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I have the question that I'm trying to solve:

At what points does the curve $r(t) = t\hat{i} + (6t-t^{2})\hat{k}$ intersect the paraboloid $z = x^{2} + y^{2}$?

So, I go about this by first breaking $r(t)$ into components:

$$x=t$$ $$z=(6t-t^{2})$$ $$y=0$$

Then, I replace the Paraboloids' parameters, as the parameters of both must be equal in order to intersect. I get the equation: $$(6t-t^{2}) = t^{2} + 0^{2}$$

Re-arranged to:

$$6t = 2t^{2}$$

And after dividing by $2t$, we can see that the point of intersection is $t=3$.

However, My professor has pointed out that this is not a valid solution. Where did I go wrong?

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    Well, perhaps he meant you must *also* take into account $\;t=0\;$ , since undoubtedly for $\;t=3\;$ you get an intersection point... – 2017-01-23

2 Answers 2

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Your equation has two solutions:

$$\;6t=2t^2\iff t(3-t)=0\implies t=\begin{cases}0\\{}\\3\end{cases}$$

Perhaps this is what your teacher meant

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I think that t=0 and t=3 are valid solutions if they stay in the domain of the curve. If the domain is for example [0,1] then t=3 is not valid. What is the domain of the curve? Often it is indicated with I and it is a real interval (it also can be I=R).

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    Very good point. +1 – 2017-01-24