This is machine translation

Translated by Microsoft
Mouseover text to see original. Click the button below to return to the English verison of the page.

Note: This page has been translated by MathWorks. Please click here
To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

Vector-Valued Functions

The toolbox supports vector-valued splines. For example, if you want a spline curve through given planar points , then the following code defines some data and then creates and plots such a spline curve, using chord-length parametrization and cubic spline interpolation with the not-a-knot end condition.

x=[19 43 62 88 114 120 130 129 113 76 135 182 232 298 ...
 348 386 420 456 471 485 463 444 414 348 275 192 106 ...
 30 48 83 107 110 109 92 66 45 23 22 30 40 55 55 52 34 20 16];
y=[306 272 240 215 218 237 275 310 368 424 425 427 428 ...
 397 353 302 259 200 148 105 77 47 28 17 10 12 23 41 43 ...
 77 96 133 155 164 157 148 142 162 181 187 192 202 217 245 266 303];

xy = [x;y]; df = diff(xy,1,2); 
t = cumsum([0, sqrt([1 1]*(df.*df))]); 
cv = csapi(t,xy);
fnplt(cv), hold on, plot(x,y,'o'), hold off

If you then wanted to know the area enclosed by this curve, you would want to evaluate the integral , with the point on the curve corresponding to the parameter value . For the spline curve in cv just constructed, this can be done exactly in one (somewhat complicated) command:

area = diff(fnval(fnint( ...
       fncmb(fncmb(cv,[0 1]),'*',fnder(fncmb(cv,[1 0]))) ...

To explain, y=fncmb(cv,[0 1]) picks out the second component of the curve in cv, Dx=fnder(fncmb(cv,[1 0])) provides the derivative of the first component, and yDx=fncmb(y,'*',Dx) constructs their pointwise product. Then IyDx=fnint(yDx) constructs the indefinite integral of yDx and, finally, diff(fnval(IyDx,fnbrk(cv,'interval'))) evaluates that indefinite integral at the endpoints of the basic interval and then takes the difference of the second from the first value, thus getting the definite integral of yDx over its basic interval. Depending on whether the enclosed area is to the right or to the left as the curve point travels with increasing parameter, the resulting number is either positive or negative.

Further, all the values Y (if any) for which the point (X,Y) lies on the spline curve in cv just constructed can be obtained by the following (somewhat complicated) command:

X=250; %Define a value of X
Y = fnval(fncmb(cv,[0 1]), ...
         mean(fnzeros(fncmb(fncmb(cv,[1 0]),'-',X))))

To explain: x = fncmb(cv,[1 0]) picks out the first component of the curve in cv; xmX = fncmb(x,'-',X) translates that component by X; t = mean(fnzeros(xmX)) provides all the parameter values for which xmX is zero, i.e., for which the first component of the curve equals X; y = fncmb(cv,[0,1]) picks out the second component of the curve in cv; and, finally, Y = fnval(y,t) evaluates that second component at those parameter sites at which the first component of the curve in cv equals X.

As another example of the use of vector-valued functions, suppose that you have solved the equations of motion of a particle in some specified force field in the plane, obtaining, at discrete times , the position as well as the velocity stored in the 4-vector , as you would if, in the standard way, you had solved the equivalent first-order system numerically. Then the following statement, which uses cubic Hermite interpolation, will produce a plot of the particle path:fnplt(spapi(augknt(t,4,2),t,reshape(z,2,2*n))).

Was this topic helpful?