# Parametric equation and vector equation

1. Oct 30, 2016

1. The problem statement, all variables and given/known data
Find Parametric equation and vector equation for the portion of parabola y = 1+(x^2) from (1,2) to (2,5)

2. Relevant equations

3. 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 ?

2. Oct 30, 2016

### PeroK

No. It's not clear to me what you are doing in your solution.

3. Oct 30, 2016

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. Oct 30, 2016

### PeroK

I don't understand any of it. Not least $(2=3t)j$.

5. Oct 30, 2016

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. Oct 30, 2016

### PeroK

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. Nov 1, 2016

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. Nov 1, 2016

### PeroK

Is a parabola a straight line?

9. Nov 1, 2016

No

10. Nov 1, 2016

### PeroK

All you are doing is finding the straight line between the end points.

11. Nov 1, 2016