Creating the Graph of a Position Function

In summary: The derivative of cos(x) is cos(x)+i*sin(x), so the derivative of cos(x) is 0 at x=pi and since the derivative of sin(x) is -sin(x)+i*cos(x), the derivative of cos(x) is 1 at x=pi.
  • #1
Love_to_Learn
13
0
1. I need to create a graph of a position function where a particle at point pi, has zero velocity, and positive acceleration.



2. Homework Equations f(x)=sin(x); f'(x)=cos(x); f''(x)=-sin(x)



3. Attempts at a Solution If I choose sin(x) as my function, then velocity is zero at pi. I am unclear if I have this next part right. Since the slope of the velocity function is increasing at pi, acceleration is therefore positive. I think I have that right. But I am not sure.
 
Physics news on Phys.org
  • #2
You're answer is right infront of you. You defined f'(x). We know f'(x) describes the slope of a curve at any given point. You want to know if the slope at the point pi is positive or negative, so compute f'(pi).
 
  • #3
Sorry for being dense, but I am just not real sure what you are getting at. If f(x) = sin(x), the value of pi at f'(x) is -1, but the slope is 0, so there is zero velocity. (Do I have it right so far?)

I am not understanding what it gets me to compute the value of f'(pi).

The value of f''(x) = 0, but function has a positive slope at that point, so the acceleration is increasing. (Do I have that right?)
 
  • #4
If you are plotting f'(x): the value at f'(pi) is the velocity and the acceleration is either the slope which you can evaluate from your plot, or which can be solved explicitly by finding f''(x) and evaluating at that value. In other words what is -sin(pi)?

The equation you have listed comes close to satisfying the required properties but not quite.
 
  • #5
Ok, so since the value of the second derivative at pi is a negative value, then the acceleration is negative.? So If I start with f(x)=cos(x), then f''(x)=1 so then the acceleration is positive. Right?
 
  • #6
check your work: what is sin (pi). (Make sure your calculator is in radians).

Let me save you some time by suggesting this: plot sin(x), cos(x),-sin(x) which are
f(x), f'(x), and f''(x) in this case. You can do the same with f(x)=cos(x) but the results will be the same.

With sines and cosines functions the first derivative will be 90 degrees phase shifted and the second derivative 180.

What this means is that the acceleration will always look like the mirror image of the position function.

There are many, many solutions to this problem, but you will have to make your initial function more complicated. One possibility would be set it up with two terms whose derivatives add up to a value of zero when x=pi, but have at least one non zero second derivatives at x=pi.
 
  • #7
By x you mean position in space?
Then the velocity is not the derivative in respect to x but in respect to time.
Maybe you mean something like this
y(t) - position function (gives position versus time)
y'(t) - velocity
y''(t) - acceleration.
Then you require that when y=pi, y'(t)=0 and y''(t)>0

Is this what you need?
 
  • #8
What we want is that f'(x) = -sin(x) so f''(x) = -cos(x). If we have this then the velocity is zero and the acceleration is positive.(It's at a maximum specifically)

What f(x) can we use to obtain this result?
 
  • #9
jegues said:
What we want is that f'(x) = -sin(x) so f''(x) = -cos(x). If we have this then the velocity is zero and the acceleration is positive.(It's at a maximum specifically)

What f(x) can we use to obtain this result?

You don't have anything like this. There is no speed or acceleration unless x varies in time.
If you actually mean "find a function of x so that the first derivative is zero and the second derivative is positive" then it makes sense. Not any derivative is a velocity.
If the question is about derivatives only, then f(x)=cos(x) satisfies the conditions;
f'(x)=-sin(x) = 0 at x=pi
f''(x)=-cos(x) =1 >0 at x=pi
 

Related to Creating the Graph of a Position Function

1. How do I create a graph of a position function?

To create a graph of a position function, you will need to plot the values of the function on a coordinate plane, where the horizontal axis represents time and the vertical axis represents position. You can use a graphing calculator or software to plot the points, or you can do it manually by inputting the values into a table and then plotting them on the graph.

2. What is a position function?

A position function is a mathematical equation that represents the position of an object as a function of time. It is typically denoted by the variable "s" and can be written as s(t) where "t" represents time. The output of the function is the position of the object at a specific time, and it is usually measured in units such as meters or kilometers.

3. What information do I need to create a graph of a position function?

In order to create a graph of a position function, you will need the equation of the function and the range of values for the independent variable (time). This information can be obtained from the problem or experiment that you are working on. You may also need to know the units of measurement for the time and position values in order to properly label the axes of the graph.

4. How can I interpret the graph of a position function?

The graph of a position function can provide valuable information about the motion of an object. The slope of the graph represents the velocity or speed of the object, while the area under the curve represents the displacement or change in position. The shape of the graph can also indicate the direction of motion (positive slope indicates movement in the positive direction, negative slope indicates movement in the negative direction).

5. Can I use a position function to predict future positions?

Yes, a position function can be used to predict future positions of an object if the function accurately represents the motion of the object. By plugging in a specific time value into the function, you can determine the position of the object at that time. However, this prediction may not be accurate if there are external factors that affect the motion of the object, such as friction or external forces.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
690
  • Introductory Physics Homework Help
Replies
28
Views
525
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
222
Replies
1
Views
224
  • Introductory Physics Homework Help
Replies
16
Views
776
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
29
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
757
Back
Top