Question

Find a vector function that represents the curve of intersection of the two surfaces. The cone $z=\sqrt{x^2+y^2}$ and the plane $z=1+y$ 

    Find a vector function that represents the curve of intersection of the two surfaces.
The cone $z=\sqrt{x^2+y^2}$ and the plane $z=1+y$
Single Variable Calculus: Early Transcendentals
Single Variable Calculus: Early Transcendentals
James Stewart,… 9th Edition
Chapter 13, Problem 51 ↓
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Find a vector function that represents the curve of intersection of the two surfaces. The cone $z=\sqrt{x^2+y^2}$ and the plane $z=1+y$
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Transcript

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00:01 So we know in this case that z equals a square root of x squared plus y squared and z equals 1 plus y.
00:09 So knowing that x and z equal r cosine theta and r sine theta, how we can do this is we can set the z values equal to 0.
00:19 So we have the square root of x squared plus y squared equaling 1 plus y.
00:25 So then solving for y, we end up getting that y is equal to x squared minus 1 over 2...
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