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受欢迎的微积分 >

tangent (sin(t))/(2+2cos(t))

解答

切线

\frac{sin ( t )}{2 + 2 cos ( t )}

解答

y = \frac{cos ( a _{0} ) ( 2 + 2 cos ( a _{0} ) ) + 2 sin ^{2} ( a _{0} )}{( 2 + 2 cos ( a _{0} ) ) ^{2}} t + \frac{2 ( sin ( a _{0} ) ( cos ( a _{0} ) + 1 ) - a _{0} ( sin ^{2} ( a _{0} ) + cos ( a _{0} ) ( cos ( a _{0} ) + 1 ) ) )}{( 2 + 2 cos ( a _{0} ) ) ^{2}}

求解步骤

计算在一般点

t = a_{0}

的切线

找到切点:

(a_{0}, \frac{sin ( a _{0} )}{2 + 2 cos ( a _{0} )})

计算

y = \frac{sin ( t )}{2 + 2 cos ( t )}

的斜率:

\frac{d y}{d t} = \frac{cos ( t ) ( 2 + 2 cos ( t ) ) + 2 sin ^{2} ( t )}{( 2 + 2 cos ( t ) ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{cos ( a _{0} ) ( 2 + 2 cos ( a _{0} ) ) + 2 sin ^{2} ( a _{0} )}{( 2 + 2 cos ( a _{0} ) ) ^{2}}

找到斜率 m=

\frac{cos ( a _{0} ) ( 2 + 2 cos ( a _{0} ) ) + 2 sin ^{2} ( a _{0} )}{( 2 + 2 cos ( a _{0} ) ) ^{2}}

且穿过

(a_{0}, \frac{sin ( a _{0} )}{2 + 2 cos ( a _{0} )})

的直线:

y = \frac{cos ( a _{0} ) ( 2 + 2 cos ( a _{0} ) ) + 2 sin ^{2} ( a _{0} )}{( 2 + 2 cos ( a _{0} ) ) ^{2}} t + \frac{2 ( sin ( a _{0} ) ( cos ( a _{0} ) + 1 ) - a _{0} ( sin ^{2} ( a _{0} ) + cos ( a _{0} ) ( cos ( a _{0} ) + 1 ) ) )}{( 2 + 2 cos ( a _{0} ) ) ^{2}}

y = \frac{cos ( a _{0} ) ( 2 + 2 cos ( a _{0} ) ) + 2 sin ^{2} ( a _{0} )}{( 2 + 2 cos ( a _{0} ) ) ^{2}} t + \frac{2 ( sin ( a _{0} ) ( cos ( a _{0} ) + 1 ) - a _{0} ( sin ^{2} ( a _{0} ) + cos ( a _{0} ) ( cos ( a _{0} ) + 1 ) ) )}{( 2 + 2 cos ( a _{0} ) ) ^{2}}

作图

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