Find the polar equation for the cartesian curve x^2-y^2 = sqrt(x^2+y^2)

Mathematics

1 answer:

cestrela7 [59]3 years ago

8 0

ANSWER:

r = $\frac{1}{cos2(theta)}$

Explaination:

Convert the given curve into the the polar form.

x = rcosθ

y = rsinθ

in f(x,y) = (x²-y²) - √(x²+y²) = 0

put the values of x & y in given curve equation.

We get at,

g(r,θ) = (r²cos²θ - r²sin²θ) - √(r²cos²θ + r²sin²θ) = 0

g(r,θ) = r²(cos²θ - sin²θ) - √r² = 0

We know that,

cos²θ - sin²θ = cos2θ

g(r,θ) = r²(cos2θ) - r = 0

Solve for r

Finally we get:

r = $\frac{1}{cos2(theta)}$

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5. Based on recent results, scores on the SAT test are normally distributed with a mean of 1511 and a standard deviation of 312.

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Answer:

The actual SAT score is 2024.

The equivalent ACT score is 29.49.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

SAT score that is in the 95th percentile

Scores on the SAT test are normally distributed with a mean of 1511 and a standard deviation of 312, which means that $\mu = 1511, \sigma = 312$