Did I Find the Correct Vector Function for the Curve Intersection?

[1st1]

Homework Statement

Find a vector function that represents the curve of intersection of the two surfaces:

The paraboloid z = 4x^2 + y^2
The parabolic cylinder y = x^2

Homework Equations

z = 4x^2 + y^2
y = x^2

The Attempt at a Solution

Combining the two equations:

z = 4(sqrt(y))^2 + y^2
z = 4y + y^2

Choose a parameter:
Let y = t
z = 4t + t^2

Therefore:

x = sqrt(t)
y = t

I get the equation:

r(t) = (sqrt(t))i + (t)j + (4t + t^2)k

Is this correct or did I do something wrong?
Any help appreciated, thanks!

 

[gabbagabbahey]

Hi 1st1, welcome to PF!

Combining the two equations:

z = 4(sqrt(y))^2 + y^2
z = 4y + y^2

Careful,

$$\sqrt{y^2}=|y|=\left\{ \begin{array}{lr} y & , y \geq 0 \\ -y & , y<0 \end{array} \right.$$

Instead of introducing this $\pm$ sign problem, try writing $z$ in terms of $x$ instead.

 

[1st1]

Hey gabba, thanks for the welcome.

I see what you mean but sqrt(y) is being squared so regardless of a positive or negative output it will be squared afterwards which would give me positive y either way. Correct me if I am overlooking something.

Anyway, here is the work in terms of x:

z = 4x^2 + (x^2)^2
z = 4x^2 + x^4

Let x = t
z = 4t^2 + t^4
x = t
y = t^2

r(t) = (t)i + (t^2)j + (4t^2 + t^4)k

Is this correct?

Thanks again.

 

[gabbagabbahey]

The reason that sqrt(y^2)=|y|, is that, by definition, the sqrt() function always returns a positive value.

Other than that, your solution looks good to me!

 

[1st1]

Sounds good, thank you.