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

• MHB
• WMDhamnekar
In summary: Comparing the coefficients of A and B m(m - 1) t^{m - 2} = k t^m \implies m(m - 1) = k t^2andn(n - 1) t^{n - 2} = k t^n \implies n(n - 1) k t^2The "obvious" way says that 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.
WMDhamnekar
MHB
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:
Dhamnekar Winod said:
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

topsquark said:
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?

Dhamnekar Winod said:
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

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?

Dhamnekar Winod said:
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.

Dhamnekar Winod said:
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

## What is vector calculus and why is it important?

Vector calculus is a branch of mathematics that deals with the study of vector fields and their derivatives. It is used to analyze and model physical phenomena that involve quantities with both magnitude and direction, such as velocity, force, and electric fields. It is important because it provides a powerful framework for solving real-world problems in fields such as physics, engineering, and economics.

## What are some common applications of vector calculus?

Vector calculus has a wide range of applications, including fluid dynamics, electromagnetism, computer graphics, and optimization. It is used to model the motion of fluids, such as air and water, and to analyze the behavior of electric and magnetic fields. It is also used in computer graphics to create realistic 3D images and animations. In optimization, vector calculus is used to find the maximum or minimum values of functions with multiple variables.

## What are the basic operations in vector calculus?

The basic operations in vector calculus are vector addition, scalar multiplication, and differentiation and integration of vector functions. Vector addition involves adding two or more vectors to obtain a resultant vector. Scalar multiplication involves multiplying a vector by a scalar quantity, which changes the magnitude of the vector but not its direction. Differentiation and integration of vector functions are used to find the derivatives and integrals of vector fields, respectively.

## How is vector calculus related to other branches of mathematics?

Vector calculus is closely related to other branches of mathematics, such as linear algebra, multivariable calculus, and differential equations. It uses concepts from linear algebra, such as vectors and matrices, to study vector fields. It also builds upon the principles of multivariable calculus, which deals with functions of multiple variables. Differential equations are used in vector calculus to model and solve physical phenomena.

## What are some resources for learning vector calculus?

There are many resources available for learning vector calculus, including textbooks, online courses, and video tutorials. Some popular textbooks on the subject include "Vector Calculus" by Jerrold E. Marsden and Anthony J. Tromba, and "Vector Calculus" by Susan J. Colley. Online courses on platforms like Coursera and edX also offer comprehensive lessons on vector calculus. Additionally, there are many YouTube channels and websites that provide free video tutorials and practice problems for learning vector calculus.

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