# 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).

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).
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Fatema Sutton
$x=f\left(t\right),y=g\left(t\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}z=h\left(t\right)$
The vector:
$r\left(t\right)=f\left(t\right)i+g\left(t\right)j+h\left(t\right)k$
Consider the equation of cone surface:
$z=\sqrt{{x}^{2}+{y}^{2}}$ (a)
Substitute z=5+y to (a)
$5+y=\sqrt{{x}^{2}+{y}^{2}}$
${\left(5+y\right)}^{2}={\left(\sqrt{{x}^{2}+{y}^{2}}\right)}^{2}$
$25+10y+{y}^{2}={x}^{2}+{y}^{2}$
$25+10y={x}^{2}$
$10y={x}^{2}-25$
$y=\frac{1}{10}\left({x}^{2}-25\right)$
$y=\frac{1}{10}{x}^{2}-\frac{5}{2}$
Consider the equation for plane surface:
z=5+y (b)
Substitute $y=\frac{1}{10}{x}^{2}-\frac{5}{2}$ to (b)
$z=5+\frac{1}{10}{x}^{2}-\frac{5}{2}$
$z=\frac{1}{10}{x}^{2}+\frac{5}{2}$
To define each variable in terms of the parameter t, set x=t
$y=\frac{1}{10}{t}^{2}-\frac{5}{2}$
and $y=\frac{1}{10}{t}^{2}-\frac{5}{2}$
Thus,
$r\left(t\right)=\left(t\right)i+\left(\frac{1}{10}{t}^{2}-\frac{5}{2}\right)j+\left(\frac{1}{10}{t}^{2}+\frac{5}{2}\right)k$