⇒dx=dtx=t \Rightarrow dx = dt and y=t2+1⇒dy=2tdty=t^2+1 \Rightarrow dy= 2 t dt

Ok,thanks for this

"Also the reason you don't get the right answer is because in \frac{d \vec{T}}{ds} you didn't take the derivatives of the unit-tangent vector, just a tangent vector...you used \vec{T} = \hat{i} + 2t \hat{j} \Rightarrow |\vec{T}|= \sqrt{1+4t^2} \ne 1...
If you do the derivative with the unit-tangent vector correctly (I did it) you will get the result you ask for."
Ok.As you said I did it and I understand.I want to ask one question what's the realtionship between $\vec{T}$ and $k$and $\vec{N}$.I thought or see $d\vec{T}/dt=k\vec{N}$ ?

 

[ChrisVer]

\frac{d \vec{T}}{ds} = k \vec{N}

 

[RyanH42]

I understand it thanks the magnitude will give us k but in the vector form this.This vector calculus is really hard.Thanks again for help.