x^2 + y^2 = a^2 (a circle)
更新1:
SORRY~~我講漏左...x,y,z are vector~~ so,,i want to find normal vector~!
更新2:
x,y are vector,,no z
partial differentiation3
2011-01-17 2:12 am
Find the normal to the following curve or surfaces:
x^2 + y^2 = a^2 (a circle)
x^2 + y^2 = a^2 (a circle)
回答 (2)
2011-01-20 7:48 am
2011-01-18 4:12 am
Let the point (b,c) be a point on x^2 + y^2 = a^2.
d(x^2 + y^2)/dx = d(a^2)/dx
2x + 2y dy/dx = 0
2y dy/dx = -2x
dy/dx = -2x/2y
= -x/y
Slope of tangent on point (b,c) = -b/c
Slope of normal on point (b,c) = -1 / (-b/c) = c/b
The equation of the normal on (b,c) is given as:
y - c = c/b (x - b)
b (y - c) = c (x - b)
by - bc = cx - bc
by = cx
y = c/b x
Also note that the normal always passes the origin for any points on the circle.
We may just get the equation by y = (slope of normal) x, i.e. y = c/b x
d(x^2 + y^2)/dx = d(a^2)/dx
2x + 2y dy/dx = 0
2y dy/dx = -2x
dy/dx = -2x/2y
= -x/y
Slope of tangent on point (b,c) = -b/c
Slope of normal on point (b,c) = -1 / (-b/c) = c/b
The equation of the normal on (b,c) is given as:
y - c = c/b (x - b)
b (y - c) = c (x - b)
by - bc = cx - bc
by = cx
y = c/b x
Also note that the normal always passes the origin for any points on the circle.
We may just get the equation by y = (slope of normal) x, i.e. y = c/b x
收錄日期: 2021-04-23 23:22:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110116000051KK00898