What is the equation of the tangent line of #r=cot^2theta - sintheta# at #theta=pi/4#?

Question

1995 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-13T01:06:13

#y~~-4.707x+4.707pi/4+0.293#

#(dr)/(d theta)=-2cot(theta)csc^2(theta)-cos(theta)=r'(theta)#

#r'(pi/4)=-2/tan(pi/4)*1/sin^2(pi/4)-cos(pi/4)#

#=-4-sqrt(2)/2~~-4.707#

This is our instantaneous rate of change (slope of the tangent) when #theta=pi/4#

Then, we can just a point-slope formula to figure out the equation of that tangent line by plugging in values into the original function:

#y-y_1=m(x-x_1)#

#r(pi/4)=cot^2(pi/4)-sin(pi/4)=1-sqrt(2)/2~~0.293#

Thus:

#y-0.293=3.293(x-pi/4)#

#y~~-4.707x+4.707pi/4+0.293#