What is the equation of the tangent line of #r=2cos(theta-(3pi)/4) +2sin(-theta+(3pi)/2)# at #theta=(-5pi)/12#?

1 Answer
Jun 3, 2016

#y = 2.17303 -0.767327(x+0.582262)#

Explanation:

Given #r(theta)# we have

#{ (x(theta)=r(theta)cos(theta)), (y(theta)=r(theta)sin(theta)) :}#

# { (dx/(d theta) = -r(theta)sin(theta)+cos(theta)((dr)/(d theta))), (dy/(d theta) = r(theta)cos(theta)+sin(theta)((dr)/(d theta))) :} #

but

#r(theta) = 2 cos(theta - 3 pi/4) + 2 sin(-theta + 3 pi/2)#

then

# { (dx/(d theta) = 2 (sin(2 theta) + sin(pi/4 + 2 theta))), (dy/(d theta)=-2 (cos(2 theta) + cos(pi/4+ 2 theta))) :} #

but

#(dy)/(dx) = (((dy)/(d theta)))/(((dx)/(d theta)))#

at point #theta_0 = -(5pi)/12# we have

#p_0 ={r(theta_0)cos(theta_0),r(theta_0)sin(theta_0)}={-0.582262,2.17303}#

and

#m_0 = ((dy)/(dx) )_{theta_0} = -0.767327#

The tangent straight reads

#y = 2.17303 -0.767327(x+0.582262)#

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