Apakah bentuk Cartesian r-theta = -2inin ^ 2theta-cot ^ 3theta?

Jawapan:

Set:

# x = rcosθ #

# y = rsinθ #

Jawapannya ialah:

#sqrt (x ^ 2 + y ^ 2) -arccos (x / sqrt (x ^ 2 + y ^ 2)) = - 2x ^ 2 / (x ^ 2 + y ^ 2) -x ^ 3 / y ^ #

Mengikut gambar berikut:

Set:

# x = rcosθ #

# y = rsinθ #

Jadi kami mempunyai:

# cosθ = x / r #

# sinθ = y / r #

# θ = arccos (x / r) = arcsin (y / r) #

# r = sqrt (x ^ 2 + y ^ 2) #

Persamaan menjadi:

# r-θ = -2sin ^ 2θ-cot ^ 3θ #

# r-θ = -2sin ^ 2θ-cos ^ 3θ / sin ^ 3θ #

#sqrt (x ^ 2 + y ^ 2) -arccos (x / r) = - 2x ^ 2 / r ^ 2 (x ^ 3 / r ^ 3) / (y ^ 3 / r ^ 3)

#sqrt (x ^ 2 + y ^ 2) -arccos (x / r) = - 2x ^ 2 / r ^ 2-x ^ 3 / y ^ 3 #

(x ^ 2 + y ^ 2) -arccos (x / sqrt (x ^ 2 + y ^ 2)) = - 2x ^ 2 / sqrt (x ^ 2 + y ^ 2) ^ 2-x ^ y ^ 3 #

#sqrt (x ^ 2 + y ^ 2) -arccos (x / sqrt (x ^ 2 + y ^ 2)) = - 2x ^ 2 / (x ^ 2 + y ^ 2) -x ^ 3 / y ^ #

<30, 70, 140> When adding vectors, simply add the coordinates. A+B=<20, 30, 50> + <10, 40, 90> =<20+10, 30+40, 50+90> = <30, 70, 140>

(X, y) => (rcostheta, / 4) = 2sqrt2 (-4, - (3pi) / 4) -> (2sqrt2,2sqrt2)

2 "theta" - cos ^ 2 (theta) (theta) - cos ^ 2 (theta) = sin ^ 2 (theta) - (1 - sin ^ 2 (theta)) = 2 sin ^ 2 (theta) - 1