If a particle follows the path defined by $$r(t) = (2t^{3/2},2t+1, \sqrt{5} t )$$ and starts at $t=0$, at what time will the particle have traveled a total of $14$ units ?
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well, I tried to find the derivative of r(t) but I am not sure if that is the right way to start – user131040 Jun 16 '14 at 23:46
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1That is in fact the first step. So you've found the velocity vector. Now find the magnitude of the velocity vector, which gives you the speed as a function of time. – David H Jun 16 '14 at 23:51
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and what is the next step ? – user131040 Jun 16 '14 at 23:52
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@user131040: what is the relationship between speed and distance traveled, in terms of time? – Alex Wertheim Jun 16 '14 at 23:58
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I am not sure if I am getting it, could you be more clear please – user131040 Jun 17 '14 at 00:02
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1@user131040, suppose you are travelling at a constant $50$ mph for $3$ hours. Then you have traveled $150$ miles. How might you go about finding how far you have traveled if your speed wasn't constant? I.e. if you were given some graph of your speed vs. time. – Kaj Hansen Jun 17 '14 at 00:08
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the acceleration ? – user131040 Jun 17 '14 at 00:09
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I just edited the equation $$r(t) = (2t^{3/2},2t+1, \sqrt{5} t )$$ – user131040 Jun 17 '14 at 00:34
3 Answers
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The formula is a mild generalization of the one for arclength in two dimensions. Take the derivatives, find the sum of their squares, take the square root, and integrate.
We get "lucky," the sum of the squares of the derivatives is simple, it is $9t+9$. So if we travel from time $0$ to time $\tau$, the path length is $$\int_0^\tau \sqrt{9t+9}\,dt.$$ Integrate, set the result equal to $14$, and solve for $\tau$.
André Nicolas
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Why could we not just evaluate $\sqrt{r_x^2+r_y^2+r_z^2}$ and set this equal to 14, and solve for $t$ (since distance is just the norm of displacement)? – beep-boop Jun 17 '14 at 00:55
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1@alexqwx: Your suggestion correctly finds the straight line distance from the beginning to the end. But we took a path that is not a straight line. It is a matter of interpretation. If I go from $A$ to $B$, a distance of $200$ km, and then back again, is the total distance travelled $400$, or is it $0$? – André Nicolas Jun 17 '14 at 01:03
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The distance $d(t)$ traveled between time $t=0$ and time $t=t$ is given by:
$$d^2(t)=|r(t)-r(0)|^2 = t(4t^2+4t+5)$$
Setting
$$d^2(t)=14^2$$ we obtain:
$$t(4t^2+4t+5)=14^2$$
The only real solution of this equation is given by:
$$t=\frac{1}{6}(-2-11\omega^{-1}+\omega)=3.25119...$$ $$\omega=\left(5329+18(87653)^{1/2}\right)^{1/3}$$
mike
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$$\vec{r}(t)=\left(2t^\frac32,2t+1,\sqrt{5}t\right)$$ You've got a vector of the position $r$ in terms of $t$. What you want to do next is find the velocity vector in terms of $t$, because "14 units" is the area swept by the magnitude of the velocity vector from $t=0$. This is done by simply differentiating: $$\vec{v}(t)=\left(3t^\frac12,2,\sqrt 5\right)$$
Now you want to find the function for the magnitude of this vector. You probably know how to do that, for vector $\vec{A}=(a, b, c)$, the magnitude is just $\big|\vec{A}\big|=\sqrt{a^2+b^2+c^2}$. Hence:
$$\begin{align}\biggr|\vec{v}(t)\biggr|&=\sqrt{\left(3t^\frac12\right)^2+\left(2\right)^2+\left(\sqrt 5\right)^2} \\ &=\sqrt{9t+4+5}\end{align}$$
Now the distance covered is the integral of that. This is a magnitude so it won't be negative, but you have just velocity, then you'd take the absolute value of it (for distance traveled rather than displacement).
Hence we've got: $$\int_{t=0}^{t=t_0} \sqrt{9t+9}=14$$
That evaluates to: $$\frac{1}{13.5}(9t+9)^\frac32\Biggr|_{t=0}^{t=t_0}=14$$
You find that $t_0=3$
Shahar
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