tangent 5x^3-x^2y+y^3+1=0,(1,-2)

解答

切线

5 x^{3} - x^{2} y + y^{3} + 1 = 0, (1, - 2)

y = - \frac{19}{11} x - \frac{3}{11}

求解步骤

计算

5 x^{3} - x^{2} y + y^{3} + 1 = 0

的斜率:

\frac{d y}{d x} = \frac{- 15 x ^{2} + 2 x y}{3 y ^{2} - x ^{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{19}{11}

找到斜率 m=

- \frac{19}{11}

且穿过

(1, - 2)

的直线:

y = - \frac{19}{11} x - \frac{3}{11}

y = - \frac{19}{11} x - \frac{3}{11}

t an g e n t y = t^{3} - 4 t - 3, (2, - 3)

x \to 2 + lim (\frac{3}{( x - 2 ) ^{3}})

\int (y e^{y}) d y

\int_{0}^{5} π (2 5 y)^{2} d y

\int \frac{e ^{x}}{36 - e ^{2 x}} d x