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Alenkinab [10]
3 years ago
14

Find the exact length of the polar curve. r = θ2, 0 ≤ θ ≤ 3π/2

Mathematics
2 answers:
Pepsi [2]3 years ago
7 0
Hello my friend it is the 21 and 31
Butoxors [25]3 years ago
7 0

Answer:

Length of polar arc =

\displaystyle\frac{1}{3}\bigg[ \bigg(\frac{9\pi^2}{4} + 4\bigg)^\frac{3}{2} - 8 \bigg]

Step-by-step explanation:

We have to find the arc length of the polar curve.

We are given the following:

r(\theta) = f(\theta) = \theta^2,\\\\0 \leq \theta \leq \displaystyle\frac{3\pi}{2}

Length of polar curve =

\displaystyle\int_b^a \sqrt{r^2 + \bigg(\displaystyle\frac{dr}{d\theta}\bigg)^2}~d\theta

Putting the values:

\displaystyle\frac{dr}{d\theta} = \frac{d(\theta^2)}{d \theta} = 2\theta\\\\\text{Length of polar curve} = \displaystyle\int_b^a \sqrt{r^2 + \bigg(\displaystyle\frac{dr}{d\theta}\bigg)^2}~d\theta\\\\= \displaystyle\int^\frac{3\pi}{2}_0 \sqrt{\theta^4 + (2\theta)^2}~d\theta\\\\= \displaystyle\int^\frac{3\pi}{2}_0 \sqrt{\theta^4 + 4\theta^2}~d\theta\\\\= \displaystyle\int^\frac{3\pi}{2}_0 \theta\sqrt{\theta^2+4}~d\theta \\\\= \bigg|\frac{(\theta^2 + 4)^\frac{3}{2}}{3}\bigg|_0^\frac{3\pi}{2}

= \displaystyle\frac{1}{3}\bigg[ \bigg(\frac{9\pi^2}{4} + 4\bigg)^\frac{3}{2} - (4)^\frac{3}{2}\bigg]\\\\= \displaystyle\frac{1}{3}\bigg[ \bigg(\frac{9\pi^2}{4} + 4\bigg)^\frac{3}{2} - 8 \bigg]

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5 0
3 years ago
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4 0
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5 0
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Verdich [7]
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