Calculating Curvature at Point P on f(t) in [0;2] for t=(0,1)

[teng125]

Find the curvature at the point P:

f : [0;2] IR^2 , f (t) = 2t,4 −2t^3 , P(2,2)

i subs 2x=0 then x=0 and 2y=2 then y=1

t=(0,1)

then i perform the curvature calculation.however,i'm confuse that in this case i have to subs t=1 to get a value instead of zero.
If the value t=(2 ,3) or others, which one should i choose??

 

[HallsofIvy]

Find the curvature at the point P:

f : [0;2] IR^2 , f (t) = 2t,4 −2t^3 , P(2,2)

i subs 2x=0 then x=0 and 2y=2 then y=1

Why substitute 2x= 0? Since x= 2 at P, you should have
x= 2t= 2 so t= 1. Unfortunately, if t= 1 then y= 4 -2(2)³= 4- 16= -12, not 2. Are you sure you have copied the problem correctly? (2,2) is not on the curve f(t)= (2t, 4- 2t³)!

t=(0,1)

What does this mean? t is number, not an interval or a set.

then i perform the curvature calculation.however,i'm confuse that in this case i have to subs t=1 to get a value instead of zero.
If the value t=(2 ,3) or others, which one should i choose??

Again, I don't know what you mean by "t= (2, 3)". As I said before, this problem is stated incorrectly. The point (2,2) is not on the curve f(t)= (2t, 4- 2t³).

 

[teng125]

but the problem stated P(2,2)

 

[VietDao29]

Why substitute 2x= 0? Since x= 2 at P, you should have
x= 2t= 2 so t= 1. Unfortunately, if t= 1 then y= 4 -2(2)³= 4- 16= -12, not 2. Are you sure you have copied the problem correctly? (2,2) is not on the curve f(t)= (2t, 4- 2t³)!

Uhmm, in fact, (2, 2) is on the curve, since:
x = 2t = 2 <=> t = 1
Plug t = 1 in, we have: y = y= 4 - 2(1)³ (It's 1, not 2 ) = 2.
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@teng125:
Ok, open up your textbook, or notes. Can you find the formula to find the curvature of a curve given parametrically?

 

[HallsofIvy]

Uhmm, in fact, (2, 2) is on the curve, since:
x = 2t = 2 <=> t = 1
Plug t = 1 in, we have: y = y= 4 - 2(1)³ (It's 1, not 2 ) = 2.
-----------------

Well, we can just kind of ignore that, can't we!

@teng125:
Ok, open up your textbook, or notes. Can you find the formula to find the curvature of a curve given parametrically?

 

[HallsofIvy]

To repeat, correctly this time (I hope): There is no reason to take "2x= 0". The point is x= 2, y= 2 so x= 2t= 2 which gives t= 1. Then, y= 4- 2(1)³= 4- 2= 2. Yes! (2,2) is a point on the line! Your textbook probably has a number of formulas for curvature.

I still don't understand what you mean by "t= (2,3)". If you meant "what if the point was P(2,3)" then what I said before would be correct: (2, 3) is not on the curve so the question makes no sense. You can't just pick points at random: they have to be on the curve in order that the question "what is the curvature at that point?" to make sense. If on the other hand the point were given as P(6, -14) then you would calculate x= 2t= 6 so t= 3. Now check: y= 4- 2(3)²= 4- 18= -14. Yes, that point is on the graph. Use your formula with t= 3 for the curvature at P(6, -14).

 

[teng125]

ooo...okok i understand already.thanx
ya i just assume "t= (2,3)" which i simply picked.