# What is the arclength of #r=costheta-3sin2theta # on #theta in [-pi/4,pi]#?

By a numerical method we obtain

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To find the arc length of the polar curve ( r = \cos(\theta) - 3\sin(2\theta) ) on the interval ( \left[-\frac{\pi}{4}, \pi\right] ), we use the formula for arc length in polar coordinates:

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

where ( r ) is the function defining the polar curve, ( \alpha ) and ( \beta ) are the starting and ending angles respectively.

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

[ \frac{dr}{d\theta} = -\sin(\theta) - 6\cos(2\theta) ]

Now, we plug ( r ) and ( \frac{dr}{d\theta} ) into the arc length formula and integrate over the given interval:

[ L = \int_{-\frac{\pi}{4}}^{\pi} \sqrt{(\cos(\theta) - 3\sin(2\theta))^2 + (-\sin(\theta) - 6\cos(2\theta))^2} , d\theta ]

Solving this integral will give us the arc length of the polar curve over the interval ( \left[-\frac{\pi}{4}, \pi\right] ).

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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.

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- What is the slope of the tangent line of #r=5theta+cos(-theta/3-(pi)/2)# at #theta=(-5pi)/6#?

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