# find the value of $t$ if any at which the tangential component of acceleration is equal to the normal component

A point on a rifle bullet travels according to the position function $$r(t) = \langle5t^2,\cos(t^2), \sin(t^2)\rangle$$ Find the value of $t$ if any at which the tangential component of acceleration is equal to the normal component.

I have found Velocity , Acceleration and the Normal and Tangential Components of acceleration it is a ton of work to get to this point so not expecting anyone to actually go through this just a hint on what to do next.

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I'm assuming you are saying that you've found the parametric functions for the normal and tangential acceleration vectors. It sounds like they want you to find the parameter value at which the magnitudes are equal. Since the magnitudes are always positive or zero, you can just compare the squares of the magnitudes, which lets you dispose of radicals... – RecklessReckoner Nov 8 '13 at 1:16
Yes, i found the normal and tangential parametric functions for the acceleration vector. What reason would i have to square each function? I knew i had to set them equal to each other but im not quite following the logic here. Magnitudes of the Tang. and Norm components represent what exactly? – Achilles Nov 8 '13 at 1:24
Not squaring the functions, but rather the magnitudes; but it's really the same as just comparing the arguments of the radicals, as user40615's answer illustrates. (The magnitudes are the speeds along the tangent and the normal to the curve at each point.) – RecklessReckoner Nov 8 '13 at 6:33
I see, thank you very much man i appreciate you taking the time to help : ) – Achilles Nov 10 '13 at 0:30

$v=r'(t)=\bigg \langle 10t, -2t\sin(t^2), 2t\cos (t^2)\bigg\rangle\Rightarrow |v|=\sqrt{104}t$, and $a_T=\frac{d|v|}{dt}=\sqrt{104}.$
$a=r''(t)=\bigg\langle 10, -2\sin(t^2)-4t^2\cos(t^2), 2\cos(t^2)-4t^2\sin(t^2)\bigg\rangle\Rightarrow |a|= \sqrt{16t^4+100t^2+4}$.
$\Rightarrow a_N= \sqrt{|a|^2-a^2_T}=\sqrt{16t^4+100t^2-100}$. Thus, $a_T=a_N\Rightarrow 4t^4+25t^2-51=0\Rightarrow t^2= \frac{-25+\sqrt{1441}}{8}\Rightarrow t\approx 1,27.$