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
ANSWERED
asked 2021-09-15
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).

Expert Answers (1)

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,
\(\displaystyle{r}{\left({t}\right)}={\left({t}\right)}{i}+{\left(\frac{{1}}{{10}}{t}^{{2}}-\frac{{5}}{{2}}\right)}{j}+{\left(\frac{{1}}{{10}}{t}^{{2}}+\frac{{5}}{{2}}\right)}{k}\)
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