how to find the length of interpolated graph?

I created a graph that would go through these specific x and y values using the interp1 function. now I can't figure out how to find the length of the graph. I pasted the code for how I was trying to find the length but I recieve an error message saying Array indices must be positive integers or logical values.

x=[0 .6 1 1.3 1.4 1.8 2];

y=[0 .4 1.2 1.7 .8 1.2 1.4];

xq=0:.1:2;

vq=interp1(x,y,xq,'spline');

plot(x,y,'co',xq,vq,'k*--')

i=0:.1:2;

l=sqrt(x(i)^2+y(i).^2)

Answer 1

[Star Strider]

The length of the graph (interpolated values) is the length of the interpolation vector ‘xq’ and here is a (1x21)

vector.

x=[0 .6 1 1.3 1.4 1.8 2];
y=[0 .4 1.2 1.7 .8 1.2 1.4];
xq=0:.1:2;

vq=interp1(x,y,xq,'spline')

vq = 1×21

0 0.2431 0.3724 0.4203 0.4193 0.4017 0.4000 0.4466 0.5738 0.8142 1.2000 1.6951 1.9883 1.7000 0.8000 0.3325 0.3947 0.7595 1.2000 1.4892 1.4000

plot(x,y,'co',xq,vq,'k*--')

i=0:.1:2;
for ii = 1:numel(x)
    l(ii)=sqrt(x(ii)^2+y(ii).^2);
end
l

l = 1×7

0 0.7211 1.5620 2.1401 1.6125 2.1633 2.4413

EDIT — (16 Nov 2021 at 2:48)

The function necessary to find the size of ‘vq’ is size —

vq_size = size(vq)

vq_size = 1×2

1 21

.

[Aidan Palermo]

I'm trying to get an answer less than or equal to 4

[Star Strider]

None of the vectors here have dimensions anywhere near that range.

Please define ‘length’.

.

[Aidan Palermo]

the graph represents a trail inside of a 2 mile by 2 mile square. "length" is the distance of the path

[Star Strider]

That’s a bit more involved, since it involves computing a line integral.

Try this —

x=[0 .6 1 1.3 1.4 1.8 2];
y=[0 .4 1.2 1.7 .8 1.2 1.4];
xq=0:.1:2;

vq=interp1(x,y,xq,'spline');

dx = gradient(xq);                                              % Calculate Derivatives
dy = gradient(vq);

LineInt = cumtrapz(sqrt(dx.^2+dy.^2));                          % Interim 'Length'
TotalLength = LineInt(end)

TotalLength = 5.2681

figure
plot(x,y,'co',xq,vq,'k*--')

figure
plot(xq, LineInt)
grid
xlabel('x_q')
ylabel('Cumulative Path Length')

See the documentation section on Integration to Find Arc Length for an example. It uses the integral function, not trapz, however the apporoach is the same.

NOTE — Line integrals are actually best when used for parametric equations, and this approach works correctly to calculate the circumference of a circle —

t = linspace(0,2*pi);
x = cos(t);
y = sin(t);

circ = trapz(t,sqrt(x.^2+y.^2))

circ = 6.2832

Getting it to work here, considering that I’m not certain how ‘xq’ and ‘vq’ could be parametrically related, is a bit more of a challenge with respect to calculating it correctly.

.

[Aidan Palermo]

that worked, thank you.

[Star Strider]

As always, my pleasure!

.