Drawing a Tangent to a Spline

I am trying to draw a tangent at a point to the spline curve obtained through the interp1() function, however, I am having trouble with not being able to find the correct point in the gradient array corresponding to the point in the data array. Here is my code:

x = ([209, 907, 346, 908, 246, 457, 235, 232, 400]).^-1;   
y = log10([9.6e20, 3.57e20, 3.3e20, 3.7e20, 2.7e21, 6.9e21, 5.7e21, 3.5e21, 7.8e22]);
xq = linspace(x(1), x(9), 20000) 
    yq = interp1(x,y,xq, 'spline') 
    plot(xq,yq,'blue')
    
m = gradient(yq, xq);
tangent1=m(xq(3))*(xq-xq(3))+y(3)

I get the error: 'Array indices must a positive integer or a logical value'

Please help.

Thanks.

Answer 1

[Bjorn Gustavsson]

When you calculate the gradient:

m = gradient(yq, xq);

Then your m will have the same dimension as your xq and yq arrays. So when you want to calculate the tangent line from the point [xq(3),yq(3)] you need to use the corresponding element of your gradient array - that is the 3rd. You just have a typo in your code:

tangent1=m(xq(3))*(xq-xq(3))+y(3)

Should be:

tangent1 = m(3)*(xq-xq(3))+yq(3);

There you'll take the 3rd element out of m, and it should really be the 3rd element out of yq not out of y.

You should also avoid making it a habit of having numbered variables like tangent1 (why 1? At least make it 3?), start adapting to storing things like that in arrays that you can index to with a number later in your code instead of having numbered variables. Here you could chose between a cell-array for your tangents (+: you can use tangent-lines with different number of elements. -: slightly more cumbersome to use.) or a simple matrix (+: easier to compare different tangents, and displaying them at once. -: all better be same sized.)

HTH

[Tb]

@Bjorn Gustavsson Thank you for your kind response and advice. However, I am trying to find the gradient at the third index of the original array x, and since the array xq has 20000 elements, xq(3) != x(3). Am I misunderstanding something in your solution, or don't we have to find the element corresponding to x(3) in xq somehow?

[Bjorn Gustavsson]

Yes, either that (look at the help and documentation of find, but that requires you to have x(3) in the array xq, and I dont know you necessarily have that.) If that is not the case, you could try to use interp1 again to reinterpolate the gradient-array to get you a value of m at x(3) - since you know both xq and m, right? Since this is a homework type task I'll only give you this hint, and you'll know what to change in the tangent-line equation to go from xq(3) to x(3).

[Tb]

Thank you. However, @Bjorn Gustavsson after trying for the past hour or so, and playing around with find(), it returns a 1x0 empty vector, this is what I wrote:

k = find(m==x(3))

[Bjorn Gustavsson]

Well, then you don't have x(3) in hte xq array. Can you use xq and m to interpolate a value of the gradient at the point x(3)?

[Tb]

Oh yes that makes sense, and that works now as well. Thanks a lot.

[Bjorn Gustavsson]

You're welcome.