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What would the polar coordinate graph of r^2=81cos2θ like?

User Ebrahimi
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Answer:

an infinity sign

Explanation:

What would the polar coordinate graph of r^2=81cos2θ like?-example-1
User Tom Shen
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The equation of a polar curve r^2 = a cos 2θ describes a limaçon, which is a type of polar curve with a loop. When a is positive, the limaçon has an outer loop and an inner loop. When a is negative, the limaçon has a dimple instead of an inner loop.

In this case, a = 81, which is positive, so the limaçon will have an outer loop and an inner loop.

To graph the limaçon, we can start by considering the behavior of the curve at θ = 0, π/4, π/2, 3π/4, and π.

At θ = 0, r^2 = 81cos(2θ) = 81cos(0) = 81, so r = ±9. This means that there are two points on the curve at θ = 0, one at (9, 0) and one at (-9, 0).

At θ = π/4 and 3π/4, cos(2θ) = cos(π/2) = 0, so r^2 = 0 and r = 0. This means that there are two points on the curve at θ = π/4 and two points at θ = 3π/4, all located at the origin.

At θ = π/2, r^2 = 81cos(2θ) = 81cos(π) = -81, which is negative. Since r is always non-negative, there are no points on the curve at θ = π/2.

At θ = π, r^2 = 81cos(2θ) = 81cos(2π) = 81, so r = ±9. This means that there are two points on the curve at θ = π, one at (9, π) and one at (-9, π).

Based on this information, we can sketch the graph of the limaçon as follows:

There are two points on the x-axis, one at (9, 0) and one at (-9, 0).

There are four points at the origin.

There are two points on the line θ = π, one at (9, π) and one at (-9, π).

The curve passes through the origin at θ = π/4, π/2, and 3π/4, and has a loop that encloses the origin.

Overall, the limaçon has a symmetric, flower-like shape with a loop that encloses the origin.

User Richard Peterson
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