From 54d52ce712fac694d2893de2df3bda6af616dca6 Mon Sep 17 00:00:00 2001
From: thing1 <thing1@seacrossedlovers.xyz>
Date: Mon, 24 Mar 2025 21:01:56 +0000
Subject: finished parametrics work

---
 maths/work/rev1/Makefile       |   2 +-
 maths/work/rev1/parametrics.ms | 132 ++++++++++++++++++++++++++++++-----------
 2 files changed, 97 insertions(+), 37 deletions(-)

(limited to 'maths/work')

diff --git a/maths/work/rev1/Makefile b/maths/work/rev1/Makefile
index acac884..77969aa 100644
--- a/maths/work/rev1/Makefile
+++ b/maths/work/rev1/Makefile
@@ -1,4 +1,4 @@
 all: parametrics.ms
-	eqn parametrics.ms | groff -Tps -ms | ps2pdf - > parametrics.pdf
+	eqn parametrics.ms | groff -Gp -Tps -ms | ps2pdf - > parametrics.pdf
 
 
diff --git a/maths/work/rev1/parametrics.ms b/maths/work/rev1/parametrics.ms
index e334ffa..cd69917 100644
--- a/maths/work/rev1/parametrics.ms
+++ b/maths/work/rev1/parametrics.ms
@@ -144,86 +144,86 @@ dy over dt = 2t - 3
 
 .NH 1
 Points of intersection
-.LP 
-EX 8D Q1a
-.EQ
-x = 5 + t
-.EN
-.EQ
-y = 6 - t
-.EN
+.LP
+This is as simple as substituting in numbers for the most part although sometimes it can be harder due to trig identities showing up.
+
+.NH 2
+Example
+.LP
+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.
 .EQ
-y = 6 - (x - 5)
+t = sqrt x
 .EN
+
 .EQ
-y = 11 - x
+y = 8{sqrt x} - 5
 .EN
+
 .EQ
-0 = 11 - x
+"Then we can say"
 .EN
+
 .EQ
--11 = - x
+5x + 4 = 8{sqrt x} - 5
 .EN
 
 .EQ
-11 = x
+5x + 9 = 8{sqrt x}
 .EN
 
-.LP
-EX 8D Q12a
 .EQ
-x = 6cos(t)
+5x over 8 + 9 over 8 = sqrt x
 .EN
+
 .EQ
-y = 4sin(2t) + 2
+({5x over 8 + 9 over 8}) sup 2 = x
 .EN
+
 .EQ
--{pi over 2} < t < {pi over 2}
+{25x sup 2 over 64 + 81 over 64} = x
 .EN
 
 .EQ
-x over 6 = cos(t)
+25x sup 2 + 81 = 64x
 .EN
 
 .EQ
-y = 4(2sin(t)cos(t))
+25x sup 2 -64x + 81 = 0
 .EN
 
 .EQ
-y = 8sin(t)cos(t)
+"This has no points of intersection because x is imaginary"
 .EN
 
+.LP 
+EX 8D Q1a
 .EQ
-y over 8cos(t) = sin(t)
+x = 5 + t
 .EN
-
-Using the identity @ sin sup 2 x + cos sup 2 x = 1 # 
-
 .EQ
-1 = {y over 8cos(t)} sup 2 + {x over 6} sup 2
+y = 6 - t
 .EN
-
 .EQ
-1 = {y sup 2 over {64cos sup 2 (t)}} + {x sup 2 over 36} 
+y = 6 - (x - 5)
 .EN
-
 .EQ
-1 = {0 over {64cos sup 2 (t)}} + {x sup 2 over 36} 
+y = 11 - x
 .EN
-
 .EQ
-1 = {x sup 2 over 36} 
+0 = 11 - x
 .EN
-
 .EQ
-36 = x sup 2 
+-11 = - x
 .EN
 
 .EQ
-x = 6, -6 
+11 = x
 .EN
-
 .LP
+
 EX 8D 12b
 .EQ
 x = 6cos(t)
@@ -327,3 +327,63 @@ sqrt {-4}
 .EQ
 sqrt {-4} " has no real solutions!"
 .EN
+
+.NH 1
+Parametric with trig
+.LP
+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
+.EQ
+sin sup 2 (x) + cos sup 2 (x) = 1
+.EN
+
+An example would be the following
+
+.EQ
+x = 2sin(t)
+.EN
+
+.EQ
+y = cos(t) + 2
+.EN
+
+.EQ
+sin(t) = x over 2
+.EN
+
+.EQ
+cos(t) = y - 2
+.EN
+
+.EQ
+({x over 2}) sup 2 + (y - 2) sup 2 = 1
+.EN
+
+.EQ
+{x sup 2 over 4} + y sup 2 -4y + 4 = 1
+.EN
+
+.EQ
+{x sup 2 over 4} + y sup 2 -4y + 4 = 1
+.EN
+
+.EQ
+{x sup 2} + 4y sup 2 -16y + 16 = 4
+.EN
+
+.EQ
+{x sup 2} -16y + 16 - 4= -4y sup 2
+.EN
+
+.EQ
+{x sup 2} + 12 = -4y sup 2 + 16y
+.EN
+
+This is an oval shape.
+.G1
+frame ht 1 wid 3
+draw solid
+pi=atan2(0,-1)
+for i from 0 to 2*pi by .1 do { next at i, sin(i) }
+.G2
-- 
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