polar (2sqrt(3),2)

解答

笛卡尔坐标转换为极坐标

(23, 2)

(4, \frac{π}{6})

求解步骤

r = x^{2} + y^{2}

r = (23)^{2} + 2^{2}

(23)^{2} + 2^{2} = 4

r = 4

θ = arctan (\frac{y}{x})

θ = arctan (\frac{2}{2 3})

arctan (\frac{2}{2 3}) = \frac{π}{6}

θ = \frac{π}{6}

(23, 2)

的极坐标

(4, \frac{π}{6})

n or ma l y = x^{2} - x^{3} + x, (- 2, 10)

s im pl i f y (- 2.4) (7.3)

in t e g r a l e^{2 x}

d er i v a t i v e f (x) = - \frac{7}{x}, at x = - 3

d er i v a t i v e f (x) = (2 sin^{3} (x) - 5 x)^{\frac{5}{3}}