Find the value of the number LaTeX: a a such that the families of curves LaTeX: y=\left(x+c\right)^{-1} y = ( x + c ) − 1 and LaTeX: y\:=\:a\le​ft(x\:+\:k​\right)^{\​frac{1}{3}​} y = a ( x + k ) 1 3 are orthogonal trajectories.

Find the value of the number a such that the families of curves y=(x+c)−1and y=a(x+k)13 are orthogonal trajectories.

im trying to write htis on matlab to have it solved can you help me

[Mauricio Alvarez]

the y=a(x+k)^1/3 thats the second y equation thats incorrect on the top

[Mauricio Alvarez]

then the first is supposed to be y=(x+c)^-1

[Walter Roberson]

Find the value of the number a such that families of curves and are othogonal trajectories

Answer 1

[Pranav Verma]

Hi Mauricio,

As per my understanding, you intend to solve the above mentioned system of equations pertaining to the specified condition of trajectories being orthogonal. I'd suggest using the diff and solve function in MATLAB for achieving the above task.

Please refer to syms documentation also to use diff and solve.

Also refer to the below discussions on the same lines:

https://www.mathworks.com/matlabcentral/answers/512447-orthogonal-trajectories-in-matlab

https://www.mathworks.com/matlabcentral/answers/559403-find-the-value-of-a?s_tid=answers_rc1-1_p1_Topic

Thanks

[Gary Shelley]

As a staff member this is a horrible answer. I have been going through your forums and help sections for 8 hours trying to figure out a similar problem and there is 0 help.

[John D'Errico]

@Gary Shelley Sorry, but this is NOT a horrible answer. The question posted was clearly homework, and we try not to do homework for people on this forum. In fact, this was a good answer, in that it points the person in a reasonable direction.

[Walter Roberson]

Orthoganal implies that the angle between angle between the two is 90 degrees. The angle of the original item is arctan of the tangent. Tangent is derivative. Derivative in MATLAB is diff() . So you have to start with diff() of the function, arctan, add pi/2, tan .

If you have calculated the derivative as y/x then tan(atan(y/x) + pi/2) works out as -x/y so you can skip the tan and arctan and go directly to taking the diff and calculating -1/diff as the angle of the orthoganal.

Now you equate the -1/diff and the second function and solve() for the parameter.

The essential MATLAB functions to use are diff() and solve() ... just like @Pranav Verma said.

... Unless you want to count multiplying by negative 1 or division as an essential MATLAB function that people need to call out in their volunteer explanations ?

[Carlos Guerrero García]

I know that the following code is only a visual interpretation, buit I think it would be nice for a visual approach to the problem

[x,y]=meshgrid(-5:0.1:5); % An adecuate region for the visualization
for a=-3:0.01:3; 
    contour(x,y,x-1./y,20,'r'); % Plotting the curves of the first family
    hold on; % For do not clear the figure
    contour(x,y,(y/a).^3-x,[-10:10],'b'); % Plotting the curves of the second family for each value of the parameter
    axis([-5 5 -5 5]); axis equal; % For the same scale
    title(['a= ',num2str(a)]); 
    drawnow;
    hold off;
end

So, while you're visualizing the animation, try to find the answer to the following cuestion: WHEN the blue lines are orthogonal to the red ones ?