# A x ds = v x dv or a • ds = v • dv?

• A
• Iqminiclip
In summary: This is because in curvilinear motion, the acceleration is not necessarily perpendicular to the velocity, so the dot product takes into account the component of acceleration in the direction of displacement. In summary, the relation ads = vdv is not applicable in 3D curvilinear motion. Instead, the dot product equality a • ds = v • dv must be used to accurately describe the relationship between acceleration, displacement, and velocity. This is because in curvilinear motion, the acceleration may not be perpendicular to the velocity, so the dot product accounts for this.

#### Iqminiclip

I have this question for 3D curvilinear motion of particles. I know that the relation ads = vdv is commonly used in rectilinear physics, but what about in 3D curvilinear motion?

Would it be a x ds = v x dv or a • ds = v • dv for describing curvilinear motion?

It will be the latter - a dot product. Consider uniform circular motion, in which the acceleration is always perpendicular to the velocity. ds will be zero because the speed is constant, but dv will not, as the velocity is changing direction. So the simple multiplicative equality will not hold because the LHS is zero but the RHS is not. But the dot product equality will hold, as both sides will be zero (because dv is perpendicular to v).

andrewkirk said:
It will be the latter - a dot product. Consider uniform circular motion, in which the acceleration is always perpendicular to the velocity. ds will be zero because the speed is constant, but dv will not, as the velocity is changing direction. So the simple multiplicative equality will not hold because the LHS is zero but the RHS is not. But the dot product equality will hold, as both sides will be zero (because dv is perpendicular to v).
Hi sir, I was wondering why ds would be zero? Wouldn't a particle in circular motion have some sort of change in displacement over time? (e.g at 90° or so)

Iqminiclip said:
Hi sir, I was wondering why ds would be zero? Wouldn't a particle in circular motion have some sort of change in displacement over time? (e.g at 90° or so)
Ah, I was interpreting your s as meaning 'speed' (=|v|) but based on your response, I see you mean it to refer to displacement.

In that case the equation has to be a dot product, not a multiplication, as one cannot multiply vectors. We have an equation for each of the three dimensions:

\begin{align}
a_1\,ds_1 &= v_1\,dv_1\\
a_2\,ds_2 &= v_2\,dv_2\\
a_3\,ds_3 &= v_3\,dv_3
\end{align}

Adding them together, we can represent this as:
$$\vec a\cdot \vec {ds} = \vec v\cdot \vec {dv}$$

## 1. What is the meaning of the equation a x ds = v x dv or a • ds = v • dv?

This equation is known as the Fundamental Theorem of Calculus, and it represents the relationship between the rate of change of a variable (a or v) and the total change of another variable (ds or dv).

## 2. How is this equation used in science?

This equation is used in many scientific fields, such as physics, engineering, and biology, to calculate important quantities like velocity, acceleration, and displacement.

## 3. What do a, ds, v, and dv stand for in this equation?

In this equation, a represents acceleration, ds represents displacement, v represents velocity, and dv represents change in velocity.

## 4. Is this equation only applicable to one-dimensional motion?

No, this equation can be used for both one-dimensional and multi-dimensional motion. In multi-dimensional motion, the variables would represent the components in each direction (e.g. x, y, and z).

## 5. Can this equation be derived from other mathematical principles?

Yes, this equation can be derived from the chain rule of calculus, which states that the derivative of a composite function is equal to the product of the derivatives of its individual components.