Question

Given the cone z=sqrt(x^2+y^2) and the plane z=5+y. Represent the curve of intersection of the surfaces with a vector function r(t).

Trigonometric functions
Given the cone $$\displaystyle{z}=\sqrt{{{x}^{{2}}+{y}^{{2}}}}$$ and the plane z=5+y.
Represent the curve of intersection of the surfaces with a vector function r(t).

2021-09-16
$$\displaystyle{x}={f{{\left({t}\right)}}},{y}={g{{\left({t}\right)}}},{\quad\text{and}\quad}{z}={h}{\left({t}\right)}$$
The vector:
$$\displaystyle{r}{\left({t}\right)}={f{{\left({t}\right)}}}{i}+{g{{\left({t}\right)}}}{j}+{h}{\left({t}\right)}{k}$$
Consider the equation of cone surface:
$$\displaystyle{z}=\sqrt{{{x}^{{2}}+{y}^{{2}}}}$$ (a)
Substitute z=5+y to (a)
$$\displaystyle{5}+{y}=\sqrt{{{x}^{{2}}+{y}^{{2}}}}$$
$$\displaystyle{\left({5}+{y}\right)}^{{2}}={\left(\sqrt{{{x}^{{2}}+{y}^{{2}}}}\right)}^{{2}}$$
$$\displaystyle{25}+{10}{y}+{y}^{{2}}={x}^{{2}}+{y}^{{2}}$$
$$\displaystyle{25}+{10}{y}={x}^{{2}}$$
$$\displaystyle{10}{y}={x}^{{2}}-{25}$$
$$\displaystyle{y}=\frac{{1}}{{10}}{\left({x}^{{2}}-{25}\right)}$$
$$\displaystyle{y}=\frac{{1}}{{10}}{x}^{{2}}-\frac{{5}}{{2}}$$
Consider the equation for plane surface:
z=5+y (b)
Substitute $$\displaystyle{y}=\frac{{1}}{{10}}{x}^{{2}}-\frac{{5}}{{2}}$$ to (b)
$$\displaystyle{z}={5}+\frac{{1}}{{10}}{x}^{{2}}-\frac{{5}}{{2}}$$
$$\displaystyle{z}=\frac{{1}}{{10}}{x}^{{2}}+\frac{{5}}{{2}}$$
To define each variable in terms of the parameter t, set x=t
$$\displaystyle{y}=\frac{{1}}{{10}}{t}^{{2}}-\frac{{5}}{{2}}$$
and $$\displaystyle{y}=\frac{{1}}{{10}}{t}^{{2}}-\frac{{5}}{{2}}$$
Thus,
$$\displaystyle{r}{\left({t}\right)}={\left({t}\right)}{i}+{\left(\frac{{1}}{{10}}{t}^{{2}}-\frac{{5}}{{2}}\right)}{j}+{\left(\frac{{1}}{{10}}{t}^{{2}}+\frac{{5}}{{2}}\right)}{k}$$