What is the slope of the tangent line of #r=-2sin(2theta)+cos(3theta)# at #theta=(-11pi)/8#?

1 Answer
Mar 11, 2018

#(3sqrt(2-sqrt(2))-4sqrt(2))/2#

Explanation:

To find the slope of a tangent line, take the derivative of the function. Here, #r=f(theta)=cos(3theta)-2sin(2theta)#

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

#d/(d theta)(cos(3theta)-2sin(2theta))#

Since #(f+-g)'=f'+-g'#, we can write:

#d/(d theta)cos(3theta)-2d/(d theta)sin(2theta)#

According to the chain rule, #(f(g(x))'=f'(g(x))*g'(x)#

First, we compute #d/(d theta)cos(3theta)#:

Here, #f(eta)=coseta# where #eta=3theta#

#d/(d eta)coseta*d/(d theta)3theta#

#-sineta*3#

Since #eta=3theta#:

#-3sin(3theta)#

Next, we compute #d/(d theta)sin(2theta)#:

Here, #f(eta)=sineta# where #eta=2theta#

#d/(d eta)sineta*d/(d theta)2theta#

#coseta*2#

Since #eta=2theta#, we write:

#2cos(2theta)#

So we have:

#d/(d theta)(cos(3theta)-2sin(2theta))=-3sin(3theta)+2(2cos(3theta))#

#=4cos(2theta)-3sin(3theta)#

This is the general formula for the slope of the tangent line for #r#. However. we are asked to find the slope at #theta=-(11pi)/8#. Inputting:

#4cos(2(-(11pi)/8))-3sin(3(-(11pi)/8))#

#4cos(-(11pi)/4)-3sin(-(33pi)/8)#

Since #sin(-x)=-sin(x)# and #cos(-x)=cos(x)#, we write:

#4cos((11pi)/4)+3sin((33pi)/8)#

Since #cos((11pi)/4)=-sqrt(2)/2# and #sin((33pi)/8)=(sqrt(2-sqrt(2)))/2#, we write:

#4(-sqrt(2)/2)+3((sqrt(2-sqrt(2)))/2)#

#-2sqrt(2)+(3sqrt(2-sqrt(2)))/2#

#(3sqrt(2-sqrt(2))-4sqrt(2))/2#

In crude decimals, that is, to four decimal places, #-1.6804#.

Graphing the function:
graph{cos(3x)-2sin(2x) [-7.017, 0.778, -0.654, 3.244]}

We see that this is true.