Graphing a function V.S. another function

  • Thread starter Thread starter GreenPrint
  • Start date Start date
  • Tags Tags
    Function Graphing
GreenPrint
Messages
1,186
Reaction score
0

Homework Statement



Hi,

I'm really use to graphing a function vs x, were the x-axis would be the input to the function and the output would be the vertical function, were at each point on the graph would be the output evaluated at that particular x value

So I'm learning how to use MATLAB for my degree and the below is a exercise from my book

x_x(t) = t V_0 cos(theta)
x_y(t) = t V_0 sin(theta) + (a t^2)/2

were
x_x(t) = horizontal displacement of a projectile as a function of time in
x_y(t) = vertical displacement of a projectile as a function of time
t = 0:.01:20 s (all the values that occur by increments of .01 from 0 to 20)
theta = pi/4
a = -9.8 m/s^2

the particular exercise had me plotting x_y(t) V.S. x_x(t) on a graph, this is when I realized I had never done such a thing before were the x-axis is not the input and the y-axis is the output of some function evaluated at that particular x value... So I was wondering what exactly is the meaning of this graph, it's the vertical displacement that occurs at a particular horizontal displacement? Sense there evaluated at the same time values? but I was wondering if there was any way to come up with the function that describes the plot?

Homework Equations


The Attempt at a Solution

 
Physics news on Phys.org
These are called parametric equations.

Square both equations. Then add them. Simplify using the pythagorean identity. Try to recognize the resulting single equations as a circle = something.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top