# Vector calculus applications

#### Dhamnekar Winod

##### Active member
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?

Last edited:

#### topsquark

##### Well-known member
MHB Math Helper
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?

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

#### Dhamnekar Winod

##### Active member
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

##### Well-known member
MHB Math Helper
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

#### Dhamnekar Winod

##### Active member
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

##### Well-known member
MHB Math Scholar
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

##### Well-known member
MHB Math Helper
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