# Help with parametric equations? What does slope of velocity mean?

• April 25th 2013, 03:34 PM
LLLLLL
Help with parametric equations? What does slope of velocity mean?
I'm working on parametric equation and motion along curve right now and I get confused by the concept of slope of velocity.
My book says that (dy/dt)/(dx/dt) is the slope of velocity. And it's also the slope of the parametric curve (position curve?) I got confused by the phrase "slope of velocity". From what I've read, it doesn't seem to be the same thing as "slope of velocity curve". Can someone please explain this concept to me?
Also, if slope of velocity = slope of position = dy/dx, what's gonna be the slope of velocity curve? d^2y/d^2x? Or d^2y/dx^2?
If it is said that velocity and acceleration are perpendicular, does it mean that their slopes multiply to -1? Or the slopes of their curve multiply to -1?

• April 25th 2013, 04:28 PM
majamin
Re: Help with parametric equations? What does slope of velocity mean?
Quote:

Originally Posted by LLLLLL
I'm working on parametric equation and motion along curve right now and I get confused by the concept of slope of velocity.
My book says that (dy/dt)/(dx/dt) is the slope of velocity. And it's also the slope of the parametric curve (position curve?) I got confused by the phrase "slope of velocity". From what I've read, it doesn't seem to be the same thing as "slope of velocity curve". Can someone please explain this concept to me?
Also, if slope of velocity = slope of position = dy/dx, what's gonna be the slope of velocity curve? d^2y/d^2x? Or d^2y/dx^2?
If it is said that velocity and acceleration are perpendicular, does it mean that their slopes multiply to -1? Or the slopes of their curve multiply to -1?

"Slope of velocity" implies acceleration. The vector $\left(\frac{dx}{dt},\frac{dy}{dt}\right)$ is the velocity vector, and $\frac{\frac{dx}{dt}}{\frac{dy}{dt}}$ is the slope of the tangent of the parametric curve at time t. The magnitude of velocity $|\vec{v}(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy} {dt}\right)^2}$. If we want to calculate acceleration, then $\vec{a}(t)=\left(\frac{d^2x}{dt^2},\frac{d^2y}{dt^ 2}\right)$. In general, velocity and acceleration are not orthogonal (perpendicular), but the unit vector of velocity and its derivative (the component of acceleration perpendicular to velocity) is. In this case, the product of the slopes would be -1 (true for any perpendicular slopes).
• April 25th 2013, 11:31 PM
LLLLLL
Re: Help with parametric equations? What does slope of velocity mean?
Quote:

Originally Posted by majamin
"Slope of velocity" implies acceleration. The vector $\left(\frac{dx}{dt},\frac{dy}{dt}\right)$ is the velocity vector, and $\frac{\frac{dx}{dt}}{\frac{dy}{dt}}$ is the slope of the tangent of the parametric curve at time t. The magnitude of velocity $|\vec{v}(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy} {dt}\right)^2}$. If we want to calculate acceleration, then $\vec{a}(t)=\left(\frac{d^2x}{dt^2},\frac{d^2y}{dt^ 2}\right)$. In general, velocity and acceleration are not orthogonal (perpendicular), but the unit vector of velocity and its derivative (the component of acceleration perpendicular to velocity) is. In this case, the product of the slopes would be -1 (true for any perpendicular slopes).

Hi majamin, thanks for you rely!
I'm still a little confused on some points. When you say "the product of the slopes would be -1 ", do you mean velocity's derivative and acceleration's derivative, since you said "Slope of velocity" implies acceleration"?
Regarding the slope of velocity, I had the same idea as you but my book seems to have a different say. Here's how it explains it (rephrased):
Let's say v = (3,5) = 3i + 5j
Graph on a paper 3 units right and 5 units up. The slope = Δy/Δx = 5/3.
So dy/dx = slope of velocity as well as slope of the curve
• April 26th 2013, 07:47 AM
majamin
Re: Help with parametric equations? What does slope of velocity mean?
Quote:

Originally Posted by LLLLLL
Hi majamin, thanks for you rely!
I'm still a little confused on some points. When you say "the product of the slopes would be -1 ", do you mean velocity's derivative and acceleration's derivative, since you said "Slope of velocity" implies acceleration"?
Regarding the slope of velocity, I had the same idea as you but my book seems to have a different say. Here's how it explains it (rephrased):
Let's say v = (3,5) = 3i + 5j
Graph on a paper 3 units right and 5 units up. The slope = Δy/Δx = 5/3.
So dy/dx = slope of velocity as well as slope of the curve

If y = f(x), then dy/dx is both the derivative and the slope. Velocity and acceleration of a parametrized path are not perpendicular in general. Velocity is calculated from the dervative of position, but it is still a vector, and that's why we calculate $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ to get dy/dx (the slope of the tangent to the path at time t). I do not mean the derivative of velocity (which is acceleration) and the derivative of acceleration (which is not typically considered). If you are given a position vector $\vec{u}(t)$, then the velocity $\vec{u}^\prime (t)$ is not a slope. It is a vector that gives you the horizontal and vertical components of velocity. The slope of this vector is indeed $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. Acceleration is $\vec{u}^{\prime\prime} (t)$, but this is not necessarily perpendicular to velocity (if the path is a circle, for instance, then it is). If they are, then, yes, the product of the slopes is -1.