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Parametric equations
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Lucas Standen
.AI
QMC

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delim @@
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delim @#
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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

.NH 1
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 #.

.NH 2
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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.NH 1
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.

.NH 2
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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.NH 1
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.

.NH 2
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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.NH 1
Points of intersection
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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 Q12a
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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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x over 6 = cos(t)
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y = 4(2sin(t)cos(t))
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y = 8sin(t)cos(t)
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y over 8cos(t) = sin(t)
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Using the identity @ sin sup 2 x + cos sup 2 x = 1 # 

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1 = {y over 8cos(t)} sup 2 + {x over 6} sup 2
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1 = {y sup 2 over {64cos sup 2 (t)}} + {x sup 2 over 36} 
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1 = {0 over {64cos sup 2 (t)}} + {x sup 2 over 36} 
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1 = {x sup 2 over 36} 
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36 = x sup 2 
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x = 6, -6 
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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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