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At what point of the curve given by $f (t) = (3t^2 − 6, 2t + 3, 7t^2 + 4t + 1)$ does the tangent line intersect the $z$-axis?

I tried the following: If $M$ is the desired point then $M = f(t_0)$ for some $t_0$.
$f'(t) = (6t,2,14t+4) \implies f'(t_0)=(6t_0,2,14t_0+4)$. I have no idea after this.

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    $\begingroup$ If this were in $2D$ you could do it, yeah? You have the slope at the point $t_0$ and you have the point $t_0$ so you can get the equation of the line. How would you extrapolate this to $3D$? $\endgroup$
    – epiliam
    Oct 9 '20 at 3:58
  • $\begingroup$ If one of the answers below answered your question, the way this site works works, you'd "accept" the answer, more here: What should I do when someone answers my question?. But only if your question really has been answered. If not, consider adding more details to the question. $\endgroup$
    – Darsen
    Oct 28 '20 at 17:21
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The directional vector of the line is given by $f'(t)$. Then your tangent line at a point $t_0$ will be given by the equation

$g(\lambda)=f(t_0)+\lambda f'(t_0)=(3t_0^2-6,2t_0+3,7t_0^2+4t_0+1)+\lambda(6t_0,2,14t_0+4)=(3t_0^2+6\lambda t_0-6,2t_0+2\lambda+3,7t_0^2+4t_0+14\lambda t_0+4\lambda+1)$.

To see when this equation intersects the Z-axis we have to set the first and second components equal to $0$. So

$3t_0^2+6\lambda t_0-6=0$

$2t_0+2\lambda+3=0\Rightarrow \lambda=\dfrac{-3-2t_0}{2}$

So substituting in the first equation we get

$3t_0^2+6\left(\dfrac{-3-2t_0}{2}\right)t_0-6=-3t_0^2-9t_0-6=0\Rightarrow t_0=-2,t_0=-1$.

And thus $\lambda=\dfrac{1}{2},\lambda=-\dfrac{1}{2}$.

So the points where the tangent line of the curve intersects the Z-axis are

$(0,0,9)$ and $(0,0,9)$, which is obviously just one point.

There are two distinct tangent lines (each one of them tangent to the curve at two distinct points) that intersect the Z-axis, but the two of them do it at the same point.

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  • $\begingroup$ $f'(t)$ only describes the slope at the point $t$. The equation for the tangent line is different. Your answer is wrong $\endgroup$
    – epiliam
    Oct 9 '20 at 3:56
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    $\begingroup$ You were right. I gotta get some sleep. I corrected it already. $\endgroup$
    – Darsen
    Oct 9 '20 at 4:20
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We can write the equation of tangent a value of parameter $t_o$ as:

$$ r(t_o, \lambda) = f(t_o) +f'(t_o) \lambda$$

Where $\lambda $ is a scalar constant.

To find the tangents which pass through the origin, simply set $r(t_o, \lambda) = 0$,

$$ f(t_o) + f'(t_o) \lambda = 0$$

Now, for the above equation there are two kinds of solution as I will show below:

  1. Setting the parameter as zero:

If you were to set $\lambda = 0$, then $ f(t_o) = 0$ are solutions to the problem statement. So, the parameter value for which the curve cuts the origin also gives tangents which cut the origin at the same point.

  1. $\lambda $ is non zero

If you had the condition that $\lambda$ were non zero, then what you could do is find the summed expression of $ f(t_o) + f'(t_o) \lambda$ , call that $q$. Now the $t_o$ values that send all components of $q$ to zero give you another set of liens that fulfill the conditions.

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