Are $\vec{r}$ and $\frac{d^2\vec{r}}{dt^2}$ Parallel When m+n=1?

[WMDhamnekar]

Given $\vec{r}=t^m* \vec{A} +t^n*\vec{B}$ where $\vec{A}$ and $\vec{B}$ are constant vectors,

How to show that if $\vec{r}$ and $\frac{d^2\vec{r}}{dt^2}$ are parallel vectors , then m+n=1, unless m=n?

I don't have any idea to answer this question. If any member knows the answer to this question, may reply with correct answer to this thread.

 

[topsquark]

Given $\vec{r}=t^m* \vec{A} +t^n*\vec{B}$ where $\vec{A}$ and $\vec{B}$ are constant vectors,

How to show that if $\vec{r}$ and $\frac{d^2\vec{r}}{dt^2}$ are parallel vectors , then m+n=1, unless m=n?

I don't have any idea to answer this question. If any member knows the answer to this question, may reply with correct answer to this thread.

How far have you gotten?

Try this. Find \(\displaystyle \vec{r}''\). If \(\displaystyle \vec{r}''\) is parallel to \(\displaystyle \vec{r}\) what can we say about a relationship to them? (ie. what would make them parallel?)

Work with that for a couple of moments and let us know how it goes.

-Dan

 

[WMDhamnekar]

How far have you gotten?

Try this. Find \(\displaystyle \vec{r}''\). If \(\displaystyle \vec{r}''\) is parallel to \(\displaystyle \vec{r}\) what can we say about a relationship to them? (ie. what would make them parallel?)

Work with that for a couple of moments and let us know how it goes.

-Dan

Hello,
If $\vec{r}$ and $ \frac{d^2\vec{r}}{dt^2}$ are parallel, then their cross product is zero vector $\vec{0}$. But question is how to prove that?

 

[topsquark]

Hello,
If $\vec{r}$ and $ \frac{d^2\vec{r}}{dt^2}$ are parallel, then their cross product is zero vector $\vec{0}$. But question is how to prove that?

Try looking at it this way: if two vectors \(\displaystyle \vec{p}\) and \(\displaystyle \vec{q}\) are parallel then then we know that \(\displaystyle \vec{q} = k \vec{p}\) where k is some scalar constant. (Theoretically we have to say that k is positive. If k were negative then they are "anti-parallel.")

So we have that
\(\displaystyle m(m - 1) t^{m - 2} \vec{A} + n(n - 1) t^{n - 2} \vec{B} = k t^m \vec{A} + k t^n \vec{B}\)

What can you do with this?

-Dan

 

[WMDhamnekar]

I shall put m*(m-1)=k, n*(n-1)=k. so $m=\frac{k}{m-1} \Rightarrow m=\frac{n*(n-1)}{m-1},\rightarrow m*(m-1)=n*(n-1)$. Then what to do next to show that if $\vec{r}$ and $\vec{r'}$ are parallel, then m+n=1, unless m=n?

 

[caffeinemachine]

I shall put m*(m-1)=k, n*(n-1)=k. so $m=\frac{k}{m-1} \Rightarrow m=\frac{n*(n-1)}{m-1},\rightarrow m*(m-1)=n*(n-1)$. Then what to do next to show that if $\vec{r}$ and $\vec{r'}$ are parallel, then m+n=1, unless m=n?

Just a typographical comment. Instead of writing $m*(m-1)$ etc, one should simply write $m(m-1)$. Aslo, one should typeset all the math using TeX. You have left some math in plain text.

 

[topsquark]

I shall put m*(m-1)=k, n*(n-1)=k. so $m=\frac{k}{m-1} \Rightarrow m=\frac{n*(n-1)}{m-1},\rightarrow m*(m-1)=n*(n-1)$. Then what to do next to show that if $\vec{r}$ and $\vec{r'}$ are parallel, then m+n=1, unless m=n?

Comparing the coefficients of A and B
\(\displaystyle m(m - 1) t^{m - 2} = k t^m \implies m(m - 1) = k t^2\)

and
\(\displaystyle n(n - 1) t^{n - 2} = k t^n \implies n(n - 1) k t^2\)

The "obvious" way says that \(\displaystyle m(m - 1) = k t^2 = n(n - 1)\).

If we ignore the t's we get m(m - 1) = n(n - 1) and you can check that m = n or m + n = 1. This is the answer we are looking for.

The problem with this is m and n are constants. \(\displaystyle m(m - 1) = kt^2\) says that m depends on t, which we can't have. This means that \(\displaystyle m (m - 1) = kt^2\) has to be adjusted so that m does not depend on t. So take another look:
\(\displaystyle m( m - 1) t^{m - 2} = k t^m\). The exponent on the t's must be the same. So we get m - 2 = m, which is impossible.

This means that the only way that we will have \(\displaystyle \vec{r''} = \vec{r}\) with \(\displaystyle \vec{r} = t^m \vec{A} + t^n \vec{B}\) is to have \(\displaystyle \vec{r} = 0\).

-Dan