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Find the points of horizontal tangency to the polar curve.r = 2 csc(๐œƒ) + 3 ย ย ย ย  0 โ‰ค ๐œƒ < 2๐œ‹

Question

Find the points of horizontal tangency to the polar curve.

r=2cscโก(๐œƒ)+3 r = 2 \csc(๐œƒ) + 3

0โ‰ค๐œƒ<2ฯ€ 0 \leq ๐œƒ < 2\pi

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Solution

To find the points of horizontal tangency to the polar curve r = 2 csc(๐œƒ) + 3, we need to find the values of ๐œƒ for which the derivative of r with respect to ๐œƒ is zero.

Step 1: Express r in terms of sin(๐œƒ) r = 2/sin(๐œƒ) + 3

Step 2: Differentiate r with respect to ๐œƒ dr/d๐œƒ = -2cos(๐œƒ)/sin^2(๐œƒ)

Step 3: Set the derivative equal to zero and solve for ๐œƒ -2cos(๐œƒ)/sin^2(๐œƒ) = 0 This implies that cos(๐œƒ) = 0

The solutions to cos(๐œƒ) = 0 in the interval 0 โ‰ค ๐œƒ < 2๐œ‹ are ๐œƒ = ๐œ‹/2 and ๐œƒ = 3๐œ‹/2.

Step 4: Substitute these values back into the original equation to find the corresponding r values r = 2 csc(๐œ‹/2) + 3 = 2 + 3 = 5 r = 2 csc(3๐œ‹/2) + 3 = -2 + 3 = 1

So, the points of horizontal tangency to the polar curve are (5, ๐œ‹/2) and (1, 3๐œ‹/2).

This problem has been solved

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