A curve is given by the parametric equations #x= c (1 - sinA)# and #y= c (1 + cosA)#, what is #dy/dx# when #A= π/3#?

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I dont understand what i am suppose to find

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Answer 1 · 2018-05-12T13:34:28

You must find #dy/dx# using #dy/dx = ((dy)/(dA))/((dx)/(dA))# then evaluate at #A = pi/3#

Given: #y= c (1 + cos(A))#

Compute #dy/(dA)#:

#dy/(dA) -c sin(A)#

Given: #x= c (1 - sinA)#

Compute #dx/(dA)#:

#dx/(dA) -c cos(A)#

Compute #dy/dx = ((dy)/(dA))/((dx)/(dA))# :

#dy/dx = (-csin(A))/(-c cos(A)#

#dy/dx = tan(A)#

Evaluate at #A = pi/3#:

#dy/dx = tan(pi/3)#

#dy/dx = sqrt3#