What is the slope of the tangent line of #r=2theta-cos(5theta-(2pi)/3)# at #theta=(-7pi)/4#?

Question

Cameron Alexander · Accepted Answer

The slope, #m ~~ -1.72#

To find the slope, m, of the tangent line, we must compute #dy/dx# in terms of #theta# and then evaluate it at #theta = (-7pi)/4#.

Here is a reference Tangents with Polar Coordinates that will give us the general equation for #dy/dx# in terms of #theta#:

#dy/dx = ((dr)/(d theta)sin(theta)+rcos(theta))/((dr)/(d theta)cos(theta)-rsin(theta))" [1]"#

We are given r:

#r=2theta-cos(5theta-(2pi)/3)" [2]"#

We must compute #(dr)/(d theta)#:

#(dr)/(d theta) = 2 -5sin(5theta-(2pi)/3)" [3]"#

Substitute the right sides of equations [2] and [3] into equation [1]:

#dy/dx = ((2 -5sin(5theta-(2pi)/3))sin(theta)+(2theta-cos(5theta-(2pi)/3))cos(theta))/((2 -5sin(5theta-(2pi)/3))cos(theta)-(2theta-cos(5theta-(2pi)/3))sin(theta))" [4]"#

When we evaluate equation [4] at #theta = (-7pi)/4# the left side becomes the slope, m:

#m = ((2 -5sin(5(-7pi)/4-(2pi)/3))sin((-7pi)/4)+(2(-7pi)/4-cos(5(-7pi)/4-(2pi)/3))cos((-7pi)/4))/((2 -5sin(5(-7pi)/4-(2pi)/3))cos((-7pi)/4)-(2(-7pi)/4-cos(5(-7pi)/4-(2pi)/3))sin((-7pi)/4))" [5]"#

#m ~~ -1.72#