Question

find a vector function, r(t), that represents the curve of intersection of the two surfaces. the cone z = \sqrt{x^{2}+y^{2}} and the plane z = 3 + y
r(t) =

Explanation:

Step1: Set the two equations equal

Set $\sqrt{x^{2}+y^{2}}=3 + y$. Square both sides to get $x^{2}+y^{2}=(3 + y)^{2}=9 + 6y+y^{2}$.

Step2: Simplify the equation

Subtract $y^{2}$ from both sides of $x^{2}+y^{2}=9 + 6y+y^{2}$, resulting in $x^{2}=9 + 6y$. Then solve for $y$: $y=\frac{x^{2}-9}{6}$.

Step3: Parameterize $x$

Let $x = t$. Then $y=\frac{t^{2}-9}{6}$.

Step4: Find $z$

Since $z = 3 + y$, substitute $y=\frac{t^{2}-9}{6}$ into it. So $z=3+\frac{t^{2}-9}{6}=\frac{18 + t^{2}-9}{6}=\frac{t^{2}+9}{6}$.

Answer:

$\mathbf{r}(t)=t\mathbf{i}+\frac{t^{2}-9}{6}\mathbf{j}+\frac{t^{2}+9}{6}\mathbf{k}$