The position function of a particle is given by $r(t) = \langle-5t^2, -1t, t^2 + 1t\rangle$. At what time is the speed minimum?
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2Write an expression for the speed, then find the minima of that. You need to show what you have done and where you are stuck... – Macavity Oct 31 '13 at 02:52
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I found the velocity vector = <-10t, -1, 2t + 1>. Then I got the magnitude as sqrt (104t^2 + 4t + 2). I know I need to minimize this magnitude, but I always get a negative t=-52 – bbm Oct 31 '13 at 02:52
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If $t < 0$ is allowed, the minimum must be at the reciprocal of what you got. – Macavity Oct 31 '13 at 02:57
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I get $t=-\frac{1}{52}$. – copper.hat Oct 31 '13 at 02:58
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The velocity vector is $\left<-10t,-1,2t+1\right>$. Thus the speed is $\sqrt{(-10t)^2 +(-1)^2+(2t+1)^2}$.
We want to minimize this, or equivalently we want to minimize $104t^2+4t+2$. This is a problem that can even be solved without calculus, by completing the square.
If negative $t$ is allowed, we get $t=-\frac{4}{208}$. If by nature $t\ge 0$, the minimum speed is at $t=0$.
André Nicolas
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When I minimized that function, I got t=-52, which is very misleading because time should be positive? – bbm Oct 31 '13 at 02:57
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1You should get $t=-\frac{1}{52}$. There is nothing in the problem as stated that forces $t$ to be positive. But if somewhere else in the problem it is specified that $t\ge 0$, then the min is at $t=0$, since the function is increasing in the interval $(-1/52,\infty)$. – André Nicolas Oct 31 '13 at 03:01
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$$v(t)=r'(t)=<-10t,-1, 2t+1>$$ $$|v(t)|=\sqrt{100t^2+1+4t^2+4t+1}=\sqrt{104t^2+4t+1}$$ Now you want to minimize what's underneath the square root which we will call $f(t)$.
$$f'(t)=208t+4=0$$ Which gives $$t=\frac{-4}{208}=\frac{-1}{52}$$
MITjanitor
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