I have been trying to plot two surfaces on a 3D graph using Mupad.
The first is created by the function:
(x-1)^2 / 4 + (y+2)^2 / 9 + (z+1)^2 = 9
The Second surface I have no problems with apart from some scaling issues:
4(x-1)^2 + 4(y+2)^2 = 81
Additionally, I want to find the intersection of the two surfaces.
My problem here is with the first surface as I can't get it to work using any of the 'plot' functions i am familiar with.
can anyone help me please?
first one is not a function, it is an ellipsoid, so isolate z and define the two halves as +/- sqrt(whatever)
Related
I am using Insertion and Merge sort and trying to indicate the intersection point on the graph below. I found a possible solution on rosettacode, but a bit confusing for a newbie like me... Where should I look, do you guys know how to indicate it in Julia?
The output would is like
I am trying to show something like this
Your data has some noise, which means there might be more than one intersection point (all close). Assuming no noise, just two monotonically increasing functions, this should work:
using Plots
x = 1:5
y1 = x .+ 5
y2 = 3x
idx = argmin(abs.(y1-y2))
plot(x, [y1, y2], label=["y1", "y2"], markershape=:circle)
scatter!([x[idx]], [y1[idx]], label="intersection", markershape=:circle, markersize=10)
Note that this method gives you 1/4 possible points closest to the intersection. The intersection itself is not defined for functions only defined from discrete points. By changing the order of y1 and y2 in the code above you can get all 4 points. For fine enough discretization using one of those 4 points might be good enough for your needs.
A better approximation would be to use those 4 points to fit 2 lines, for y1 and y2, and find the intersection of those two lines analytically (e.g. with the equations y1=m1*x+b1, ...). But I leave this to you ;)
I am back at college learning maths and I want to try and use some this knowledge to create some svg with d3.js.
If I have a function f(x) = x^3 - 3x^2 + 3x - 1
I would take the following steps:
Find the x intercepts for when y = 0
Find the y intercept when x = 0
Find the stationary points when dy\dx = 0
I would then have 2 x values from point 3 to plug into the original equation.
I would then draw a nature table do judge the flow of the graph or curve.
Plot the known points from the above and sketch the graph.
Translating what I would do on pen and paper into code instructions is what I really could do with any sort of advice on the following:
How can I programmatically factorise point 1 of the above to find the x-intercepts for when y = 0. I honestly do not know where to even start.
How would I programmatically find dy/dx and the values for the stationary points.
If I actually get this far then what should I use in d3 to join the points on the graph.
Your other "steps" have nothing to do with d3 or plotting.
Find the x intercepts for when y = 0
This is root finding. Look for algorithms to help with this.
Find the y intercept when x = 0
Easy: substitute to get y = 1.
Find the stationary points when dy\dx = 0
Take the first derivative to get 3x^2 - 12x + 9 and repeat the root finding step. Easy to get using quadratic equation.
I would then have 2 x values from point 3 to plug into the original
equation. I would then draw a nature table do judge the flow of the
graph or curve. Plot the known points from the above and sketch the
graph.
I would just draw the curve. Pick a range for x and go.
It's great to learn d3. You'll end up with something like this:
https://maurizzzio.github.io/function-plot/
For a cubic polynomial, there are closed formulas available to find all the particular points that you want (https://en.wikipedia.org/wiki/Cubic_function), and it is a sound approach to determine them.
Anyway, you will have to plot the smooth curve, which means that you will need to compute close enough points and draw a polyline that joins them.
Doing this, you are actually performing the first steps of numerical root isolation, with such an accuracy that the approximate and exact roots will be practically undistinguishable.
So an easy combined solution is to draw the curve as a polyline and find the intersections with the X axis as well as extrema using this polyline representation, rather than by means of more sophisticated methods.
This approach works for any continuous curve and is very easy to implement. So you actually draw the curve to find particular points rather than conversely as is done by analytical methods.
For best results on complicated curves, you can adapt the point density based on the local curvature, but this is another story.
I am trying to generate a plot which uses arrows as markers in Gnuplot. These arrows I want to turn in a specific angle which I know. So I have value triples of x1 ... xn, y1...yn, alpha1...alphan. Sorry, I wasn't able to include a pic from my hard drive to illustrate what I want to achieve.
Basically, for every (15th or so) x-y pair, the marker should be an arrow which uses a certain angle.
The measured data is tightly packed so I suppose I will have to define an increment between the markers. The length of the arrow can be the same all over.
I would appreciate your ideas.
Gnuplot has a plot mode with vectors that is what you want
Given that your file has the following format, x y angle and assuming that
your angle is in radians, you have to take into account that
with vectors requires 4 parameters, namely x y dx dy where dx
and dy are the projections of the lenght of the arrow.
this draws only the arrows, if you want a line you have to make
two passes on the data.
you want to draw an arrow for a data point over, say, 10 points.
That said, I'd proceed like this
dx(a) = 0.2*cos(a) # 0.2 is an arbitrary scaling factor
dy(a) = 0.2*sin(a)
# this draws the arrows
plot 'mydata.dat' every 10 using 1:2:(dx(a)):(dy(a)) with vectors
# this draws the line
plot 'mydata.dat'
You may want to use help plot to find the detailed explanation of all the parameters that you can apply to a with vectors plot.
Credits: An article on the gnuplotting site
t = 0:%pi/50:10*%pi;
plot3d(sin(t),cos(t),t)
When I execute this code the plot is done but the line is not visible, only the box. Any ideas which property I have to change?
Thanks
The third argument should, in this case, be a matrix of the size (length arg1) x (length arg2).
You'd expect plot3d to behave like an extension of plot and plot2d but it isn't quite the case.
The 2d plot takes a vector of x and a vector of y and plots points at (x1,y1), (x2,y2) etc., joined with lines or not as per style settings. That fits the conceptual model we usually use for 2d plots - charting the relationship of one thing as a function of another, in most cases (y = f(x)). THere are other ways to use a 2d plot: scatter graphs are common but it's easy enough to produce one using the two-rows-of-data concept.
This doesn't extend smoothly to 3d though as there are many other ways you could use a 3d plot to represent data. If you gave it three vectors of coordinates and asked it to draw a line between them all what might we want to use that for? Is that the most useful way of using a 3d plot?
Most packages give you different visualisation types for the different kinds of data. Mathematica has a lot of 3d visualisation types and Python/Scipy/Mayavi2 has even more. Matlab has a number too but Scilab, while normally mirroring Matlab, in this case prefers to handle it all with the plot3d function.
I think of it like a contour plot: you give it a vector of x and a vector of y and it uses those to create a grid of (x,y) points. The third argument is then a matrix whose dimensions match those of the (x,y) grid holding the z-coordinates of each point. The first example in the docs does what I think you're after:
t=[0:0.3:2*%pi]';
z=sin(t)*cos(t');
plot3d(t,t,z);
The first line creates a column vector of length 21
-->size(t)
ans =
21. 1.
The second line computes a 21 x 21 matrix of products of the permutations of sin(t) with cos(t) - note the transpose in the cos(t') element.
-->size(z)
ans =
21. 21.
Then when it plots them it draws (x1,y1,z11), (x1,y2,x12), (x2,y2,z22) and so on. It draws lines between adjacent points in a mesh, or no lines, or just the surface.
I am writing an regression algorithm which tries to "capture" points inside boxes. The algorithm tries to keep the boxes as small as possible, so usually the edges/corners of the boxes go through points, which determines the size of the box.
Problem: I need graphical output of the boxes in R. In 2D it is easy to draw boxes with segments(), which draws a line between two points. So, with 4 segments I can draw a box:
plot(x,y,type="p")
segments(x1,y1,x2,y2)
I then tried both the scatterplot3d and plot3d package for 3D plotting. In 3D the segments() command is not working, as there is no additional z-component. I was surprised that apparently (to me) there is no adequate replacement in 3D for segments()
Is there an easy way to draw boxes / lines between two points when plotting in three dimensions ?
The scatterplot3d function returns information that will allow you to project (x,y,z) points into the relevant plane, as follows:
library(scatterplot3d)
x <- c(1,4,3,6,2,5)
y <- c(2,2,4,3,5,9)
z <- c(1,3,5,9,2,2)
s <- scatterplot3d(x,y,z)
## now draw a line between points 2 and 3
p2 <- s$xyz.convert(x[2],y[2],z[2])
p3 <- s$xyz.convert(x[3],y[3],z[3])
segments(p2$x,p2$y,p3$x,p3$y,lwd=2,col=2)
The rgl package is another way to go, and perhaps even easier (note that segments3d takes points in pairs from a vector)
plot3d(x,y,z)
segments3d(x[2:3],y[2:3],z[2:3],col=2,lwd=2)