Solving Curve C Tangent P: (-3,-2,2)

[tifa8]

Hello, I need help for this problem

Homework Statement

There exist a curve C such that its parametric equation is (x,y,z)=(3−3t,1−t^{2},t+2t^{3}). There is a unique point P on the curve with the property that the tangent line at P passes through the point (−3,−2,2). Find the coordinates of P.

Homework Equations

(C) : (x,y,z)=(3−3t,1−t^{2},t+2t^{3})

The Attempt at a Solution

Attempt to solve it
(x',y'z')= (-3,-2t,1+6t^{2} )
since the above is the direction vector of the tangent T then I tried to express the parametric equation of the tangent in function of t which has given me
x=-3s-3
y=-2ts-2
z=(1+6t^{2})s+2

after that I tried to solve xp=x by replacing x in the line equation by the curve equation but I can't solve that ! I really don't know how to approach this exercise ...

Thank you for your help

 

[chiro]

Hello, I need help for this problem

Homework Statement

There exist a curve C such that its parametric equation is (x,y,z)=(3−3t,1−t^{2},t+2t^{3}). There is a unique point P on the curve with the property that the tangent line at P passes through the point (−3,−2,2). Find the coordinates of P.

Homework Equations

(C) : (x,y,z)=(3−3t,1−t^{2},t+2t^{3})

The Attempt at a Solution

Attempt to solve it
(x',y'z')= (-3,-2t,1+6t^{2} )
since the above is the direction vector of the tangent T then I tried to express the parametric equation of the tangent in function of t which has given me
x=-3s-3
y=-2ts-2
z=(1+6t^{2})s+2

after that I tried to solve xp=x by replacing x in the line equation by the curve equation but I can't solve that ! I really don't know how to approach this exercise ...

Thank you for your help

Hey there tifa and welcome to the forums.

Have you ever studied or covered linear interpolation? Or have you covered the equation of a line in n dimensions (or just 3)?

 

[tifa8]

thank you !

No I didn't cover yet linear interpolation but I think we will see it next week. And no didn't see equations of lines in more than 3 dimensions. What I'm covering now is curves and motion in curves.

 

[Dick]

What you know is that the difference (3−3t,1-t^2,t+2t^3)-(−3,−2,2) is parallel to your derivative direction (-3,-2t,1+6t^2). Two vectors A and B are parallel if A=k*B for some k. Can you write down an equation expressing that and solve for t?

 

[tifa8]

thank you ! I found t=3 which was correct :)