How do you find the points of horizontal tangency of #r=2csctheta+5#?

Question

Emily Ammerman · Accepted Answer

Points of horizontal tendency are #theta=pi/2# and #theta=(3pi)/2#

Let us convert #r=2csctheta+5# to Cartesian coordinates by using #rcostheta=x#, #rsintheta=y# and #r^2=x^2+y^2#. We can write #r=2csctheta+5# as

#rsintheta=2+5sintheta# or #y=2+y/sqrt(x^2+y^2)#

i.e. #y/sqrt(x^2+y^2)=y-2#

or #y^2/(x^2+y^2)=(y-2)^2#

or #(x^2+y^2)(y-2)^2=y^2#

Now horizontal tangents are there where #(dy)/(dx)=0# and hence finding #(dy)/(dx)# using implicit differentiiation, we get

#2(x^2+y^2)(y-2)(dy)/(dx)+(y-2)^2(2x+2y(dy)/(dx))=2y(dy)/(dx)#

or #(dy)/(dx)[2(x^2+y^2)(y-2)+2y(y-2)^2-2y]=-2x(y-2)^2#

and hence #(dy)/(dx)=0# when either #y=2# or #x=0#

Now #y=2# means #rsintheta=2# but as equation can be written as #rsintheta=2+5sintheta#, this means when #y=2#, we have #sintheta=0# or #theta=0#. But as function #r=2csctheta+5# s not defined at #theta=0#, we ignore this.

#x=0# means #rcostheta=0# or #costheta=0#, i.e. #theta=pi/2# or #(3pi)/2# and hence points of horizontal tendency are #theta=pi/2# and #(3pi)/2#

graph{(x^2+y^2)(y-2)^2=y^2 [-10, 10, -5, 5]}

Paisley Johnson · Answer

undefined To find the points of horizontal tangency of the polar curve $ r = 2csc(\theta) + 5 $, we need to first find the derivative of $ r $ with respect to $ \theta $, and then determine where this derivative equals zero.

The derivative of $ r $ with respect to $ \theta $ is found using the chain rule:

$$ \frac{dr}{d\theta} = \frac{d}{d\theta}(2csc(\theta) + 5) = -2csc(\theta)cot(\theta) $$

Now, to find the points of horizontal tangency, we set $ \frac{dr}{d\theta} = 0 $ and solve for $ \theta $:

$$ -2csc(\theta)cot(\theta) = 0 $$

This equation is satisfied when $ csc(\theta) = 0 $ or $ cot(\theta) = 0 $.

$ csc(\theta) = 0 $ when $ \theta = \frac{\pi}{2} $ or $ \theta = \frac{3\pi}{2} $.

$ cot(\theta) = 0 $ when $ \theta = \frac{\pi}{2} $ or $ \theta = \frac{3\pi}{2} $.

So, the points of horizontal tangency occur at $ \theta = \frac{\pi}{2} $ and $ \theta = \frac{3\pi}{2} $. To find the corresponding values of $ r $, plug these values of $ \theta $ into the equation $ r = 2csc(\theta) + 5 $.