[unusual problem]making an illustration on graphing calculator

  • Thread starter Thread starter Horror Business
  • Start date Start date
  • Tags Tags
    Calculator Graphing
AI Thread Summary
A user is seeking help to create an illustration on a TI-83 Plus graphing calculator for a pre-calculus assignment, expressing difficulty in generating creative graphs beyond basic shapes. Suggestions include using polar coordinates for visually appealing graphs, such as a four-petal flower from the function cos(2θ), and combining trigonometric functions for unique designs. The challenge of restricting domains for more complex shapes, like a person, is acknowledged as potentially complicated. Users share their past experiences with similar assignments, highlighting trial and error as a common approach. The discussion emphasizes creativity in graphing while navigating the constraints of the calculator's functions.
Horror Business
Messages
3
Reaction score
0
hello, I'm new! well anyways, I'm having trouble with this assignment. the assignment is to, on my TI-83 plus graphing calculator, create an illustration as a graph. it doesn't have to be much, it could be a person, or a flower, pretty much anything. the thing is, I'm not too good at pre-calculus and all I'm managing to make is parabolas and awkward lines. Also, after all the functions for the graph are entered and the graph is made, i have to draw it on a piece of paper, and write all the functions on it, so cheating like using the pen tool won't work. thank you VERY much. :smile:
don't get me wrong, i don't want you to do this for me. i just looking for some ideas, or a cool-looking function to graph. anyhting helps, thanks.
 
Last edited:
Physics news on Phys.org
Welcome to PF!
I think polar coordinates have nice looking graphs.. Even something as simple as cos(2\theta) produces a four-petal flower. Combining random trig functions together produces nice results but as for drawing a person that could get tough because you have to figure out how to restrict the domains of the functions so that only the desired portion of the graph is shown. For example, if you wanted to draw y = x from x = -1 to x = 1, you would have to transform y = x into y = x \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}so that way any x outside of [-1, 1] returns a nonreal result and therefore won't be graphed. I don't think you want to think quite that hard for this assignment though :)
 
Hehe, i made a :) face.
((5-5(x+3)^2)^(1/2))+2
-((5-5(x+3)^2)^(1/2))+4
((5-5(-x+3)^2)^(1/2))+2
-((5-5(-x+3)^2)^(1/2))+4
-((1-5(x+3)^2)^(1/2))+4
-((1-5(x+3)^2)^(1/2))+4
-((6-.5(x)^2)^(1/2))-3
-((5-.33(x)^2)^(1/2))-2
((5-5(-x+3)^2)^(1/2))+3
((5-5(x+3)^2)^(1/2))+3
 
i had this assignment last year. was probably one of the more fun projects i had in my calculus class. I made stewie from family guy. everyone recognized it right away, i was so proud. haha. good luck, i had mostly trial and error. friend of mine (beign much smarter) figure out his stuff on paper first and made a sailbot in the water. was impressive.
 
thank you all. and vsage, i tried your four petal flower, but it didn't work. and brettisrad, i think you typed the functions wrong or something because your face came out really delapidated. thanks for the reply though.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top