# Thread: Problems for calculus withparametric curves.

1. ## Problems for calculus withparametric curves.

1.Find the length of the polar curve given by on the interval .

I know how to do it with x and y, but what to do with this one?

Also same problem with this question

2.For what values of on the polar curve , with , are the tangent lines horizontal? Vertical?

Isn't just tangent=0 and tangent=1?

2. Originally Posted by dylan5188
1.Find the length of the polar curve given by on the interval .

I know how to do it with x and y, but what to do with this one?

Also same problem with this question

2.For what values of on the polar curve , with , are the tangent lines horizontal? Vertical?

Isn't just tangent=0 and tangent=1?
1. $\displaystyle S = \int_{\theta_1}^{\theta_2} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta$

2. vertical tangent lines occur where $\displaystyle \frac{dy}{dx}$ is undefined

horizontal tangent lines occur where $\displaystyle \frac{dy}{dx}$ equals zero

$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

where $\displaystyle y = r\sin{\theta} = \theta\sin{\theta}$

and $\displaystyle x = r\cos{\theta} = \theta\cos{\theta}$

3. Originally Posted by dylan5188
1.Find the length of the polar curve given by on the interval .
try $\displaystyle L = \int_0^{\frac{1}{6}}\sqrt{1+(3.6e^{0.4\theta})^2}~ d\theta$

4. 2. vertical tangent lines occur where $\displaystyle \frac{dy}{dx}$ is undefined

horizontal tangent lines occur where $\displaystyle \frac{dy}{dx}$ equals zero

$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

where $\displaystyle y = r\sin{\theta} = \theta\sin{\theta}$

and $\displaystyle x = r\cos{\theta} = \theta\cos{\theta}$[/quote]

So I got theta=tantheta when horizontal tangent occur
and theta=cotangent theta when vertical tangent occur. Then How to calculate them?

5. Originally Posted by dylan5188
2. vertical tangent lines occur where $\displaystyle \frac{dy}{dx}$ is undefined

horizontal tangent lines occur where $\displaystyle \frac{dy}{dx}$ equals zero

$\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

where $\displaystyle y = r\sin{\theta} = \theta\sin{\theta}$

and $\displaystyle x = r\cos{\theta} = \theta\cos{\theta}$

So I got theta=tantheta when horizontal tangent occur
and theta=cotangent theta when vertical tangent occur. Then How to calculate them?[/quote]

correction ... horizontal tangent occurs when $\displaystyle \theta = -\tan{\theta}$

have to solve these with a calculator.

6. Still dont know how to calculate

7. Originally Posted by dylan5188
Still dont know how to calculate
graph the equation $\displaystyle y = x + \tan{x}$ with your graphing calculator and look for the zeros between $\displaystyle 0$ and $\displaystyle 2\pi$