Examine how well a cubic spline duplicates a circle. In making this examination generate the x and y coordinates of one-half of a circle on a 12° increment starting with 0° and ending with 180° (0°is at the bottom of the circle). Pass the cubic spline through these generated coordinates, and then

Question

Examine how well a cubic spline duplicates a circle. In making this examination generate the x and y coordinates of one-half of a circle on a 12° increment starting with 0° and ending with 180° (0°is at the bottom of the circle). Pass the cubic spline through these generated coordinates, and then on a 5° increment compare the interpolated x with the actual x. Also numerically integrate the cubic spline to determine the areas and wetted perimeters (i.e., the circumferences extending on both sides starting at the bottom of the circle) at each of the latter 5° increments and compare these with the actual areas and wetted perimeters. Use a radius of R = 10 in computing the values requested.

Accepted Answer

In solving this problem we note that the actual x and y coordinates are obtained from x = R sin(β) and y = R(1 − cos β). The actual area and wetted perimeter are obtained from A = R²(β − cos β sin β) and P = 2Rβ = Dβ. Simpson’s one-third rule, as implemented in SIMPR, will be used to carry out numerical integrations by noting that the area A can be obtained from A = 2[latex]\int{x}[/latex] dy and the perimeter P = 2[latex]\int{\left\{ d x^{2} + d y^{2}\right\}^{{1}/{2}}}[/latex]dy. Using the properties of a cubic spline as developed previously in this section,

[latex]dx=\left\{\frac{\Delta x}{\Delta y}+\frac{\Delta y}{6}\left[-\left(3a^{2}-1\right)X^{\prime \prime }_{j}+\left(3b^{2}-1\right)X^{ \prime \prime}_{j} +1\right] \right\}dy=Zdy[/latex]

where a and b are the linear cubic spline weighting functions as given earlier, but since y is the independent variable they become a = ([latex]Y_{j+1}[/latex] − y)/([latex]Y_{j+1} − Y_{j}[/latex]) and b = 1 − a, and Δx =[latex] X_{j+1} − X_{j}[/latex]and Δy = [latex]Y_{j+1} − Y_{j}[/latex], and [latex]X^{\prime\prime}_{j}[/latex] and [latex]X^{\prime\prime}_{j+1}[/latex] are the second derivates d²x/dy² at the two points j and j + 1 as generated by solving the tridiagonal system of equations as described in the cubic spline method with x considered the dependent variable and y the independent variable. Using the above to substitute for dx in the integral for the perimeter P gives

[latex]P= 2 \int{\left\{Z^{2}+1 \right\}}^{{1}/{2}} dy[/latex]

where Z= {Δx/Δy+Δy/6[-(3a²-1)[latex]X^{\prime \prime }_{j}[/latex]+(3b²-1)[latex]X^{ \prime \prime}_{j+1}[/latex]]}.

Program CIRCSPLI.FOR obtains the numbers to compare the requested results from using a cubic spline for interpolation with the actual values. The table below the program listing is the output from running the program. Notice Program CIRCSPLI call on the subroutine SPLINESU to obtain the second derivatives needed to interpolate using a cubic spline, as well as SIMPR to carry out the numerical integrations. You should note the logical in function subprograms EQUAT and EQUAT1 (as well as the main program), that supply the arguments for the numerical integration of the area and wetted perimeters, respectively, in order to use the two original points that bracket the interval in which the interpolation is being done. In a sense using a cubic polynomial to approximate a circle raises some interesting questions since the derivative dx/dy is infinite at the bottom and top where β = 0 and β = π, and dy/dx is infinite at the sides where β = π/2 and β = 3π/2. The second derivatives are undefined at these points. Since the cubic spline equates first and second derivatives (in this problem dx/dy and d²x/dy²) at the given points it is clear that the cubic splines do not duplicate the circle. In the program the natural boundary condition at the ends, which sets d²x/dy² = 0 at β = 0 and β = π is, used.

Question B.6: Examine how well a cubic spline duplicates a circle. In maki...

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