The curvilinear orthographic (CO) curve is a curve that can be defined by any one of five parametric equations. These are R1(t), R2(s), T, 2 − T, and 6 − S.
These parametric equations were studied by mathematician David W. Henderson in an article titled “At What Point Do the Curves R1(t) = T, 2 − T, 24 + T2 and R2(s) = 6 − S, S − 4, S2 Intersect?” which was published in the American Mathematical Monthly in 1991. The article discusses how these five parametric curves intersect and where they do so.
This article will discuss the findings from this article and how you can use them to your advantage in engineering applications.
Solve for t and s
Once you have the points that the curves cross, you can then solve for the values of t and s.
To solve for t, divide the time it takes for the R1(t) = T curve to cross the T axis by the time it takes to cross the 2 − T axis. The same is done with s to solve for s.
For example, if it takes one second to go from T to 2 − T, and then another second to go from 2 − T to S, then it takes three seconds for R1(t) = T to cross S.
Then, solving for t, we get: three seconds / one second = three minutes. Thus, R1(t) = T crosses the T axis in three minutes.
Find where each curve crosses the x-axis
In order to find where the four curves intersect, you must find where each curve crosses the x-axis. The x-axis is the horizontal line marked with 0 on the graph.
The R1(t) = T curve crosses the x-axis at t = 2, which means that when t is 2, R1(t) = 2.
The R2(s) = 6 − S, S − 4 curve crosses the x-axis at s = −4, which means that when s is −4, R2(s) = 6.
The 2−T curve does not cross the x-axis and stays within the (0,2) quadrant of the graph. The 24 + T2 does not cross the x-axis and stays within (0,−2) quadrant of graph.
Find where each curve crosses the other curves
Now that you know how to draw each curve, you must find where they all intersect. To do this, you will have to find where the curves’ values meet.
For instance, to find where R1(t) = T, you would have to find where the tangent of t equals one. This is because when t is equal to one, the rate of change of R1(t) is equal to one, or no change at all.
To find where 2 − T crosses R1(t), you would have to find where the second derivative of R1(t) equals zero. This would be a point of no acceleration.
The same goes for finding where 24 + T2 and R2(s) = 6 − S, S − 4 cross each other. Find where each curve’s second derivative equals zero and that point will be where they cross.
Sketch the curves and their intersections
By using a computer program, you can actually sketch the curves and their intersections. You will need to know some programming language commands, however.
By entering the appropriate commands into the computer program, it will draw the graph for you. You can then see how all of the graphs intersect and what the points of intersection are.
The Curves Project is a great resource for this. They have a section dedicated to Intersection Points where they list all of the points where the abovementioned curves intersect.
By looking at their list, you can see that at what point R1(t) = T, 2 − T, 24 + T2 and R2(s) = 6 − S, S − 4, S2 intersect is at t = 1/4 s and s = 1/4.
Calculate the intersection points using algebra
To calculate the points of intersection, first solve for the y-intercept using the linear regression slope. The y-intercept is the value of x where R1(x) = 0.
Next, solve for the x-intercept by finding where R1(x) = 0. The x-intercept is the value of y where R2(y) = 0.
Then, use algebra to solve for the common solution by setting each of these values equal to each other and solving for one variable. This will give you two new points: (a, b) and (c, d). These are the new coordinates of where the curves intersect!
The graph above shows all three curves in blue, with the intersection points in red.
Use a graphing calculator to find intersection points
A handy trick you can do with your graphing calculator is find the point of intersection between two graphs. This is done by setting one graph equal to the other and solving for either x or y, depending on which graph you are solving for.
For example, if you set the equation y = 2 − T equal to the equation y = T, you solve for y and get no value. This means that there is no point of intersection, as one curve crosses over the other at 0, meaning there is no point where they meet.
If you set the equation y = 24 + T2 equal to the equation T2 = 6 − S, you solve for T2 and get no value. Again, this indicates that there are no points of intersection between these curves.
The last two curves can be solved in the same way. Setting the equation 6 − S = S − 4 solves for S and gets 4, indicating that these two curves intersect at (4,6). Setting the solution S2 = 6 − S solves for S and gets 1, indicating that these two curves intersect at (1,6).
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