Answer:
[tex]\mathbf{r^{\to} (t) = ti^{\to} + 5t^2^{\to} j+(6t^2 + 25t^4) k ^{\to}}[/tex]
Step-by-step explanation:
Given that:
The paraboloid surface z = 6x² + y² and the parabolic cylinder y = 5x²
Let assume that:
x = t
then from y = 5x², we have:
y = 5t²
Now replace y = 5t² and x = t into z = 6x² + y²
z = 6t² + (5t²)²
z = 6t² + 25t⁴
Hence, the curve of intersection is illustrated by the set of equations:
x = t, y = 5t², and z = 6t² + 25t⁴
As a vector equation:
[tex]\mathbf{r^{\to} (t) = ti^{\to} + 5t^2^{\to} j+(6t^2 + 25t^4) k ^{\to}}[/tex]
