Derivative notation uniformality

In summary: I don't think it was meant to be difficult to understand. If you are not sure where something comes from in the equation, you can always look it up. There is a guide to Newtonian notation that might make it easier to understand. It is called a "guide to uber correct and uniform notation." However, I am not sure where you would find this.
  • #1
aeb2335
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0

Homework Statement



I have elected to put this in the "math" section because it is primarily a math question however, please read to problem knowing that it came from an engineering dynamics textbook.

Given:

(as written)

A particle moving along y= x - (x^2 / 400) where x and y are in ft. If the velocity component in the x direction remains constant at Vx= 2 ft/s , determine the magnitudes of velocity and acceleration when x = 20 ft.


Homework Equations



V = ds/dt
A = dv/dt

The Attempt at a Solution



Ok, at first I became a little confused so I drew a Cartesian co-ordinate system and plotted the curve Y going up X horizontal. I then realized that its a path equation so it needs to be differentiated.

The problem becomes I don't really understand the notation

It should possible differentiate the path saying f(x,y)'= the velocity function

What I don't get is how you can use Leibniz's notation for this problem to solve it. After being frustrated at the lack of clarity I looked through the book and at the solution and it appears the book's notes on how to solve this problem and the solution uses Newtonian notation.

I apologize for not knowing how to make the dot character

Saying that (first line of solution)

ydot = xdot - (2x (xdot) / 400 )

The problem is that I tend to drop terms like the underlined xdot above because I am not really sure where it comes from. It seems to me that there is some implicit magic going on. I would have no problem in Leibniz notation and the reason this problem gave me pause was because it didn't appear to be easy to express in Leibinz notation.

Can the question be properly framed by saying given a curve S(x,y) such that

S(x,y)= x-(x^2/400)-y

then saying:

S(x,y)ds/dt=(x-(x^2/400)-y)ds/dt
V(x,y) = x ds/dt - (x^2 ds/dt / 400) - y ds/dt ?

would it be even better to say

S(x,y)= x(i) -(x^2(i)/400)-y(j) ?

Is there a guide or reference to Newtonian notation (other that wikipedia etc.) that makes it easy to understand?

Does Newtonian notation in this example treat X and Y as functions themselves and not variables which is why the x dot term exists? And if so would X and Y Always notate a function and directions would be explicitly in i,j,k ? Furthermore why would you treat X and Y as a functions in this equation when it also corresponds directly to the inputs of that function?

Also is there a guide to uber correct and uniform notation and/or an explanation of the mathematical and physical symbols.

Thanks so much
 
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  • #2
The problem says that the particle moves along a path with coordinates related according to y(t)=x(t) - (x(t)^2 / 400) (t is the time).
The components of velocity are the time derivatives of the coordinates:
[itex]v_x=\dot x[/itex], [itex] v_y=\dot y[/itex].
vx is given, vx=2 ft/s, and it is constant.
You get vy applying the chain rule for (y(x(t)). You did it already and got
[itex]v_y=\dot x-2x \dot x/400=v_x-2xv_x/400[/itex].

Substitute x=20 ft and vx=2 ft/s and you get vy at x=20 ft.

Do you know how to determine the magnitude of velocity or of any vector from the components?

The acceleration is the time derivative of the velocity. ax=0 as vx =constant. Differentiate [itex]v_y=v_x-2x v_x /400[/itex] to get ay.

ehild
 
  • #3
Everything makes sense except for the problem I guess. I get the "root sum square" concept etc. but I think what is confusing me is that the problem states that the particle moves along the path but does not explicitly define the fact that there is a t parameter in the equation. You yourself added the y(t) (which makes everything much much clearer and was not in the problem statement). Should I always assume that when given a path there is always a parameter of t? Or is this problem just written poorly?

thanks
 
  • #4
aeb2335 said:
what is confusing me is that the problem states that the particle moves along the path but does not explicitly define the fact that there is a t parameter in the equation. You yourself added the y(t) (which makes everything much much clearer and was not in the problem statement). Should I always assume that when given a path there is always a parameter of t? Or is this problem just written poorly?

If the particle moves it does in time and space. There is no motion without passing time. Whenever there is some reference to velocity or speed, time is involved.
If the path is given as a curve, relation between coordinates, time is not necessarily involved. You can find the distance between two points of the path, or the tangent of the path, without having any object moving along.

Yes, you can say that the problem was written poorly. But it happens in Physics problems quite often. Read the text carefully, and try to read the mind of the problem writer.

ehild
 
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  • #5
Thanks for your replies its starting to make sense I guess I am just lost in a sea of math/physics/engineering notation.

Would this be the correct way of expressing this problem in terms of leibniz notation?

Its over rigorous but I really want to understand the notation and what meaning it carries.

My attempt at reconciling the notation is below would it be

d(y(x))/dt= (x(t) -(x(t)^2/400))dx/dt ?

then how would you evaluate x(t) dx/dt and the other terms?

would x(t)dx/dt turn into dx/dt via a chain rule? which could then be expressed as xdot or simply just v?

then x(t)^2/400 would turn into 2x dx/dt /400 which simplifies to 2x dx/dt

to give an overall expression of

d(y(x)/dt = dx/dt -( 2x(dx/dt) /400) ?
 
  • #6
aeb2335 said:
Would this be the correct way of expressing this problem in terms of leibniz notation?

Its over rigorous but I really want to understand the notation and what meaning it carries.

My attempt at reconciling the notation is below would it be

d(y(x))/dt= (x(t) -(x(t)^2/400))dx/dt ?

d(y(x))/dt=d( x - (x^2 / 400))/dt

The derivative of a sum is the sum of the derivatives, d( x - (x^2 / 400))/dt=dx/dt-d(x^2/400)/dt.

The derivative of x(t)^2/400 is (2x(t)/400) dx/dt, so

dy/dt=dx/dt-(2x/400) dx/dt.

Isolate dx/dt:

so dy/dt=[1-(x/200)]dx/dt.

You need to calculate dy/dt when dx/dt , the velocity in the x direction is 2 ft/s and x=20 ft. ehild
 
  • #7
Ok so what this is saying is that you are differentiating a function y(x) w.r.t to time

But then it appears that you are treating x as a function of time and not as a variable so why wouldn't

y(x) = x(t) -(x(t)^2/400)

d(y(x))/dt= (x(t) -(x(t)^2/400))dx/dt

be correct?

Because at least to me d(y(x))/dt=d( x - (x^2 / 400))/dt seems like you are differentiating to some other variable not represented in the equation. Are we not using t as the parameter and therefore what we are differentiating with respect to?
 
  • #8
aeb2335 said:
Ok so what this is saying is that you are differentiating a function y(x) w.r.t to time

But then it appears that you are treating x as a function of time and not as a variable so why wouldn't

y(x) = x(t) -(x(t)^2/400)

d(y(x))/dt= (x(t) -(x(t)^2/400))dx/dt

be correct?

Why should it be correct?

x is a function of the time. Otherwise dx/dt has no sense.

y(t)=x(t)-x2(t)/400 is a composite function dependent on x, which is function of t.

I think you have learned about chain rule. What is the derivative of the function f(g(x)) = sin(x2), for example?

In the problem in your post, t is the independent variable and y=y(x(t)).

dy/dt=(dy/dx) (dx/dt).

dy/dx=1-2x/400. So dy/dt=(1-2x/400)(dx/dt).

ehild
 
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  • #9
ok got you thanks
 

FAQ: Derivative notation uniformality

1. What is derivative notation uniformality?

Derivative notation uniformality refers to the consistent use of notation when representing derivatives in mathematics. This includes using the same symbols and conventions for differentiation, such as using prime notation or Leibniz notation.

2. Why is derivative notation uniformality important?

Derivative notation uniformality is important because it allows for clear and consistent communication in mathematical equations. It also helps to avoid confusion and errors when performing calculations or solving problems involving derivatives.

3. How does derivative notation uniformality impact the understanding of calculus?

Derivative notation uniformality is a fundamental aspect of calculus, as it allows for a standardized way of representing and manipulating derivatives. This consistency in notation makes it easier for students to understand and apply concepts in calculus.

4. Are there any exceptions to derivative notation uniformality?

While there are generally accepted conventions for derivative notation, there may be variations or exceptions in certain contexts or specific fields of mathematics. For example, in physics or engineering, different notation may be used for derivatives to represent specific physical quantities.

5. How can one ensure derivative notation uniformality in their work?

To ensure derivative notation uniformality, one should be familiar with the commonly accepted notation for derivatives and use it consistently in their work. It can also be helpful to refer to established textbooks or resources for guidance on notation. Additionally, proofreading and double-checking for consistent notation can help to catch any errors or deviations.

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