# Curvature calculation

1. Jul 18, 2006

### 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??

2. Jul 18, 2006

### HallsofIvy

Staff Emeritus
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)!

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

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).

3. Jul 18, 2006

### teng125

but the problem stated P(2,2)

4. Jul 18, 2006

### VietDao29

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?

5. Jul 18, 2006

### HallsofIvy

Staff Emeritus
Well, we can just kind of ignore that, can't we! :uhh: :uhh:

6. Jul 18, 2006

### HallsofIvy

Staff Emeritus
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).

7. Jul 19, 2006