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In this video, weβll learn how we can use our understanding of polar and Cartesian coordinates to change between equations in polar and Cartesian form.
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Weβll consider how these techniques might help us to recognise graphs of equations written in polar form by converting to Cartesian or rectangular form and interpreting from there.
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Remember, the polar coordinate system is a way of describing points in a plane by their distance from the origin or the pole and the angle the line joining this point of the origin makes with the positive horizontal axis measured in a counterclockwise direction.
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These are of the form ππ, where π is the distance from the origin to that point and π is that angle.
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We convert from polar to Cartesian form using the formulae π₯ equals π cos π and π¦ equals π sin π.
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And these equations are suitable for all values of π and π.
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The converse formulae are π squared equals π₯ squared plus π¦ squared and tan π is equal to π¦ divided by π₯.
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Now in this case, we need to be a little bit careful establishing the value of π because it works beautifully for coordinates plotted in the first quadrant.
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But for other quadrants, the calculator can give us an incorrect value.
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And we do have a set of rules that we can follow for calculating the exact value of π.
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However, we donβt really need this formula in this video.
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Weβre looking to establish how we can convert between polar equations, one where π is some function of π, and Cartesian or rectangular equations, where π¦ is some function of π₯.
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We do, however, use the other three formulae to perform these conversions.
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Letβs see what that might look like.
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Convert the equation π₯ squared plus π¦ squared equals 25 into polar form.
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Remember, we convert polar coordinates to Cartesian or rectangular coordinates using the formulae π₯ equals π cos π and π¦ equals π sin π.
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And these are suitable for all values of π and π.
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In our original equation, weβve got π₯ squared and π¦ squared.
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So letβs use our formulae for π₯ and π¦ to generate expressions for π₯ squared and π¦ squared in terms of π and π.
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Since π₯ is equal to π cos π, it follows that π₯ squared must be π cos π all squared, which we can distribute and say that π₯ squared is equal to π squared times cos squared π.
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Similarly, we see that π¦ squared must be equal to π sin π all squared, which is equal to π squared sin squared π.
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Now our original equation says that the sum of these is equal to 25.
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So we can say that π squared cos squared π plus π squared sin squared π equals 25.
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Our next step is to factor π squared on the left-hand side of this equation.
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So π squared times cos squared π plus sin squared π equals 25.
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But why did we do this?
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Well, here is where itβs useful to know some of our trigonometric identities by heart.
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We know that cos squared π plus sin squared π is equal to one for all values of π.
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So we can replace cos squared π plus sin squared π in our equation with one.
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So π squared times one equals 25.
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Well, we donβt need this one.
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π squared is simply equal to 25.
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We solve this equation by taking the square root of both sides.
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And we find that π is equal to five.
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Remember, we would usually take both the positive and negative square root of 25.
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But since π represents a length, we donβt need to.
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π₯ squared plus π¦ squared equals 25 is the same as π equals five in polar form.
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Now if we think about what we know about the equation π₯ squared plus π¦ squared equals 25 and polar coordinates, this makes a lot of sense.
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The equation π₯ squared plus π¦ squared equals 25 represents a circle whose centre lies at the origin and whose radius is the square root of 25, or five.
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We can also think about what the equation π equals five means in polar form.
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Well, it must be all points whose distance from the origin is five units.
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Now of course, if we think right back to what we know about locus or loci, it follows that this form is a circle whose centre is the origin and whose radius is five.
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Letβs now look at converting a polar equation into rectangular form.
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Convert the polar equation π equals four cos π minus six sin π to the rectangular form.
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Remember, we convert from polar coordinates to Cartesian coordinates or rectangular coordinates using the following formulae.
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π₯ is equal to π cos π and π¦ is equal to π sin π.
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And these are suitable for all values of π and π.
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Our aim is going to be to manipulate each of our equations so that we have an equation for cos π and sin π.
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Well, if we divide both sides of our first equation by π, we see that cos π is equal to π₯ over π.
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Similarly, dividing through by π in our second equation, and we find sin π equals π¦ over π.
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We can then replace cos π with π₯ over π and sin π with π¦ over π in our original polar equation.
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And we see that π is equal to four times π₯ over π minus six times π¦ over π.
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This simplifies to four π₯ over π minus six π¦ over π.
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Weβre next going to multiply everything through by π.
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And we find that π squared equals four π₯ minus six π¦.
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Now weβre obviously not quite done.
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We want to convert to rectangular form.
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This is usually of the form π¦ is equal to some function of π₯, although weβre essentially looking for an equation with π₯ and π¦ as its only variables.
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So we recall the other conversion formulae that we use to convert Cartesian to polar coordinates.
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Thatβs π squared equals π₯ squared plus π¦ squared.
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We should be able to see now that we can replace π squared with π₯ squared plus π¦ squared.
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So π₯ squared plus π¦ squared equals four π₯ minus six π¦.
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Weβre almost there.
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You might recognise this equation.
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Weβre going to manipulate it by completing the square.
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We subtract four π₯ from both sides and add six π¦.
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Then weβre going to complete the square for π₯ and π¦.
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We halve the coefficient of π₯ β that gives us negative two β and then subtract negative two squared.
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So we subtract four.
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Similarly, we halve the coefficient of π¦ to get three and then subtract three squared, which is nine.
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And of course, this is all equal to zero.
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Negative four minus nine is negative 13.
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So we add 13 to both sides of our equation.
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And in rectangular form, our equation is π₯ minus two all squared plus π¦ plus three all squared equals 13.
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Now this is actually a really useful technique as it tells us something about the shape of the graph.
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We canβt easily spot what the shape of the graph given by π equals four cos π minus six sin π looks like.
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But we do know that a circle whose centre lies at π, π and whose radius is π has the equation π₯ minus π all squared plus π¦ minus π all squared equals π squared.
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So our polar equation, which also has the rectangular form π₯ minus two all squared plus π¦ plus three all squared equals 13, must be a circle whose centre is at two, negative three and whose radius is the square root of 13.
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Letβs have a look at a similar example.
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Consider the rectangular equation π₯ squared minus π¦ squared equals 25.
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Convert the given equation to polar form.
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The second part of this question asks us, which of the following is the sketch of the equation?
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So we begin by recalling that we can convert from polar to Cartesian coordinates using the formulae π₯ equals π cos π and π¦ equals π sin π.
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Our equation contains π₯ squared and π¦ squared.
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So letβs square each of these formulae.
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And when we do, we find that π₯ squared equals π squared cos squared π and π¦ squared equals π squared sin squared π.
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We know the difference of these is equal to 25.
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Thatβs our rectangular equation.
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So π squared cos squared π minus π squared sin squared π equals 25.
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We can then factor π squared.
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So π squared times cos squared π minus sin squared π equals 25.
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But we know that cos two π is equal to cos squared π minus sin squared π.
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So letβs replace cos squared π minus sin squared π with cos two π.
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That tells us that π squared times cos two π equals 25.
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And we can then divide both sides of this equation by cos two π.
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Now of course, one over cos π is sec π.
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So we find that π squared is equal to 25 sec two π.
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For part two, we need to identify which of the following is a sketch of the equation.
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Now we wouldnβt be particularly easy to sketch the graph of π squared equals 25 sec two π.
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But we do know the general form of the graph whose equation is π₯ over π all squared minus π¦ over π all squared equals one.
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Itβs a standard hyperbola, centred at the origin with vertices at plus or minus π, zero and covertices at zero, plus or minus π.
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Letβs rearrange our equation to equate it to one.
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To do that, we divide everything through by 25.
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And since 25 is five squared, we can write this as π₯ over five all squared minus π¦ over five all squared equals one.
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We know we have a standard hyperbola with vertices at plus or minus five, zero.
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And in fact, thereβs only one graph that satisfies this criteria.
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Itβs graph A.
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Of course, it is useful to know that if we were struggling, we could even try substituting some values of π₯ or π¦ in and sketching the order pairs.
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Letβs now have a look at another example which involves sketching the graph.
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Sketch the graph of π equals two csc π.
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Here we have a polar equation.
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And itβs not particularly easy to spot what the graph of this function might look like.
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So instead, weβre going to convert into rectangular form first.
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We recall that csc π is one over sin π.
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We also know that one of the formulae we use to convert from polar to Cartesian form is the formula π¦ equals π sin π.
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Dividing through by π, and we obtain the second formula is equivalent to saying sin π equals π¦ divided by π.
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So csc π must be equivalent to one over π¦ over π.
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Well, when we divide by a fraction, we multiply by the reciprocal of that fraction.
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So we can actually say that csc π must be equal to π over π¦.
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By replacing csc π with π over π¦ in our original equation, we find that π is equal to two times π over π¦.
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Letβs divide through by π.
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We get one equals two over π¦.
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Next, weβll multiply by π¦.
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And we find in rectangular form the equation is π¦ equals two.
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And of course, we can now easily sketch this.
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Itβs simply the horizontal line that passes through the π¦-axis at two.
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This is a great demonstration of when converting to rectangular form can make sketching a graph given in polar form much more simple.
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In this video, weβve seen that, by using the formulae for converting between polar and Cartesian coordinates, we can convert quite easily between polar and rectangular equations.
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We also saw that this process can help us sketch more complicated graphs given in polar form.