Spooky Spooky Spooky

Level pending

x = t sin ( t ) , y = 1 cos ( t ) + 0.5 t , 3 π < t < 3 π x = t - \sin(t), y = 1 - \cos(t) + 0.5\left|t\right|, \hspace{.2cm} -3\pi < t < 3\pi

x = t sin ( t ) , y = 4.1 + 0.5 cos ( t ) + 0.33 t , 3 π < t < 3 π x = t - \sin(t), y = 4.1 + 0.5\cos(t) + 0.33\left|t\right|, \hspace{.2cm} -3\pi < t < 3\pi

y = 0.5 x + 6.2 , 1 < x < 1 y = 0.5\left|x\right| + 6.2, \hspace{.2cm} -1 < x < 1

y = 2 x 4 + 4.7 , 1 < x < 1 y = 2x^4 + 4.7, \hspace{.2cm} -1 < x < 1

What creepy shape do the graphs of the equations above trace out? You might want to use a graphing tool for this one!

Ghost Mummy Witch Bat

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3 solutions

Andrew Ellinor
Oct 29, 2015

Here's what a plot of this guy looks like:

If that's not a scary looking bat \boxed{\text{bat}} , then I don't know what is.

Plot this:

( ( x 7 ) 2 x 3 x 3 + ( x 3 ) 2 y + 3 33 7 y + 3 33 7 1 ) ( x 2 ( 3 33 7 112 ) x 2 3 + 1 ( x 2 1 ) 2 y ) ( 9 ( x 1 ) ( x . 75 ) ( 1 x ) ( x . 75 ) 8 x y ) ( 3 x + . 75 ( x . 75 ) ( x . 5 ) ( . 75 x ) ( x . 5 ) y ) ( 2.25 ( x . 5 ) ( x + . 5 ) ( . 5 x ) ( . 5 + x ) ) ( 6 10 7 + ( 1.5 . 5 x ) x 1 x 1 6 10 14 4 ( x 1 ) 2 y ) = 0 \left( \left(\frac{x}{7}\right)^2 \sqrt{\frac{||x|-3|}{|x|-3}} + \left(\frac{x}{3}\right)^2 \sqrt{\frac{|y+\frac{3\sqrt{33}}{7}|}{y+\frac{3\sqrt{33}}{7}}} -1 \right) \cdot \left( |\frac{x}{2}| - \left(\frac{3\sqrt{33}-7}{112}\right)x^2 -3 + \sqrt{1-(||x|-2|-1)^2}-y \right) \\ \cdot \left( 9 \sqrt{\frac{|\left(|x|-1\right)\left(|x|-.75 \right)|}{\left(1-|x|\right) \left(|x|-.75\right)}} -8|x|-y \right) \cdot \left( 3|x|+.75 \sqrt{\frac{|\left(|x|-.75\right)\left(|x|-.5\right)|}{\left(.75-|x|\right)\left(|x|-.5 \right)}}-y \right) \\ \cdot \left(2.25 \sqrt{\frac{|\left(|x|-.5\right)\left(|x|+ .5\right)|}{\left(.5-|x|\right)\left(.5+|x|\right)}} \right) \cdot \left( \frac{6\sqrt{10}}{7} + \left(1.5 - .5|x|\right) \sqrt{\frac{||x|-1|}{|x|-1}} - \frac{6\sqrt{10}}{14} \sqrt{4-\left(|x|-1\right)^2}-y \right) = 0

Patrick Engelmann - 5 years, 7 months ago
Arnab Sen
Nov 1, 2015

matlab code : t=-3 pi:0.001:3 pi; x1=t-sin(t); y1=1-cos(t)+0.5 abs(t); x2=x1; y2=4.1+0.5 cos(t)+0.33 abs(t); x=-1:0.001:1; y3=0.5 abs(x)+6.2; y4=(2*x.^4)+4.7; plot(x1,y1,x2,y2,x,y3,x,y4)

Axel Ordóñez
Nov 1, 2015

You need to put a lvl cause it gives only 10 points xD it took me some minutes to solve it.

Yeah but you could easily graph it, no sense in free points...

Hobart Pao - 5 years, 7 months ago

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