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find the unit tangent vector to the curve at their points of intersection?

y=x^2
y=x^(1/3)

I am not exactly sure what it is asking of me
thank you jen

2007-01-21 04:56:11 · 3 answers · asked by Jen H 1 in Science & Mathematics Mathematics

3 answers

I know you can find the rect tangent to the curve by derivation:

case 1
y=x^2
y'=(x^2)'---> derivation
y'=2x------->equation of the tangent to the curve

case 2
y=x^(1/3)
y'=[x^(1/3)]'
y'=1/3x^-2/3

enjoy

2007-01-21 05:06:21 · answer #1 · answered by nexusdhr84 2 · 0 0

To find the points of intersection


x^2 = x^1/3 so x^6 =x and x^6-x= x(x^5-1)=0
so x=0 or x=1 so the points are (0,0) and (1,1)
the tg to the first curve at (0,0) is y=0 and at (1,1) y-1 =2(x-1)
y= 2x-1 .Calling i and j the unit vectors along x axis and y axis

v= i at(0,0) and v= (1/2 *i - j) /Sqrt (1/4+1)

Exp. If you take tintercept witx ox and 0y you get x=1/2 and y=-1

the vector is 1/2 i-j and its length sqrt(1/4+1) for the second curve at (0,0) the tangent is x=0 so v=j
At (1,1) the tg is y-1=1/3 ( x-1)
y= 1/3x +2/3 The intercepts are x=-2 y= 2/3 and

v=(-2i+2/3 j) sqrt (4+4/9)
I hope this is what you asked for

2007-01-21 14:21:32 · answer #2 · answered by santmann2002 7 · 0 0

I think it sounds like: find the unit tangent vector to the curveS at their points of intersection?
Then y=xx and z=x^(1/3); intersect: y=z or xx=x^(1/3) or x^(2 –1/3) =1, hence x=1 and y=1;
Now y’(x)=2x and y’(1) =2 =tan(u), sin(u)= tan(u)/sqrt(1+(tan(u))^2) = 2/sqrt(5), cos(u) = 1/sqrt(5); thus unit tangent vector to parabola y=xx is e=(1/sqrt5, 2/sqrt5); Now z’(x) =(1/3)*x^(1/3 –1) =(1/3)*x^(-2/3) and z’(1) =(1/3)*1 =1/3 =tan(v); thus unit tangent vector to curve y=x^(1/3) is f=(cos(v), sin(v)) = (3/sqrt10, 1/sqrt10);
So e =/= f ☺.

2007-01-21 14:53:00 · answer #3 · answered by Anonymous · 0 0

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