Parametric equation of a basic cubic function

Click For Summary
SUMMARY

The discussion focuses on finding a parametric form for the cubic curve defined by the equation y = x³, constrained between the points (2, 8) and (0, 0). The initial solution proposed was r(t) = (t, t³) for t in [0, 2], which correctly represents the curve from (0, 0) to (2, 8). However, an alternative solution r(t) = (2 - t, 8 - t³) was incorrectly interpreted as the same curve, leading to confusion. The correct approach to reverse the direction is to substitute t with (2 - t) in the original parametric equation.

PREREQUISITES
  • Understanding of parametric equations
  • Familiarity with cubic functions
  • Basic knowledge of graphing functions
  • Ability to manipulate algebraic expressions
NEXT STEPS
  • Study the concept of parametric equations in depth
  • Learn about transformations of functions, specifically reflections and translations
  • Explore cubic functions and their properties
  • Practice graphing parametric equations using software tools like Desmos or GeoGebra
USEFUL FOR

Students studying calculus, mathematics educators, and anyone interested in understanding parametric equations and cubic functions.

pondzo
Messages
168
Reaction score
0

Homework Statement



Find a parametric form for the part-cubic curve with equation y = x3, 0 ≤ y ≤ 8; starting point (2, 8),

The Attempt at a Solution



The question beforehand was the exact same but the starting and ending points reversed. My answer for that was; r(t) = (t, t^3) for t an element of [0,2], ill call this equation A.

So for this question is set up a few linear equations and solved them and came up with the parametric vector function; r(t) = (2-t, 8-t^3), ill call this equation B. B satisfies the starting and ending point conditions but, to me, it looks like A and B are describing different curves. When i graph them, they look different. And I am not sure how to interpret it. Is it that my answer is wrong? or is it that when you reverse the order of the start and end points as you do in B, its almost as if you you are looking at the graph upside down and considering (2,8) as the origin (if that makes any sense at all). Thank you for your time.
 
Physics news on Phys.org
By "starting point" you mean the point given by y= 8? And by "the starting and ending points reversed" you mean starting at y= 0 and ending at y= 8?

So for that first problem, starting at y= 0 and ending at y= 8, you got \vec{r}(t)= (t, t^3) for t in [0, 2]. Yes, that is correct. Note that if t= 0, you have (0, 0), if t= 1 you have (1, 1), and for t= 2, (2, 8).

If you change that to \vec{r}(t)= (2- t, 8- t^3) then it certainly is true that for t= 0 you get (2, 8), for t= 2 you get (0, 0) but for t= 1 you get (1, 7) not (1, 1). No, that is not the same:

In the first case you are using \vec{r}= (t, t^3) and in the second case you are using (2- t, 8- t^3)= (2, 8)- \vec{r}(t). No, that is NOT the same curve.

Instead just reverse t: replace t, rather than x, with 2- t: instead of (t, t^3), use (2- t, (2- t)^3).
 
Thats exactly how the question was stated in the tutorial, but you got the right interpretation. Thanks for the reply that makes sense now.

If you don't mind, I have another question on parametric equations for polynomials. Could I post it on this thread?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
1K