Question

Prove that if a particle moves along a curve at a constant speed, then the velocity and acceleration vectors are orthogonal.

Short answer

Ans:

$\frac{d}{dt}\left({||r^{\text{'}}||}^2\right)=\frac{d}{dt}\left(r^{\text{'}}·r^{\text{'}}\right)=0$

Step-by-step solution

Step 1. GIven information: 

Consider a twice differentiable vector function $r\left(t\right)$.let a particle moves along the curve at a constant speed.

Step 2. Theory:

Here we assume that the velocity and acceleration vectors are orthogonal and prove that the particles move along $r\left(t\right)$at a constant speed. That is, we show that $\left|r^{\text{'}}\left(t\right)\right|$is constant. Since the velocity and acceleration vectors are orthogonal, their dot product is zero. That is we have

$v\left(t\right)·a\left(t\right)=0\Rightarrow r^{\text{'}}\left(t\right)·r^{\text{'}\text{'}}\left(t\right)=0\dots \dots .\left(1\right)$

Step 3. Proving: 

Consider

$\frac{d}{dt}\left({||r^{\text{'}}||}^2\right)$

$=\frac{d}{dt}\left(r^{\text{'}}·r^{\text{'}}\right)$

$=r^{\text{'}}\frac{d}{dt}\left(r^{\text{'}}\right)+r^{\text{'}}\frac{d}{dt}\left(r^{\text{'}}\right)$Derince $\Vert r{\Vert }^2=r·r$

$=r^{\text{'}}·r^{\text{'}\text{'}}+r^{\text{'}}·r^{\text{'}\text{'}}\frac{d}{dt}\left(r^{\text{'}}\right)=r^{\text{'}\text{'}}$

$=2r^{\text{'}}-r^{\text{'}\text{'}}$ add from equations(1) product of 2 function

$=2\left(0\right)$multiply

$=0$

$\frac{d}{dt}\left({||r^{\text{'}}||}^2\right)=0$.that means $||r^{\text{'}}||$ is a constant. That is, the magnitude of the velocity vector is

constant.

Thus the particle movesalong $r\left(t\right)$ at a constant speed.