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

1 Answer
Dec 27, 2017

(2 +pi/2 +5sin ((11pi)/12)- cos((11pi)/12))/(2-pi/2 +5sin((11pi)/12) +cos ((11pi)/12))

Explanation:

Hand written answer is now replaced by typed answer, with an error in calculating(dr)/(d(theta) rectified:

First of all 5theta+(5pi)/3 is equivalent to 5theta + 2pi -pi/3 which is equivalent to 5theta -pi/3. Hence r=2theta -cos(5theta-pi/3)->(dr)/(d (theta))= 2+5sin(5theta-pi/3)

Formula for the slope of a polar curve is dy/dx=( (dr)/(d(theta)) sin theta + r cos theta)/((dr)/(d(theta)) cos theta -rsin theta)

Slope is required at theta= (-7pi)/4 which is equivalent to pi/4 -2pi -> pi/4. For this theta, sine and cosine values would be both 1/sqrt2. Hence dy/dx = ((dr)/(d(theta)) + r)/((dr)/(d(theta)) -r)

Now at theta=pi/4, r= pi/2 -cos((5pi)/4-pi/3)= pi/2 -cos((11pi)/12) and (dr)/(d(theta)) =2+5sin((11pi)/12)

Substituting the values of (dr)/(d(theta)) and r, slope would be

(2 +pi/2 +5sin ((11pi)/12)- cos((11pi)/12))/(2-pi/2 +5sin((11pi)/12) +cos ((11pi)/12))