问题描述
我有一个二次贝塞尔曲线，我想计算给定点的切线斜率.例如，将其设为二次贝塞尔曲线的中点，因此t = 0.5(请参见下面的链接以获取有关此图片).我已经为二次贝塞尔曲线计算了公式的一阶导数；但是我得到的斜率值为400，尽管应该为0.也许我以错误的方式使用了一阶导数?我知道我也可以使用三角函数来计算切线；但是我想使用一阶导数来做，这不可行吗?谢谢您的提示！
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!
为澄清起见，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，不仅是要获得起点和终点的切线.
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.
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
A picture of my problem including the text above:http://cid-0432ee4cfe9c26a0.skydrive.live.com/self.aspx/%c3%96ffentlich/Quadratic%20Bezier%20Curve.pdf
非常感谢您提供任何提示！
Thank you very much for any hint!
推荐答案
使用B'(t)的公式，在t=1/2处求值，我们得到
Using your formula for B'(t), evaluated at t=1/2, we get
B'(1/2) = -P0 + P2
从图形的外观来看，P0 =(0,0)和P2 =(400,0).所以
From the look of your graph, P0 = (0,0) and P2 = (400,0). So
B'(1/2) = (400,0).
这是在t = 1/2处沿B(t)传播的点的速度".
This is the "velocity" of a point traveling along B(t) at t=1/2.
(400,0)是水平向量，大小为400.
(400,0) is a horizontal vector, with magnitude 400.
一切都应该如此.由于B'(t)是水平的，所以它的斜率"确实为0.
So all is as it should be. Since B'(t) is horizontal, it does have "slope" 0.
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