# Vector calculus velocity question

• lonewolf219
In summary, In an example my professor gave in class, a particle moves (in a straight line) from point A to point B (starting from t=0) and traveling at 6 m/s. The solution included a unit vector. Does that mean all vector equations describing velocity must include a unit vector? No, they don't.
lonewolf219
Hi, I am not fully understanding how to express velocity as a vector equation. In an example my professor gave in class, a particle moves (in a straight line) from point A to point B (starting from t=0) and traveling at 6 m/s. The solution included a unit vector. Does that mean all vector equations describing velocity must include a unit vector?

The solution was (initial point)+[<vector>/unit vector](speed)

No, they don't. From what I've perceived from your example:

$\vec{r}$=$\vec{r}$0+$\vec{v}$t → $\vec{r}$=$\vec{A}$+6t$\hat{r}$

Because when t=0, $\vec{r}$0=$\vec{A}$.

And $\vec{r}$B=$\vec{A}$+6tB$\hat{r}$, with $\vec{r}$B=$\vec{B}$. Taking one derivative:

$\vec{v}$=6$\hat{v}$

Being $\hat{v}$ a unit vector (normed vector), like $\hat{r}$.
It's not obligatory to represent, in a vector equation describing velocity, the unit vector $\hat{v}$. You can always replace it by $\frac{\vec{v}}{|\vec{v}|}$, which is the same thing.
However, it's advised to use them because they facilitate geometrical description of equations.
For example, the Newton's law of gravitational force can be written like this:

$\vec{F}$g=-G(m1m2)/(r)3 $\vec{r}$

Instead of the usual:

$\vec{F}$g=-G(m1m2)/(r)2 $\hat{r}$=|$\vec{F}$g|$\hat{r}$

The first one may lead you to think that $\vec{F}$g $\propto$ r-3 but it's not, as you certainly know.
I hope I've cleared your doubt about unit vectors, if not other users may, or else explain yourself better.

Thanks for your explanation, Mathaholic. If A=(1, 2, 3), and the vector parallel to motion is <4, 5, 6>, and the speed is 6m/s, is it incorrect to express the velocity as

r'(t)=A+<4, 5, 6>6t

If there is an initial point, a speed and a vector, what am I missing if this is incorrect?

lonewolf219 said:
Thanks for your explanation, Mathaholic. If A=(1, 2, 3), and the vector parallel to motion is <4, 5, 6>, and the speed is 6m/s, is it incorrect to express the velocity as

r'(t)=A+<4, 5, 6>6t

If there is an initial point, a speed and a vector, what am I missing if this is incorrect?

If |$\vec{v}$| is constant through time and $\vec{v}$=$\vec{const}$ (straight line, as you said before), then the vector velocity is expressed as:

$\vec{v}$=|$\vec{v}$|$\hat{v}$

Now, if there's a vector parallel to the movement (thus, to the vector velocity) and the speed is 6 m/s. Firstly, you note that the magnitude of (4,5,6) is not equal to 6, otherwise, that'd be your vector velocity. But if (4,5,6) is parallel to the velocity, then:

λ(4,5,6)=$\vec{v}$

You need to determine the x,y,z coordinates of $\vec{v}$ using that expression and knowing its magnitude (6 m/s) and most importantly, calculating λ.
And then you'll be able to express the vector velocity as:

$\vec{v}$=vx$\hat{x}$+vy$\hat{y}$+vz$\hat{z}$

And yes, it's incorrect to express the vector velocity as:

$\vec{r'(t)}$=A+(4,5,6)6t

The RHS has units of m/s and LHS has at least units of m (but never m/s!). Secondly, you cannot sum points (A) with vectors of velocity, you can only do that with vectors of position ($\vec{r}$).

Hi Mathoholic if you're still around! Would the correct vector equations be

r'(t)=[<4,5,6>/square root77](6t)? That would make the equation v(t)=speed(velocity unit vector)?

r(t)= (1,2,3)+[<4,5,6>/square root77](6t)

Hope I've got it?

lonewolf219 said:
Hi Mathoholic if you're still around! Would the correct vector equations be

r'(t)=[<4,5,6>/square root77](6t)? That would make the equation v(t)=speed(velocity unit vector)?

r(t)= (1,2,3)+[<4,5,6>/square root77](6t)

Hope I've got it?

The vector position is well written. As for the vector velocity, not so much. If the speed is constant through time, then the vector velocity doesn't have the parameter time (t), as it is also the 1st derivative of the position. As for the rest, it's all good.

Thanks, yes, definitely, the t would not be there if I just took the derivative... appreciate your patience! See you around!

## 1. What is vector calculus?

Vector calculus is a branch of mathematics that deals with the study of vector fields and their derivatives. It involves the use of vector operations such as dot product, cross product, and gradient to solve problems related to motion and forces.

## 2. How does vector calculus relate to velocity?

Velocity is a vector quantity that describes the rate of change of an object's position. Vector calculus is used to calculate the velocity of an object by taking the derivative of its position vector with respect to time.

## 3. What is the difference between scalar and vector fields?

A scalar field is a function that assigns a single value (scalar) to every point in space, while a vector field assigns a vector to every point in space. Velocity is an example of a vector field, while temperature is an example of a scalar field.

## 4. How is the gradient used in vector calculus?

The gradient is a vector operation that represents the rate of change of a scalar field. In vector calculus, it is used to find the direction and magnitude of the steepest ascent or descent of a scalar field at a specific point.

## 5. What are some real-world applications of vector calculus?

Vector calculus has many applications in various fields, including physics, engineering, and computer graphics. It is used to study motion and forces, electric and magnetic fields, fluid dynamics, optimization problems, and 3D modeling and animation, to name a few.

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