cartesian (6,(5pi)/4)

解答

极坐标转换为笛卡尔坐标

(6, \frac{5 π}{4})

(- 32, - 32)

求解步骤

x = r \cdot cos (θ)

x = 6 cos (\frac{5 π}{4})

6 cos (\frac{5 π}{4}) = - 32

x = - 32

y = r \cdot sin (θ)

y = 6 sin (\frac{5 π}{4})

6 sin (\frac{5 π}{4}) = - 32

y = - 32

(6, \frac{5 π}{4})

的笛卡尔坐标

(- 32, - 32)

p o l a r (- 2, - 2)

f = 1

d er i v a t i v e y = x

c a r t es ian (4, \frac{3 π}{2})

x^{2} + y^{2} - 4 x = 0