# Parameterization of line

## Homework Statement

Consider the curve of intersection of the surface x2 + y2 + x + z = 1 and
the plane z - x + 2 = 0. Find parametric equations for the tangent line to the intersection curve at the point (1; 0;-1).

## The Attempt at a Solution

i solve for intersection curve by getting rid of z

i get 1-x2-y2-x = x-2
3-y2 = x2+2x
-y2 = x2+2x-3

let x = t to parameterize

so r(t) = [t , sqrt(-t2-2t+3) , t-2]

r'(t) = (1 , 0.5(-t2-2t+3)-0.5(-2t-2) , 1)

at point (1,0,-1) my t = 1

therefore r'(t) = (1 , 0 , 1) this is a vector parallel to the line right?

therefore the line parameterization at (1,0,-1) is

x = 1 + t
y = 0
z = -1 + t

is this correct?

somethings wrong..... i solve it using a different method and got a diff ans :(

∇f = (2x+1, 2y , 1)
∇g = (-1 , 0, 1)

∇f(1,0,-1) = (3,0,1)
∇g(1,0,-1) = (-1,0,1)

so ∇f x ∇g gives me a vector parallel to the line at (1,0,-1) right?
so ∇f x ∇g = 0i -4j + 0k

so parameterized line is
x = 1
y= -4t
z = -1

??? which is totally different from my 1st post answer

the answer is x=1, y=2t, z=-1

does anyone know what went wrong?

thanks

HallsofIvy
Homework Helper

## Homework Statement

Consider the curve of intersection of the surface x2 + y2 + x + z = 1 and
the plane z - x + 2 = 0. Find parametric equations for the tangent line to the intersection curve at the point (1; 0;-1).

## The Attempt at a Solution

i solve for intersection curve by getting rid of z

i get 1-x2-y2-x = x-2
3-y2 = x2+2x
-y2 = x2+2x-3

let x = t to parameterize
That will work for a single point but doesn't actually give a paremeterization for the entire curve- y is not a function of x. You took only the positive square root and so get only part of the curve. What I would have done was complete the square- $x^2+ 2x+ 1+ y^2= 4$ so $(x+ 1)^2+ y^2= 2^2$ so the curve projectts down to a circle in the xy-plane with center at (-1, 0) and radius 2. Use the standard sine and cosine parameterization of such a circle for x and y and then use z= x- 2 to get the parameterization for z. However, since (1, 0, -1) is in the part of the curve your parameterization gives, your method almost works.

so r(t) = [t , sqrt(-t2-2t+3) , t-2]

r'(t) = (1 , 0.5(-t2-2t+3)-0.5(-2t-2) , 1)

at point (1,0,-1) my t = 1

therefore r'(t) = (1 , 0 , 1) this is a vector parallel to the line right?
Are you ready for this? At t= 1, $(-t^2- 2t+ 3)^{1/2}= 0$ but you have the negative 1/2 power. That is not 0, it "blows up" to infinity. This particular parameterization does not work at this particular point!

therefore the line parameterization at (1,0,-1) is

x = 1 + t
y = 0
z = -1 + t

is this correct?

somethings wrong..... i solve it using a different method and got a diff ans :(

∇f = (2x+1, 2y , 1)
∇g = (-1 , 0, 1)

∇f(1,0,-1) = (3,0,1)
∇g(1,0,-1) = (-1,0,1)

so ∇f x ∇g gives me a vector parallel to the line at (1,0,-1) right?

so parameterized line is
x = 1
y= -4t
z = -1

??? which is totally different from my 1st post answer

the answer is x=1, y=2t, z=-1

does anyone know what went wrong?

thanks
Yes, that works. However, one line may have many different parameterizations. For example, the first method above, using the parameterization $x= 2cos(\theta)- 1$, $y= 2sin(\theta)$, $z= 2cos(\theta)- 3$ gives derivatives $x= -2sin(\theta)$, $y'= 2 cos(\theta)$, $z'= -2sin(\theta)$ and, at $\theta= 0$ (which gives the point (1, 0, -1)) $x'= 0$, $y'= 2$, $z'= 0$ which gives the parameterization x= 1, y= 2s, z= -1 (as in your answer key) which is exactly the same line as x= 1, y= 4t, z= -1 except that my s is equal to 2 times your t. This line is simply the line parallel to the y-axis with x= 1, z= -1 at every point and x= 1, y= f(t), z= -1 is a parameterization for every monotone function f(t).

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ah... so my prof purposely set it at 0 to blow up my method :(

thanks!