Find out the answer to "Given the cone z=x2+y2 and the plane z=5+y...."

babeeb0oL

Answered question

2021-09-15

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).

Fatema Sutton

Skilled2021-09-16Added 88 answers

x = f (t), y = g (t), and z = h (t)

The vector:

r (t) = f (t) i + g (t) j + h (t) 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}}

{(5 + y)}^{2} = {(\sqrt{x^{2} + y^{2}})}^{2}

25 + 10 y + y^{2} = x^{2} + y^{2}

25 + 10 y = x^{2}

10 y = x^{2} - 25

y = \frac{1}{10} (x^{2} - 25)

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 (t) = (t) i + (\frac{1}{10} t^{2} - \frac{5}{2}) j + (\frac{1}{10} t^{2} + \frac{5}{2}) k

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