Given the position of a particle, for example, $<4t, 5t^2>,$ you are asked to find the speed at $t = 2.$ The correct answer is to find the particle's velocity (the derivative of its position) then take the magnitude.
Why is it incorrect to find the particle's distance $(\sqrt{16t^2 + 25t^4})$ then find the derivative of that at $t=2?$
I ask this because in a previous problem, given a velocity vector, to find the distance traveled by an object, they first found its speed then took the integral, instead of finding its position (the integral of velocity) then taking the magnitude.
To me, the two seem like they are asking for the reverse action, but I do not understand why they have different approaches when solving.
Best Answer
That's because distance travelled during $\boldsymbol{[t_1,t_2]}$ = time elapsed $\times$ average speed = $\color\red{\displaystyle\int_{t_1}^{t_2}|\boldsymbol v|\,\mathrm dt}.$
On the other hand, $\displaystyle\int_{t_1}^{t_2}\boldsymbol v\,\mathrm dt$ is the displacement from $\boldsymbol{t_1}$ to $\boldsymbol{t_2}$, so $\color\red{\left|\displaystyle\int_{t_1}^{t_2}\boldsymbol v\,\mathrm dt\right|}$ is the (shortest) distance between $\boldsymbol{t_1}$ and $\boldsymbol{t_2}.$ Observe that:
Yes: speed = $\color\red{\left|\dfrac{\mathrm d\boldsymbol r}{\mathrm dt}\right|}.$
$|\boldsymbol r|=\sqrt{16t^2 + 25t^4}$ is the (shortest) distance to the reference point, so its derivative $\color\red{\dfrac{\mathrm d|\boldsymbol r|}{\mathrm dt}}$ is the rate of change of distance to the reference point. (The particle starts at the reference point iff $\boldsymbol r(t{=}0)=\boldsymbol 0.$) Discarding $\boldsymbol r$ after taking $|\boldsymbol r|$ loses information: for example, given circular motion about the reference point $3$ units away and no additional information, $\left|\dfrac{\mathrm d\boldsymbol r}{\mathrm dt}\right|$ cannot be determined but we know that $\dfrac{\mathrm d|\boldsymbol r|}{\mathrm dt}=0\ne\left|\dfrac{\mathrm d\boldsymbol r}{\mathrm dt}\right|.$