- #1
tifa8
- 14
- 0
Hello, I need help for this problem
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.
(C) : (x,y,z)=(3−3t,1−t[tex]^{2}[/tex],t+2t[tex]^{3}[/tex])
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
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