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
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 2π / 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)] |
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