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The position vector at time $t$ of another point, $Q$, is given by $OQ = (t^2, 1-t, 1-t^2)$, $t \ge 0$. Find the value of $t$ for which the distance from $Q$ to the origin is minimum.

This is where I have reached

$d(t) = \sqrt{(2t^4-t^2-2t+2)}$. How do I continue from here to get the value of t?

Ok based on your suggestions I solved for the first derivative and got this

$d'(t)= 4t^3-t-1/\sqrt{(2t⁴-t²-2t+2)}$. I guess I need to solve this equation $4t^3-t-1=0$. It's eluding me

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  • $\begingroup$ Do you know how to calculate the distance of a point $(x, y, z)$ to the origin? If so, do the same thing substituting $x=t^2, y=1-t, z=t-t^2$. $\endgroup$ – FullofDill Sep 13 '17 at 16:13
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    $\begingroup$ You can always minimise the square of the distance ... $\endgroup$ – Mark Bennet Sep 13 '17 at 16:15
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Let $d(t)$ be the distance between $Q$ and the origin. Then $$d(t)=\sqrt{(t^2-0)^2+(1-t-0)^2+(1-t^2-0)^2}$$ To find the minimum distance between the origin and $Q$, we want to minimize $d(t)$. This can be done by solving $$\left.\frac{\text{d}}{\text{d}t} d(t)\right|_{t=t_0}=0$$for $t_0$, which will be the time at which the distance is the minimum. Of course, you can use the second derivative to verify that this critical point is indeed a minimum.

Note, that minimizing $d(t)$ can be done by minimizing $d^2(t)$ since $d(t)\geq 0$.

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  • $\begingroup$ I solved for the first derivative and got d'(t)= 4t³-t-1/√(2t⁴-t²-2t+2). Looks like solving for t is eluding me $\endgroup$ – Ashalley Samuel Sep 13 '17 at 16:54
  • $\begingroup$ This looks correct. So now you set that equation equal to zero and solve for $t$. As I remark in my answer, it may be easier to find the first derivative of $$d^2(t)=2t^4-t^2-2t+2$$ since this eliminates the pesky square root. $\endgroup$ – Dave Sep 13 '17 at 17:08
  • $\begingroup$ Yea my difficulty now is in solving the 4t³-t-1 for the value of t $\endgroup$ – Ashalley Samuel Sep 13 '17 at 17:56
  • $\begingroup$ Yes, I plugged this into Wolfram Alpha and the roots are rather nasty. Are you sure that you have the correct expression for $OQ$ in terms of $t$? Also, where'd you get this question? $\endgroup$ – Dave Sep 13 '17 at 18:05
  • $\begingroup$ Yea from the IB questionbank, 3rd edition, higher level. $\endgroup$ – Ashalley Samuel Sep 13 '17 at 18:12
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$f(t)=4 t^3-t-1$

As $f(0.5)=-1$ and $f(1)=+2$ for the intermediate value theorem there exists a root in $(0.5,1)$

Now apply Newton's method to find an approximate solution

Let's start with $t_0=0.5$ and define for $n>0$

$t_{n+1}=t_n-\dfrac{f(t_n)}{f'(t_n)}$

as $f'(t)=12t^2-1$ we have

$t_{n+1}=t_n-\dfrac{4 t_n^3-t_n-1}{12t_n^2-1}$

after a few iterations we find the root

$t\approx 0.76069$

which is the requested minimum where

$d\approx 0.754745$

Hope this helps

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  • $\begingroup$ Interesting...is there any other way apart from the value theorem method? $\endgroup$ – Ashalley Samuel Sep 13 '17 at 17:54
  • $\begingroup$ Thanks man. It has been very helpful $\endgroup$ – Ashalley Samuel Sep 13 '17 at 23:54
  • $\begingroup$ I got the value of t after the 7th iteration. Was long but worth it. Thanks man $\endgroup$ – Ashalley Samuel Sep 14 '17 at 7:22

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