Parametric equation and vector equation

  • Thread starter chetzread
  • Start date
  • #1
801
1

Homework Statement


Find Parametric equation and vector equation for the portion of parabola y = 1+(x^2) from (1,2) to (2,5)

Homework Equations




The Attempt at a Solution


m = (2-1) i +(5-2)j = i +3j
x = t , y = 1+t^2
r(t) = ti + (2=3t)j, where 0<t<1
, but the ans given is r(t) = ti + (1+t^2) j , 1<t<2

Is my ans acceptable ?
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
17,164
8,963

Homework Statement


Find Parametric equation and vector equation for the portion of parabola y = 1+(x^2) from (1,2) to (2,5)

Homework Equations




The Attempt at a Solution


m = (2-1) i +(5-2)j = i +3j
x = t , y = 1+t^2
r(t) = ti + (2=3t)j, where 0<t<1
, but the ans given is r(t) = ti + (1+t^2) j , 1<t<2

Is my ans acceptable ?

No. It's not clear to me what you are doing in your solution.
 
  • #3
801
1
No. It's not clear to me what you are doing in your solution.
Which part you don't understand ?
My ans is
r(t) = ti + (2=3t)j, where 0<t<1
, but the ans given is r(t) = ti + (1+t^2) j , 1<t<2
 
  • #4
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
17,164
8,963
Which part you don't understand ?
My ans is
r(t) = ti + (2=3t)j, where 0<t<1
, but the ans given is r(t) = ti + (1+t^2) j , 1<t<2

I don't understand any of it. Not least ##(2=3t)j##.
 
  • #5
801
1
I don't understand any of it. Not least ##(2=3t)j##.
My ans is
r(t) = ti + (2+3t)j, where 0<t<1
, but the ans given is r(t) = ti + (1+t^2) j , 1<t<2
 
  • #6
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
17,164
8,963
My ans is
r(t) = ti + (2+3t)j, where 0<t<1
, but the ans given is r(t) = ti + (1+t^2) j , 1<t<2

For ##t = 0##, you have ##r(0) = 2j##, which is not correct.

Moreover, what you have is a straight line ##y = 3x + 2##. Not a parabola.
 
  • #7
801
1
For ##t = 0##, you have ##r(0) = 2j##, which is not correct.

Moreover, what you have is a straight line ##y = 3x + 2##. Not a parabola.
Find a vector equation for the line segment that joins A(1, -1 , 2) and B (4,1,7)

My working is :
m = (4-1)i + (1+1 ) j + (7-2)k , so m = 3i + 2j +5k

so , x= 1+2t , y = -1+ 2t z = (2+5t)k
so t = (x-2)/2 = (3-y)/6 = (z+1)/3

So , i use the same concept to do the question in post #1 , is it wrong ? why cant i do so ?
 
  • #8
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
17,164
8,963
Find a vector equation for the line segment that joins A(1, -1 , 2) and B (4,1,7)

My working is :
m = (4-1)i + (1+1 ) j + (7-2)k , so m = 3i + 2j +5k

so , x= 1+2t , y = -1+ 2t z = (2+5t)k
so t = (x-2)/2 = (3-y)/6 = (z+1)/3

So , i use the same concept to do the question in post #1 , is it wrong ? why cant i do so ?

Is a parabola a straight line?
 
  • #10
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
17,164
8,963
No

All you are doing is finding the straight line between the end points.
 
  • #11
801
1
All you are doing is finding the straight line between the end points.
Ok
 

Related Threads on Parametric equation and vector equation

  • Last Post
Replies
1
Views
2K
Replies
1
Views
525
  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
16
Views
21K
  • Last Post
Replies
7
Views
7K
Replies
5
Views
1K
Replies
15
Views
271
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
3
Views
2K
Top