# What is the arclength of the polar curve #f(theta) = csc^2theta-cot^2theta # over #theta in [pi/3,pi/2] #?

Hi there!

Let's start off with the formula for the arc length of a polar curve whereby:

Notice anything about this function? A trig identity perhaps?

Therefore, this function would simplify down to r = 1! This makes everything much easier!

Substituting everything into the formula we get:

Simplifying everything:

Integrating a constant with respect to theta we get:

Now substitute the bounds in:

Hopefully everything was clear and concise! If you have any questions, feel free to ask! :)

By signing up, you agree to our Terms of Service and Privacy Policy

To find the arc length of the polar curve ( f(\theta) = \csc^2(\theta) - \cot^2(\theta) ) over ( \theta ) in ( \left[\frac{\pi}{3}, \frac{\pi}{2}\right] ), we use the formula for arc length of a polar curve:

[ L = \int_{\alpha}^{\beta} \sqrt{[r(\theta)]^2 + \left(\frac{dr}{d\theta}\right)^2} , d\theta ]

Where ( r(\theta) ) is the polar function and ( \frac{dr}{d\theta} ) is the derivative of ( r(\theta) ) with respect to ( \theta ).

Given ( f(\theta) = \csc^2(\theta) - \cot^2(\theta) ), we need to rewrite this in terms of ( r(\theta) ):

[ r(\theta) = \csc(\theta) - \cot(\theta) ]

Now, find ( \frac{dr}{d\theta} ):

[ \frac{dr}{d\theta} = -\csc(\theta)\cot(\theta) + \csc^2(\theta) ]

Now, plug these into the formula:

[ L = \int_{\pi/3}^{\pi/2} \sqrt{(\csc(\theta) - \cot(\theta))^2 + (-\csc(\theta)\cot(\theta) + \csc^2(\theta))^2} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} \sqrt{\csc^2(\theta) - 2\csc(\theta)\cot(\theta) + \cot^2(\theta) + \csc^2(\theta) - 2\csc^3(\theta) + \csc^4(\theta)} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} \sqrt{2\csc^2(\theta) - 2\csc(\theta)\cot(\theta) - 2\csc^3(\theta) + \csc^4(\theta)} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} \sqrt{2\csc^2(\theta)(1 - \cot(\theta)) - 2\csc(\theta)\cot(\theta)(1 + \csc(\theta))} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} \sqrt{2\csc^2(\theta)(1 - \cot(\theta)) - 2\csc(\theta)\cot(\theta)(\csc(\theta) + 1)} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} \sqrt{2\csc^2(\theta)(1 - \cot(\theta)) - 2\csc(\theta)(\csc(\theta)\cot(\theta) + \cot(\theta))} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} \sqrt{2\csc^2(\theta) - 2\csc(\theta) - 2\csc^2(\theta)} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} \sqrt{-2\csc(\theta)} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} \sqrt{\frac{-2}{\sin(\theta)}} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} \frac{\sqrt{-2}}{\sqrt{\sin(\theta)}} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} \frac{\sqrt{-2}}{\sqrt{2\sin(\theta/2)\cos(\theta/2)}} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} \frac{\sqrt{-1}}{\sqrt{\sin(\theta/2)\cos(\theta/2)}} , d\theta ]

[ L = \int_{\pi/3}^{\pi/2} i , d\theta ]

[ L = i\left(\frac{\pi}{2} - \frac{\pi}{3}\right) ]

[ L = i\left(\frac{\pi}{6}\right) ]

[ L = \frac{i\pi}{6} ]

Where ( i ) is the imaginary unit. Therefore, the arc length of the polar curve is ( \frac{\pi}{6} ).

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7