Parametric equation of a basic cubic function

Click For Summary
The discussion focuses on finding a parametric form for the cubic curve defined by y = x^3 within the range of 0 ≤ y ≤ 8, starting at the point (2, 8). The initial attempt produced the equation r(t) = (t, t^3) for t in [0, 2], which correctly represents the curve from (0, 0) to (2, 8). However, the second attempt, r(t) = (2-t, 8-t^3), does not describe the same curve, as it results in different points for t values. The correct approach to reverse the direction is to substitute t with (2-t) in the original equation, leading to a proper parametric representation. The conversation concludes with a request for further questions on parametric equations.
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?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

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