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I have the vector-valued function $\vec{r}(t) = t \hat{i}+2t\cos(t) \hat{j} + 2t \sin(t) \hat{k}$, and the cone $4x^2=y^2+z^2$.

I figured, that by showing I could re-write the equation for the cone as the vector-valued function, it would be sufficient proof that the vector-valued function lies on the cone.

So, I let $y = 2x\cos(t)$ and $z=2x\sin(t)$, and, finally, let $x=t$

This shows that the cone can be written as $\vec{r}(t) = t \hat{i}+2t\cos(t) \hat{j} + 2t \sin(t) \hat{k}$

However, the answer is: enter image description here

Did I correctly answer this question? I don't think the answer wrote out enough steps.

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it is clear that the coefficient of $k$ needs a $t$ and $i$ needs a $2$. See the first line you have written. –  B. S. Feb 27 '13 at 13:35
    
I believe I edited correctly, in accordance to your comment. –  Mack Feb 27 '13 at 13:37
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Not quite, @Eli. Babak was talking about the formula in your first line. –  Cameron Buie Feb 27 '13 at 13:40
    
I just fixed that one too. –  Mack Feb 27 '13 at 13:42

2 Answers 2

up vote 1 down vote accepted

It's worth noting that the cone, itself, cannot be written in the form $$\vec r(t)=t\hat i+2t\cos(t)\hat j+2t\sin(t)\hat k.$$

The question at hand is, does that curve lie on the cone? In particular, all we need to show is that for $\langle x,y,z\rangle$ on that curve--that is, $x=t,y=2t\cos(t),z=2t\sin(t)$ for some $t$--we have $y^2+z^2=4x^2$, and the book wrote out precisely enough steps to show that.

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Okay, I believe I see. Can the cone be represented by any vector-valued function? –  Mack Feb 27 '13 at 13:46
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We can only parameterize the cone in $2$ variables, since it is a surface. For example, we could parameterize it by $$\vec r(u,v)=u\hat i+2u\cos(v)\hat j+2u\sin(v)\hat k.$$ Your curve, then, is simply the portion of this cone where $u=v$. –  Cameron Buie Feb 27 '13 at 13:49
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Yes, but since it's a surface (rather than a curve) it will be vector-valued function of two variables. –  bubba Feb 27 '13 at 13:50

Writing it another way, the vector valued function (the curve) can be written as $(x,y,z) = (t, 2t\cos t, 2t\sin t)$. Note the equation for $z$ -- I think you have it wrong in the question. Writing the coordinates separately, this becomes:

$$ x = t \quad ; \quad y = 2t\cos t \quad ; \quad z = 2t\sin t $$

Now you just have to check that the point $(x,y,z)$ given by these equations lies on the cone for all values of $t$. In other words, you have to check that $4x^2 = y^2 + z^2$ for all values of $t$. That's what the book answer does.

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