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.
