# What is the arclength of #r=2sin(theta) -3theta# on #theta in [(-3pi)/8,(7pi)/8]#?

I used WolframAlpha

From the reference Arc Length with Polar Coordinates

Square the function:

Compute the derivative of the function:

Square the derivative:

Compute the argument under the radical:

Combine like terms:

Substitute into the integral:

I used WolframAlpha to evaluate the integral:

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To find the arc length of the curve defined by ( r = 2\sin(\theta) - 3\theta ) on the interval ( \theta \in \left[-\frac{3\pi}{8}, \frac{7\pi}{8}\right] ), we use the formula for polar arc length:

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

Substitute ( r = 2\sin(\theta) - 3\theta ) and ( \frac{dr}{d\theta} = 2\cos(\theta) - 3 ) into the formula and integrate over the given interval. Then calculate the definite integral.

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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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