二次贝塞尔曲线：计算正切
我有一条二次贝塞尔曲线，我想计算给定点的切线斜率。例如，让它成为二次贝塞尔曲线的中点，因此 t=0.5（请参阅下面的链接以获取相关图片）。我已经计算了二次贝塞尔曲线公式的一阶导数；然而我得到的斜率值是 400，尽管它应该是 0。也许我以错误的方式使用一阶导数？我知道我还可以使用三角函数计算切线；但是我想使用一阶导数来做到这一点，这应该不可能吗？感谢您的任何提示！
为了澄清/请注意：我对获取二次贝塞尔曲线上任意给定点的斜率的通用方法感兴趣，而不仅仅是获取起点和终点的切线。
我的问题的图片，包括上面的文字： http://cid-0432ee4cfe9c26a0。 skydrive.live.com/self.aspx/%c3%96ffentlich/Quadratic%20Bezier%20Curve.pdf
非常感谢您的任何提示！
I have a quadratic bezier curve and I want to calculate the slope of the tangent in a given point. For example, let it be the middlepoint of the quadratic bezier curve, therefore t=0.5 (please see the link below for a picture of this). I've calculated the first derivative of the formula for the quadratic bezier curve; however I get 400 as value for the slope, though it should be 0. Maybe I'm using the first derivative in a wrong way? I know I could also calculate the tangents using trigonometric functions; however I'd like to do it using the first derivative, shouldn't this be possible? Thanks for any hint!
For clarification / please note: I'm interested in a general way to get the slope in a arbitrary given point on a quadratic bezier curve, not only to get the tangent in the start- and end point.
A picture of my problem including the text above:
http://cid-0432ee4cfe9c26a0.skydrive.live.com/self.aspx/%c3%96ffentlich/Quadratic%20Bezier%20Curve.pdf
Thank you very much for any hint!
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使用您的
B'(t)公式，在t=1/2进行评估，我们得到从您的图表来看，P0 = (0,0) 和 P2 = (400,0)。所以
这是在 t=1/2 时沿 B(t) 行进的点的“速度”。
(400,0) 是一个水平向量，其大小为 400。
所以一切都是它应该的样子。由于 B'(t) 是水平的，因此它的“斜率”确实为 0。
Using your formula for
B'(t), evaluated att=1/2, we getFrom the look of your graph, P0 = (0,0) and P2 = (400,0). So
This is the "velocity" of a point traveling along B(t) at t=1/2.
(400,0) is a horizontal vector, with magnitude 400.
So all is as it should be. Since B'(t) is horizontal, it does have "slope" 0.
贝塞尔曲线的导数
Derivatives of a Bézier Curve