How do you convert #3=(x+4y)^2+3x# into polar form?

Question

Delilah Aguilar · Accepted Answer

Please see the explanation for a detailed description of how the conversion is done.

Expand the square, using the F.O.I.L. method: #3 = x^2 + 8xy + 16y^2 + 3x# Substitute #rcos(theta) # for x and #rsin(theta)# for y: #3 = (rcos(theta))^2 + 8(rcos(theta))(rsin(theta)) + 16(rsin(theta))^2 + 3(rcos(theta))# This can be written as a quadratic in r: #0 = (cos^2(theta) + 8cos(theta)sin(theta) + 16sin^2(theta))r^2 + 3cos(theta)r - 3# The coefficient for #r^2# factors into a square: #0 = (cos(theta) + 4sin(theta))^2r^2 + 3cos(theta)r - 3# Use the quadratic formula: #r = (-b +-sqrt(b^2 - 4(a)(c)))/(2a)# where: #a = (cos(theta) + 4sin(theta))^2# #b = 3cos(theta)# #c = -3# Also, we must change the #+-# to only #+#, because negative values for r do not make sense. #r = (-3cos(theta) +sqrt(9cos^2(theta) + 12(cos(theta) + 4sin(theta))^2))/(2(cos(theta) + 4sin(theta))^2)#

Zoe Austin · Answer

undefined To convert the equation $3 = (x + 4y)^2 + 3x$ into polar form, we use the conversion formulas $x = r \cos(\theta)$ and $y = r \sin(\theta)$.

Substitute these values into the equation and simplify:

$3 = ((r \cos(\theta)) + 4(r \sin(\theta)))^2 + 3(r \cos(\theta))$

$3 = (r \cos(\theta) + 4r \sin(\theta))^2 + 3r \cos(\theta)$

Expand the squared term:

$3 = (r^2 \cos^2(\theta) + 8r^2 \cos(\theta) \sin(\theta) + 16r^2 \sin^2(\theta)) + 3r \cos(\theta)$

Combine like terms:

$3 = r^2(\cos^2(\theta) + 8 \cos(\theta) \sin(\theta) + 16 \sin^2(\theta)) + 3r \cos(\theta)$

Simplify the expression inside the parentheses using trigonometric identities:

$3 = r^2(\cos^2(\theta) + 2\cos(\theta) \sin(\theta) + \sin^2(\theta) + 7\sin^2(\theta) + 7\cos^2(\theta)) + 3r \cos(\theta)$

$3 = r^2(1 + 7\sin^2(\theta) + 7\cos^2(\theta)) + 3r \cos(\theta)$

Since $1 = \sin^2(\theta) + \cos^2(\theta)$, we can simplify further:

$3 = r^2(1 + 7) + 3r \cos(\theta)$

$3 = 8r^2 + 3r \cos(\theta)$

So, the equation in polar form is $3 = 8r^2 + 3r \cos(\theta)$.