# What is the slope of the tangent line of #r=theta/3+sin((3theta)/8-(5pi)/3)# at #theta=(7pi)/6#?

##### 1 Answer

#### Explanation:

First, let's go ahead and find the values of

#r = theta/3 + sin((3theta)/8-(5pi)/3)#

#= (7pi)/18 + sin((21pi)/48 - (5pi)/3)#

#~~ 1.881#

#(dr)/(d theta) = 1/3 + 3/8 cos((3theta)/8 - (5pi)/3)#

#= 1/3 + 3/8cos((21pi)/48 - (5pi)/3)#

#~~ 0.0514#

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Now, the slope of the tangent line at any point is

Luckily, we can apply a version of the chain rule which states that

#dy/dx = (dy"/"d theta)/(dx"/"d theta)# We will also have to use the rectangular --> polar coordinate formulas:

#x = rcostheta#

#y = rsintheta#

Since we have expressions for

#dy/(d theta) = d/(d theta) rsintheta = (dr)/(d theta)sintheta + rcostheta#

#dx/(d theta) = d/(d theta) rcostheta = (dr)/(d theta)costheta -rsintheta#

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With these formulas, we can find the slope of the line at

#dy/dx = (dy"/"d theta)/(dx"/"d theta) = ((dr)/(d theta)sintheta + rcostheta)/((dr)/(d theta)costheta -rsintheta)#

We already know that at

#sin((7pi)/6)=-0.5#

#cos((7pi)/6) = -0.866#

Now we can find our slope:

#((dr)/(d theta)sintheta + rcostheta)/((dr)/(d theta)costheta -rsintheta) = (0.0514 * (-0.5) + 1.881 * (-0.866))/(0.0514 * (-0.866) - 1.881 * (-0.5))#

#= -1.847#

*Final Answer*