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In this video, weβre going to learn how to convert between parametric and rectangular or Cartesian equations.
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Weβll learn how this process can help us to sketch the curve given by a pair of parametric equations and how to parameterize circles with centres at the origin and line segments between two points.
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Letβs begin by recalling that a Cartesian equation or a rectangular equation is one given in the variables π₯ and π¦.
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Itβs usually given as π¦ is equal to some function of π₯, although thatβs not always the case.
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A pair of parametric equations are given in terms of a third variable, usually π‘, such that π₯ is equal to some function of π‘ and π¦ is equal to some other function of π‘.
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And one method we have for converting from parametric to rectangular form is rearranging to eliminate the variable π‘.
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Letβs see what that might look like.
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Convert the parametric equations π₯ equals π‘ squared plus two and π¦ equals three π‘ minus one to rectangular form.
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Here, we have a pair of parametric equations.
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We have π₯ is equal to some function of π‘ and π¦ is equal to some other function of π‘.
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To convert parametric equations to rectangular form, we need to find a way to eliminate the π‘.
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So looking at our equations, we can see that we can rearrange the equation in π¦ to make π‘ the subject.
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We begin by adding one to both sides.
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And then, we divide through by three.
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So we see that π‘ is equal to π¦ plus one all over three.
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Now, we go back to our equation for π₯.
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We replace π‘ with π¦ plus one over three.
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And we find that π₯ equals π¦ plus one over three all squared plus two.
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And there will be certain circumstances where weβre required to distribute the parentheses and simplify.
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In this case, thatβs not necessary.
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And so, weβre finished.
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Weβve converted the parametric equations π₯ equals π‘ squared plus two and π¦ equals three π‘ minus one into rectangular or Cartesian form.
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Itβs π₯ equals π¦ plus one over three all squared plus two.
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In our next example, weβre going to look at how trigonometric identities can help us to eliminate the variable π‘.
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Find the Cartesian equation of the curve defined by the parametric equations π₯ equals two plus cos π‘ and π¦ equals four cos of two π‘.
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Remember, a Cartesian equation is one which contains only the variables π₯ and π¦.
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So weβre going to need to find a way to eliminate our third variable π‘ from our parametric equations.
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And at first glance, it doesnβt seem to be a nice way to do so.
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But we can begin by recalling some trigonometric identities.
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We have cos of two π‘ in our second parametric equation.
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And we know that cos of two π‘ is equal to two times cos squared π‘ minus one.
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This means we can rewrite our equation for π¦ as four times two cos squared π‘ minus one.
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Next, weβll look to rearrange our equation for π₯ to make cos of π‘ the subject.
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Once weβve done that, weβll be able to find an expression for cos squared π‘ in terms of π₯.
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We can subtract two from both sides.
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And we see that π₯ minus two equals cos of π‘.
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Then, by squaring both sides of this equation, we find that cos squared π‘ is equal to π₯ minus two all squared.
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And so, weβre now able to replace cos squared π‘ with π₯ minus two squared.
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That gives us π¦ equals four times two times π₯ minus two all squared minus one.
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We distribute this first pair of parentheses.
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And we find that π₯ minus two all squared is equal to π₯ squared minus four π₯ plus four.
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We distribute again by multiplying each of these terms by two and then simplifying: eight minus one is seven.
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Well, finally, we distribute one more time by multiplying each term of two π₯ squared minus eight π₯ plus seven by four.
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And we find the Cartesian equation of the curve defined by parametric equations π₯ equals two plus cos π‘ and π¦ equals four cos of two π‘ is π¦ equals eight π₯ squared minus 32π₯ plus 28.
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Now, we might be able to see that this process can be really useful in helping us to sketch curves given in parametric form.
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For example, in this case, we were given a pair of parametric equation.
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It might now be instantly obvious what the curve defined by these parametric equations looks like.
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But by writing it in Cartesian form, we can see that we have a quadratic curve with a positive leading coefficient of π₯.
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So we know weβre going to have that typical parabola shape.
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And we can use our usual methods for sketching quadratic curves to sketch the curve defined by these parametric equations.
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Letβs see what the whole process might look like.
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Convert the parametric equations π₯ equals three cos π‘ and π¦ equals three sin π‘ to rectangular form.
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Remember, the rectangular form of an equation is one which contains the variables π₯ and π¦ only.
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So weβll need to find a way to eliminate the third variable π‘ from our parametric equations.
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And at first glance, it doesnβt seem to be a nice way to do so.
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But letβs recall some trigonometric identities.
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We know that cos squared π plus sin squared π is equal to one.
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So letβs begin by simply squaring our expressions for π₯ and π¦.
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With π₯, we get π₯ squared equals three cos π‘ all squared, which is equal to nine cos squared π‘.
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And so, we can say that cos squared π‘ must be equal to π₯ squared over nine.
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Similarly, we can say that π¦ squared is equal to three sin π‘ all squared.
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We distribute the parentheses and we find that π¦ squared is nine times sin squared π‘.
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And then, we divide through by nine.
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And we see that sin squared π‘ is equal to π¦ squared over nine.
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Then, if we replace π with π‘ in our identity, remember that doesnβt change the identity.
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We see that we can replace cos squared π‘ with π₯ squared over nine.
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We can replace sin squared π‘ with π¦ squared over nine.
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And then, this is all equal to one.
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We next multiply through by nine.
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And we find that the rectangular form of our equation is equal to π₯ squared plus π¦ squared equals nine.
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Hence sketch the curve described by this pair of parametric equations.
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Well, weβve just seen that this pair of parametric equations π₯ equals three cos π‘ and π¦ equals three sin π‘ is π₯ squared plus π¦ squared equals nine in rectangular form.
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And so, we recall that the equation of a circle whose centre is at π, π and whose radius is π is π₯ minus π all squared plus π¦ minus π all squared equals π squared.
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We compare our equation β thatβs π₯ squared plus π¦ squared equals nine.
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And in doing so, we see that π and π simply have to be zero.
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We also find that π squared is equal to nine.
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Well, if we square root this, we find that π is equal to three.
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And so, we see that the parametric equations π₯ equals three cos π‘ and π¦ equals three sin π‘ give a circle centred at zero, zero β thatβs the origin β with a radius of three.
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That might look a little something like this.
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We do, however, need to work out the direction which our curve is being sketched.
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So letβs work out the value of our π₯- and π¦-coordinates when π‘ is equal to zero.
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When π‘ is equal to zero, π₯ is equal to three times cos of zero, which is simply three.
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And when π‘ is equal to zero, π¦ is equal to three sin of zero, which is zero.
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So we would begin by plotting the point with Cartesian coordinates three, zero.
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Letβs next choose π‘ is equal to π by two.
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Now, we could, of course, choose π‘ equals one.
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But since weβre working with trigonometric expressions, π by two seems to make a little more sense.
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So when π‘ is equal to π by two, π₯ is equal to three cos of π by π‘, which is zero and π¦ is equal to three times sin of π by two, which is equal to three.
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So the second point we plot has Cartesian coordinates zero, three.
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We can, therefore, see that we move in a counterclockwise direction while sketching the circle.
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Next, weβre going to look at how we can go up from rectangular into parametric form.
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Convert the rectangular equation π₯ squared plus π¦ squared equals 25 to parametric form.
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Letβs begin by recalling what we actually know about this equation.
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We know that a circle whose centre is at the origin and whose radius is π can be given by the Cartesian equation π₯ squared plus π¦ squared equals π squared.
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By rewriting our rectangular equation as π₯ squared plus π¦ squared equals five squared, we see that we have a circle whose centre is at the origin and whose radius is five units.
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And so, Iβve sketched that on the π₯π¦-plane, as shown.
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Weβre looking to convert this equation to parametric form.
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So we know that a pair of parametric equations describe the π₯- and π¦-coordinates in terms of a third parameter, π‘.
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So letβs pick a general point π₯, π¦.
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Iβm going to choose this one in the first quadrant.
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We can drop in a right triangle, whose height is π¦ units and whose width is π₯ units.
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And then, we can label the included angle π‘.
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Since the radius of the circle is five units, we know that the hypotenuse of our triangle is five.
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Then, using standard conventions, we label the sides of our triangle.
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We have the adjacent; thatβs π₯.
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We have the opposite; thatβs π¦.
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We have the hypotenuse; thatβs five.
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We also know that in rectangular trigonometry, sin π is equal to the opposite divided by the hypotenuse and cos π is equal to the adjacent over the hypotenuse.
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So we can say that sin π‘ equals π¦ over five and cos π‘ is equal to π₯ over five.
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We can multiply through by five for both of our equations.
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And we find that π¦ is equal to five sin π‘ and π₯ equals five cos π‘.
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So for any point in our circle, the π¦-coordinate is given by five sin of π‘ and the π₯-coordinate is given by five cos of π‘.
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Now moving in a counterclockwise direction from the positive horizontal axis, we see that as π₯ increases from zero, it generates our corresponding π₯- and π¦-coordinates.
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Then, our equations are π₯ equals five cos π‘ and π¦ equals five sin π‘.
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Now, in these previous two examples, weβve seen that the equation of a circle whose centre is at the origin and whose radius is π is given by the equation π₯ squared plus π¦ squared equals π squared.
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But then, we also saw we can convert this into parametric form.
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In this case, we had π¦ equals five sin π‘ and π₯ equals five cos π‘, where five is the radius.
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In general, we can state this as a result.
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We define a circle as the locus of the points that satisfy the equations π₯ equals π cos π‘ and π¦ equals π sin π‘, where π₯ and π¦ are the coordinates of any point on the circle, π is the radius, and π‘ is the parameter; itβs the angle subtended by the point at the circle centre.
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We can also use our knowledge of simple coordinate geometry to parameterize a line segment.
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Letβs see what that might look like.
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Consider the points π΄ equals negative one, one and π΅ equals four, two.
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Parameterize the segment π΄π΅, where π‘ is greater than or equal to zero and less than or equal to one.
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Letβs begin by simply sketching the diagram, showing our points and our line segment π΄π΅.
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It looks a little something like this.
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Now, we know weβre going to go from left to right generally.
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So weβll begin by letting π‘ be equal to zero at the first point, at negative one, one.
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And that means we need to let π‘ be equal to one at the other end of our interval at the point four, two.
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Now, weβre going to use the vector form for the equation of a line.
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Itβs π equals π₯ nought, π¦ nought plus π‘ times ππ.
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Now, whilst this is an equation of a line in two dimensions, this process can be extended to working with a line in three dimensions.
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Weβll next say that the vector π is of the form π₯, π¦.
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Then, weβll combine the vectors on the right to π₯ nought plus π‘π, π¦ nought plus π‘π.
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And we see that the only way for the vector on the left to be equal to the vector on the right is if its component parts are equal.
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That is, if π₯ is equal to π₯ nought plus π‘π and π¦ is equal to π¦ naught plus π‘π.
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In fact, this pair of equations is called the parametric Form of the equation of a line.
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Now, we can use these along with the information in our question to parameterize the line segment from π to π.
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Letβs go back to what we said earlier.
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We said π‘ is equal to zero at negative one, one.
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When π‘ is equal to zero, this gives us the values for π₯ nought, π¦ nought.
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So π₯ nought must be equal to negative one and π¦ nought must be equal to one.
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So we see the π₯ is equal to negative one plus π‘π and π¦ is equal to one plus π‘π.
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Weβll now use the fact that when π‘ is equal to one, π₯ is equal to four and π¦ is equal to two.
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So our first equation becomes four equals negative one plus one π.
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And our second equation becomes two equals one plus one π.
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By adding one to both sides of our first equation, we find π to be equal to five.
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And by subtracting one from both sides of our second, we find π is equal to one.
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And we have our parametric equations to describe the line segment π΄ to π΅ for values of π‘ between zero and one.
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They are π₯ equals five π‘ minus one and π¦ equals π‘ plus one.
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In this video, weβve seen that if we want to convert from a parametric equation to a rectangular equation, we need to find a way to eliminate π‘.
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We also saw that, in general, a circle whose centre is at the origin and whose radius is π is defined by the parametric equations π₯ equals π cos π‘ and π¦ equals π sin π‘.
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And finally, we saw that the parametric equation of a line is π₯ equals π₯ nought plus π‘π and π¦ equals π¦ nought plus π‘π.