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I know this is easy but I am unsure about how to answer a multivariable-calculus question as a volunteer tutor, I left a couple of students with a method but a lot of others needed help today so I had to move on, I am hoping someone can confirm my method or reveal a correct method.

Question: Find the points of intersection between the parametrized line: $$x = 1 + t$$ $$y = 3 + 2t$$ $$z = 1 + t$$

And the curve: $z = x^2 + 2y^2$

I noticed there are 4 equations and 4 unknowns. So was it correct to tell the students that it appears to be a substitution problem, and by simply plugging in the given values for $x, y, z$, we can solve for $t$. And it appears for $t$ there are two values for, which makes sense if there are two intersections. By plugging in the value of $t$ for the line one can surely find points. I assume the work can also be checked by using the equation of the curve for such points as well to determine they work in the equation of the curve.

I only have taken multivariable last term, so I am still getting comfortable with the idea here and I think I am the only tutor that can do it where I work, so verification is much appreciated.

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1 Answer 1

up vote 1 down vote accepted

The method is fine: a point of intersection must satisfy the equation $$1+t=(1+t)^2+2(3+2t)^2\;.$$ However, the line and the curve do not intersect: one finds that

$$t=\frac{-25\pm\sqrt{-23}}{18}\;.$$

In this case one might also notice that on the line we have $z=x$ and $y=2x+1$, so that we can instead find points of intersection in terms of $x$ by solving $x=x^2+(2x+1)^2$.

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Thanks for confirming, the question is probably not the same as this is from my memory. –  Leonardo Jul 8 '13 at 23:55
    
@Leonardo: You’re welcome. –  Brian M. Scott Jul 8 '13 at 23:56

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