首页 > 几何图形公式

给定点的曲线的切线和法线方程

时间:2020-12-05 20:30:25

找到以下曲线的切线和法线的方程

(1)y = x 2 -4x-5,x = -2

(2)y = x-正弦x cos x,x =π/ 2

(3)   y = 2正弦2 3x,x =π/ 6

(4)   y =(1 + sin x)/ cos x,x =π/ 4时

问题1:

y = x 2 -4x-5,x = -2时

当x = -2时

y =(-2)2 -4(-2)-5

   = 4 + 8-5

y = 7

dy / dx = 2x-4

dy / dx = 2(-2)-4

dy / dx = -8

因此,所需的点是(-2,7)

切线方程:

(yy 1)= m(xx 1

(y-7)= -8(x + 2)

 y-7 = -8(x + 2)

 y-7 = -8x-16

8x + y-7 + 16 = 0

8x + y + 9 = 0

正态方程:

(yy 1)=(-1 / m)(xx 1

(y-7)=(1/8)(x + 2)

 8(y-7)= 1(x + 2)

8y-56 = x + 2

x-8y + 58 = 0

问题2 :

y = x-正弦x cos x,x =π/ 2

解决方案:

当x =π/ 2时,y = x-sin x cos x。

y =π/ 2-正弦π/ 2cosπ/ 2

=π/ 2-(1)(0)

=π/ 2 

dy / dx = 1- [sin x(-sin x)+ cos x(cos x)]

= 1-[-sin 2 x + cos 2 x]

= 1 [cos 2 x-sin 2 x]

= 1-cos 2x

x =π/ 2处的切线斜率

= 1-cos2(π/ 2)

= 1-cosπ

= 2

因此,所需的点是(π/ 2,π/ 2)

切线方程:

(yy 1)= m(xx 1

[y-(π/ 2)] = 2 [x-(π/ 2)]

[y-(π/ 2)] = 2x-π

2x-y-π+(π/ 2)= 0

2x-y-(π/ 2)= 0

正态方程:

(yy 1)=(-1 / m)(xx 1

[y-(π/ 2)] =(-1/2)[x-(π/ 2)]

2 [y-(π/ 2)] = -1 [x-(π/ 2)]

2y-π= -x +(π/ 2)

x +2y-π-(π/ 2)= 0

x + 2y-(3π/ 2)= 0

问题3:

y = 2sin 2 3x,在x =π/ 6

解决方案:

y = 2sin 2  3x

dy / dx = 2(2sin3x)(cos3x)3

= 6(2sin3x cos3x)  

= 6sin 2(3x)

= 6sin6x

x =π/ 6处的斜率

= 6罪孽6(π/ 6)

= 6sinπ

= 6(0)

= 0

y = 2sin 2 3x

= 22 3(π/ 6)

= 2sin 2(π/ 2)

= 2

因此,所需的点是(π/ 6,2)

切线方程:

(yy 1)= m(xx 1

(y-2)= 0 [x-(π/ 6)]

 y-2 = 0

正态方程:

(yy 1)=(-1 / m)(xx 1

(yy 1)=(-1/0)(xx 1

0(y-2)= -1 [x-(π/ 6)]

0 = -1 [x-(π/ 6)]

x-(π/ 6)= 0

问题4:

y =(1 + sin x)/ cos x,x =π/ 4时

解决方案:

y =(1 + sin x)/ cos x

dy / dx = [cos x(cos x)-(1 + sin x)(-sin x)] /cos²x

= [cos 2 x + sinx + sin 2 x] / cos 2 x

=(1 + sinx)/ cos 2 x

x =π/ 4处的斜率

y =(1 +sinπ/ 4)/ cos  / 4

=(1+(1 /√2))/(1 /√2)2

= [(√2+ 1)/√2] /(1/2)

= [(√2+ 1)/√2]    (2/1)

=(√2+ 1)√2

y = 2 +2√2

y =(1 + sin x)/ cos x

=(1 +sinπ/ 4)/cosπ/ 4

= [1+(1 /√2)] /(1 /√2)

= [(√2+ 1)/√2] /(√2/ 1)

=(√2+ 1)

因此,所需的点是(π/ 4,(√2+ 1))

切线方程:

(yy 1)= m(xx 1

[y-(√2+ 1)] =(2 +2√2)[x-(π/ 4)]

正态方程:

(yy 1)=(-1 / m)(xx 1

[y-(√2+ 1)] = [-1 /(2 +2√2)] [x-(π/ 4)]


 

载入中…
点这里查看与之相关的计算

.

条评论

昵称: 需审核请等待!

密码: 匿名发表

验证码:

载入中…

.

.
分享到: