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Tangent to a curve

  • Thread starter tifa8
  • Start date
  • #1
14
0
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[tex]^{2}[/tex],t+2t[tex]^{3}[/tex]). 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[tex]^{2}[/tex],t+2t[tex]^{3}[/tex])

The Attempt at a Solution



Attempt to solve it
(x',y'z')= (-3,-2t,1+6t[tex]^{2}[/tex] )
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[tex]^{2}[/tex])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
 

Answers and Replies

  • #2
chiro
Science Advisor
4,790
131
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[tex]^{2}[/tex],t+2t[tex]^{3}[/tex]). 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[tex]^{2}[/tex],t+2t[tex]^{3}[/tex])

The Attempt at a Solution



Attempt to solve it
(x',y'z')= (-3,-2t,1+6t[tex]^{2}[/tex] )
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[tex]^{2}[/tex])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)?
 
  • #3
14
0
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.
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
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?
 
  • #5
14
0
thank you !!! I found t=3 which was correct :)
 

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