Curve.Evaluate (Double, Int32) Method

Computes the curve derivatives at parametric value u.
Public Function Evaluate( _ 
ByVal u As Double, _ 
ByVal d As Integer
) As Vector3D()
This language is not supported or no code example is available.
public Vector3D[] Evaluate( 
double u
int d 
)
This language is not supported or no code example is available.

Parameters

u
double

The parametric value

d
int

The degree of the derivative

Return Value

Vector3D[]

The derivatives array.

Example
 
The following code fragment demonstrates how to compute the first 2 derivatives of the curve at the parametric value [0.25].
 Vector3D[] CK = myCurve.Evaluate(0.25, 2);
 
 Vector3D C   = CK[0];
 Vector3D Cu  = CK[1];
 Vector3D Cuu = CK[2];
 					
This language is not supported or no code example is available.
 Dim CK() As Vector3D = MyCurve.Evaluate(0.25, 2)
 
 Dim C As Vector3D = CK(0)
 Dim Cu As Vector3D = CK(1)
 Dim Cuu As Vector3D = CK(2)
 					
This language is not supported or no code example is available.

.NET Framework

Supported in: 4.8

In this article

Definition