commit ce461971506167f017a2fca98e03c5689041e663
parent 96155de15af32262f47696751d50ffc86b52e13e
Author: thing1 <thing1@seacrossedlovers.xyz>
Date: Mon, 3 Nov 2025 18:36:13 +0000
did homework
Diffstat:
4 files changed, 164 insertions(+), 0 deletions(-)
diff --git a/.gitignore b/.gitignore
@@ -1,3 +1,4 @@
+*.ps
*.pdf
*.html
*.o
diff --git a/MP10610/hw2/Makefile b/MP10610/hw2/Makefile
@@ -0,0 +1,5 @@
+out.pdf: hw2.ms
+ magick graph.jpeg graph.ps
+ groff -ms -e -Tps hw2.ms | ps2pdf - > out.pdf
+clean:
+ rm *.ps *.pdf
diff --git a/MP10610/hw2/graph.jpeg b/MP10610/hw2/graph.jpeg
Binary files differ.
diff --git a/MP10610/hw2/hw2.ms b/MP10610/hw2/hw2.ms
@@ -0,0 +1,158 @@
+.TL
+Calculus assignment 2
+.AU
+Lucas Standen (lus53@aber.ac.uk)
+
+.EQ
+delim @@
+.EN
+.EQ
+delim @#
+.EN
+
+.2C
+
+.LP
+3)a) @f sub 1 (x) = x sup 2#
+
+@f sub 2 (x) = sqrt x#
+
+@f sub 1 (1) = 1#
+
+@f sub 2 (1) = 1#
+
+@f sub 1 (1) = f sub 2 (1)#
+
+Therefore f(x) is continuous
+
+@f' sub 1 (x) = 2x#
+
+@f' sub 2 (x) = 1 over {2 sqrt x}#
+
+@lim sub { -> 1} f' sub 1 (x) = 1#
+
+@lim sub {x -> 1} f' sub 2 (x) = 1 over 2#
+
+@lim sub {x -> 1} f' sub 2 (x) != lim sub { -> 1} f' sub 1 (x)#
+
+Therefore f(x) is not differentiable
+
+3)b)
+.I
+See end of doc.
+
+5)c) @lim sub {h -> 0} {sqrt {3 (x + h) + 1} - sqrt {3 x + 1}} over h#
+
+@lim sub {h -> 0} {sqrt {3 (x + h) + 1} - sqrt {3 x + 1}} over h {sqrt {3 (x + h) + 1} + sqrt {3 x + 1}} over {sqrt {3 (x + h) + 1} + sqrt {3 x + 1}}#
+
+@lim sub {h -> 0} {(3 (x + h) + 1) - {(3 x + 1)}} over { h sqrt {3 ( x + h ) + 1} + sqrt {3 x + 1}}#
+
+@lim sub {h -> 0} { 3h } over { h sqrt {3 ( x + h ) + 1} + sqrt {3 x + 1}}#
+
+@lim sub {h -> 0} { 3 } over { sqrt {3 ( x + h ) + 1} + sqrt {3 x + 1}}#
+
+@3 over { 2 sqrt {3 x + 1}} #
+
+6)b) @u = sqrt x#
+
+@u' = 1 over { 2 sqrt x }#
+
+@v = sin x#
+
+@v' = cos x#
+
+@ dy over dx = 1 over {2 sqrt x} sin x + sqrt x cos x#
+
+6)c) @u = 2x#
+
+@u' = 2#
+
+@v = 4 + x sup 2#
+
+@v' = 2x#
+
+@ dy over dx = {{2 (4 + {x sup 2})} - {4 x sup 2}} over {(4 + {x sup 2})} sup 2#
+
+@ dy over dx = - {2x sup 2 + 8} over {(4 + {x sup 2})} sup 2#
+
+7)d)
+
+let @f(x) = {x sup 2 + 1} over {x sup 2 - 1}#
+
+let @y = f(x) sup 3#
+
+@f' (x)#
+
+@u = x sup 2 + 1#
+
+@u' = 2x#
+
+@v = x sup 2 - 1#
+
+.I
+next page, first column
+
+.LP
+@v' = 2x#
+
+@f'(x) = {{2x (x sup 2 - 1)} - {2x (x sup 2 + 1)}} over {(x sup 2 - 1) sup 2}#
+
+@f'(x) = {{(x sup 3 - 2x)} - {(x sup 3 + 2x)}} over {(x sup 2 - 1) sup 2}#
+
+@f'(x) = {-4x} over {(x sup 2 - 1) sup 2}#
+
+@dy over dx = 3 f(x) sup 2 f'(x)#
+
+@dy over dx = -12 ({{{x sup 2} + 1} over {{x sup 2} - 1}}) sup 2 x over {{ (x sup 2 - 1) } sup 2}#
+
+8) @g'(x) = cos x - 8 sin 4x#
+
+@g''(x) = -sin x - 32 cos 4x#
+
+@g'({pi over 4}) = {sqrt 2} over 2#
+
+@g''({pi over 4}) = {64 - sqrt 2} over 2#
+
+9)c)
+Differentiate the first half
+@d over dy cos(x + y)#
+
+@y = cos t#
+
+@y' = -sin t#
+
+@t = x + y#
+
+@t' = 1 + dy over dx#
+
+@dy over dx = t' sin t#
+
+@dy over dx = -{({1 + dy over dx})}(sin(x + y))#
+
+
+Differentiate the second half
+@d over dy sin(x + y)#
+
+@y = sin t#
+
+@y' = cos t#
+
+@t = x + y#
+
+@t' = 1 + dy over dx#
+
+@dy over dx = t' sin t#
+
+@dy over dx = {({1 + dy over dx})}(cos(x + y))#
+
+Put the whole thing together
+
+@1 over 3 = (1 + {dy over dx}) (-sin (x + y) + cos (x + y))#
+
+@1 over {3 (1 + {dy over dx})} = (-sin (x + y) + cos (x + y))#
+
+@1 over {3 (-sin (x + y) + cos (x + y)) } = (1 + {dy over dx}) #
+
+@1 over {3 (cos (x + y) - sin(x + y)) } - 1 = {dy over dx} #
+
+.PSPIC graph.ps
