path: root/maths/work/rev/parametrics.ms

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Parametric equations
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Lucas Standen
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QMC

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What are they?

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Parametric equations are two equations that are linked by a common variable usually @ t #

They are written in the format 
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x = at
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y = bt
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The can also include trig functions, exponents and other parts of maths

You may be asked to, convert to Cartesian, find the range and domain, differentiate and finding points of intersection

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Converting to Cartesian
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You will often be asked to convert to a Cartesian equivalent equation. To do that you will need to rearrange one of the given equations to get it in terms of @ t #, then substitute that value into the other equation. You will most often find that rearranging @ x # is easier as this will result in an equation equal to @ y #.

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Example
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x = 2t
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y = t sup 2
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t = x over 2
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y = ({x over 2}) sup 2
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y = x sup 2 over 4
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Finding the domain and range
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The domain of the Cartesian equation is the range of the @ x # and the range of the Cartesian is the range of the @ y # equation.

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Examples
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x = t - 2
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y = t sup 2 + 1
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"where" -4 <= t <= 4
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t = x + 2
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y = (x + 2) sup 2 + 1
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y = x sup 2 + 4x + 4 + 1
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y = x sup 2 + 4x + 5
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f(x) = x sup 2 + 4x + 5
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domain = -6 <= x <= 2
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range = 1 <= f(x) <= 16
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Differentiating Parametric equations
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This process is relatively simple, and can be solved by viewing @ dy over dx # as fractions. As we can already find @ dx over dt # and @ dy over dt # we can say the following.

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dy over dx = {dy over dt} over {dx over dt}
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This is because the @ dt # will cancel out on the top and bottom.

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Examples
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x = 2t
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y = t sup 2 - 3t + 2
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dx over dt = 2
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dy over dt = 2t - 3
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{dy over dt} over {dx over dt} = 2 over {2t -3}
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Points of intersection
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This is as simple as substituting in numbers for the most part although sometimes it can be harder due to trig identities showing up.

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Example
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Lets say we have the parametric @ x = t sup 2 # and @ y = 8t - 5 # and we want to know if there is any points of
intersection with the line @ y = 5x + 4 #

To solve this we can find the Cartesian equation and then do this normally.
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t = sqrt x
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y = 8{sqrt x} - 5
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"Then we can say"
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5x + 4 = 8{sqrt x} - 5
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5x + 9 = 8{sqrt x}
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5x over 8 + 9 over 8 = sqrt x
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({5x over 8 + 9 over 8}) sup 2 = x
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{25x sup 2 over 64 + 81 over 64} = x
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25x sup 2 + 81 = 64x
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25x sup 2 -64x + 81 = 0
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"This has no points of intersection because x is imaginary"
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EX 8D Q1a
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x = 5 + t
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y = 6 - t
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y = 6 - (x - 5)
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y = 11 - x
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0 = 11 - x
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-11 = - x
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11 = x
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EX 8D 12b
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x = 6cos(t)
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y = 4sin(2t) + 2
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-{pi over 2} < t < {pi over 2}
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4 = 4sin(2t) + 2
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2 = 4sin(2t) 
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1 over 2 = sin(2t) 
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2t = arcsin({1 over 2})
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2t = pi over 6, {5 pi} over 6
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t = pi over 12, {5 pi} over 12
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EX 8D 12c

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x = 6cos(t)
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x = 3 sqrt 3, - 3 sqrt 3
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(3 sqrt 3, 4), (- 3 sqrt 3, 4)
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EX 8D 13

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x = 2t
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y = 4t sup 2 - 4t
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t = x over 2
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y = 4({x over 2}) sup 2 - 4 ({x over 2})
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y = 4({x sup 2 over 4}) - 4 ({x over 2})
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y = x sup 2 - 4 ({x over 2})
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y = x sup 2 - 2x 
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2x - 5  = x sup 2 - 2x 
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0 = x sup 2 - 4x + 5
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sqrt {-4 sup 2 - 4 (1) (5) }
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sqrt {16 - 20 }
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sqrt {-4}
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sqrt {-4} " has no real solutions!"
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Parametric with trig
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Trig often shows up in parametric equations, however using the identities we know, they can be easy to solve. The reason they are harder, is because we can't rearrange to get @ t # like in other questions.

What we have to do is to use the identities we know, such as
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sin sup 2 (x) + cos sup 2 (x) = 1
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An example would be the following

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x = 2sin(t)
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y = cos(t) + 2
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sin(t) = x over 2
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cos(t) = y - 2
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({x over 2}) sup 2 + (y - 2) sup 2 = 1
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{x sup 2 over 4} + y sup 2 -4y + 4 = 1
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{x sup 2 over 4} + y sup 2 -4y + 4 = 1
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{x sup 2} + 4y sup 2 -16y + 16 = 4
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{x sup 2} -16y + 16 - 4= -4y sup 2
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{x sup 2} + 12 = -4y sup 2 + 16y
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This is an oval shape.