In Exercises 79–80, convert each polar equation to a rectangular ...

Question

In Exercises 79–80, convert each polar equation to a rectangular equation. Then determine the graph's slope and y-intercept.r sin (θ − π/4) = 2

Accepted Answer

Hello, today we're going to be transforming the given polar equation into rectangular form, then we're going to identify the slope and Y intercept of the resulting equation. So before we take a look at the equation, one thing to understand is that if we are given an equation in polar form, the equation is in terms of R comma theta. And since we want to transform this into rectangular form, we're going to be transforming the terms of our comma theta into terms of X comma Y. We have two transformation factors to help us with that process. We have X is equal to our cosine of data and we have Y is equal to our sign of the. So let's take a look at the given equation. We have R multiplied by sine of theta minus pi over three and that is equal to four. So before we transform this equation from polar to rectangular, we're first going to need to simplify the sign of the minus pi over three term. Now we have a trigonometric identity for a sign difference identity. To help us with that we are given that sine of A minus B can be rewritten as sign of A multiplied by cosine of B minus cosine of A multiplied by sign of B, we can apply this identity to the given equation. But we first need to identify our A and B terms. Well, currently theta is in the A spot. So A is equal to theta and pi over three takes up the spot of B. So B is equal to pi over three. So if we were to use this identity, we can rewrite the left hand side of the equation as R multiplied by the quantity of sine of theta multiplied by cosine of pi over three minus cosine of theta multiplied by sine of pi over three. And that is all equal to four. So there's a bit of simplification that we can do to the current equation. First cosine pi over three will evaluate to give us one half and sine pi over three will evaluate to give us the square over 3/2. So we can rewrite the left hand side of the equation as R multiplied by the quantity of sine theta multiplied by a half minus cosine of theta multiplied by the quantity square over 3/2. And that is all equal to four. The next thing we can do is we can distribute R to each term inside of the difference. And that will give us R multiplied by sine of data multiplied by a half minus R multiplied by cosine of data. Multiplied by the square root 3/2. And that is all equal to four. The next thing we can do is we can at least eliminate the denominators of the fractions. And we can do that by multiplying every term in the equation by two, that will leave us with R multiplied by sine of the minus R multiplied by cosine of theta multiplied by the square root of three. And that is equal to four multiplied by two, which will give us eight. Now recall that earlier, we stated that X is equal to our cosine of data and Y is equal to our sign of the. And notice that currently we have our sign of data term and our cosign data of term, we can rewrite this with its respective X and Y terms. And that will leave us with Y minus X multiplied by the square root of three is equal to eight, negative X multiplied by the square root of three will give us negative square root of three multiplied by X. So we have Y minus the square root of three, X is equal to eight. And finally, if we solve for Y by adding the squared of three X to both sides, what we are left with is Y is equal to the square root of three X plus eight. This is going to be the rectangular form of the given polar equation. Now we can move on to the next part of the problem. And that asks us to identify the slope and the Y intercept. So we know that the rectangular form of the given polar equation is Y is equal to the square root of three X plus eight. This is an equation of the slope of line in point slope form. I mean slope intercept form, I mean apologies about that. And the slope intercept form of a line is Y is equal to MX plus B. Here M represents the slope and B represents the Y intercept. So currently squared of three is in the position of M and eight is in the position of B. What this means is that the slope of the equation of the line is going to equal to the square root of three. And the Y intercept of the equation of the line is equal to eight. And that is going to be the answer to the second part of the problem. So just to summarize, we converted the polar equation to rectangular form and that gave us Y is equal to the square root of three X plus eight. And the slope and Y intercept of this equation has a slope of the square root of three and the Y intercept of eight. And these are going to be the solutions to the problem. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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In Exercises 79–80, convert each polar equation to a rectangular equation. Then determine the graph's slope and y-intercept.r sin (θ − π/4) = 2

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