What is the slope of the tangent line of #r=2theta^2-3thetacos(2theta-(pi)/3)# at #theta=(-5pi)/3#?

Question

Elijah Abram · Accepted Answer

Derivative with Polar Coordinates is # (\partial x) / (\partial y) = ((\partial r) / (\partial theta) sin(theta)+ rcos(theta))/((\partial r) / (\partial theta) cos(theta)- rsin(theta)#

#(\partial r) / (\partial theta) = 4 theta-3*cos(2 theta-(pi)/(3))-3 theta*(-sin(2 theta - pi/3))*2=# #=4 theta-3*cos(2 theta-(pi)/(3))+6 theta*(sin(2 theta - pi/3))# The numerator: # (\partial r) / (\partial theta)*sin(theta) = # #=((4 theta-3*cos(2 theta-(pi)/(3))+6 theta*(sin(2 theta - pi/3)))*sin(theta)# #r*cos(theta)=(2 theta^2-3 theta cos(2 theta-(pi)/3))*cos(theta)#

The denominator: # (\partial r) / (\partial theta)*cos(theta) = # #4 theta-3*cos(2 theta-(pi)/(3))+6 theta*(sin(2 theta - pi/3))*cos(theta)# #r*sin(theta)=(2 theta^2-3 theta cos(2 theta-(pi)/3))*sin(theta)# You are calculating the slope at #theta = -(5 pi)/3#, insert #theta = -(5 pi)/3# in the formulae above:

The numerator: # (\partial r) / (\partial theta)*sin(theta) = -42.999# #r*cos(theta)=31.343#

The denominator: #(\partial r) / (\partial theta)*cos(theta) = -24.825# #r*sin(theta)=54.287#

# (\partial x) / (\partial y)(theta=-(5 pi)/3)=# #=(-42.999+31.343)/(-24.825-54.287)=0.15 #

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Emma Aldridge · Answer

undefined To find the slope of the tangent line at a given point on a polar curve, you can use the formula $ \frac{dy}{dx} = \frac{r' \cdot \sin(\theta) + r \cdot \cos(\theta)}{r' \cdot \cos(\theta) - r \cdot \sin(\theta)} $, where $ r' $ is the derivative of $ r $ with respect to $ \theta $. Plugging in the values for $ r $, $ r' $, and $ \theta $ given in the question, we get:

$ r = 2\theta^2 - 3\theta \cos(2\theta - \frac{\pi}{3}) $
$ r' = 4\theta - 3(2\theta - \frac{\pi}{3}) \sin(2\theta - \frac{\pi}{3}) $

Now substitute $ \theta = -\frac{5\pi}{3} $ into these equations to find $ r $ and $ r' $, then plug them into the slope formula to calculate the slope of the tangent line at that point.