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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 $

$\mathbf{r}(t)=t \mathbf{i}+\frac{1}{2}\left(t^{2}-1\right) \mathbf{j}+\frac{1}{2}\left(t^{2}+1\right) \mathbf{k}$

Calculus 3

Chapter 13

Vector Functions

Section 1

Vector Functions and Space Curves

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In mathematics, a vector (…

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The problem is, I kinda walk to function that a pretence, a curve of intersection. Of the two surfaces, the corn is equal to beautif X square plus y square, and the plane three is equal to one plus y. The first if we light actually call to heed on behalf. See, here's the cartoon he squire Plus Why square? So we have C square. It's an equal to a squire Asli Square And things see is going to want us. Why? So this is control. Say squire, for us seem wise. One squad, This is a courtroom squire pass. It's a square minus toothy last line. We can cancel out Esquire. So we have C Is the cultural he squire? Ask one over two. Then why is the door to C minus one? This is because of a T Square minus one over two. Now back to function. Artie, is they caught you? You square once one away, too. In Squire Cross, one over two. So this's a Wachter function that represents the curve of intersection. That was true. Surfaces

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