Deriving a particle's speed from its distance function

Question

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.

Answer 1

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,

That's because distance travelled during $\boldsymbol{[t_1,t_2]}$ = time elapsed $\times$ average speed = $\color\red{\displaystyle\int_{t_1}^{t_2}|\boldsymbol v|\,\mathrm dt}.$

instead of finding its position (the integral of velocity) then taking the magnitude.

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:

• distance travelled during $[t_1,t_2] \ge$ distance between $t_1$ and $t_2.$

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.

Yes: speed = $\color\red{\left|\dfrac{\mathrm d\boldsymbol r}{\mathrm dt}\right|}.$

Why is it incorrect to find the particle's distance $(\sqrt{16t^2 + 25t^4})$ then find the derivative of that at $t=2?$

$|\boldsymbol r|=\sqrt{16t^2 + 25t^4}$ 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 $\dfrac{\mathrm d|\boldsymbol r|}{\mathrm dt}=0\ne\left|\dfrac{\mathrm d\boldsymbol r}{\mathrm dt}\right|.$

Answer 2

The vector describing the motion of the particle gives its instantaneous position along a parametric curve $ \ x \ = \ 4t \ \ , \ \ y \ = \ 5t^2 \ \ , \ $ which traces out a parabola [blue curve in the graph above]. You have correctly given the distance of the particle from the origin [measured along the green line], but this only has an indirect relation to the particle's motion. (At $ \ t \ = \ 2 \ \ , \ $ the particle is at $ \ (8 \ , \ 20) \ \ , \ $ which puts it a distance $ \ d \ = \ \sqrt{464} \ \approx \ 21.54 \ \ $ from the origin. The derivative of $ \ d(t) \ $ would be described as the rate at which the distance of the particle from the origin is changing, but this is not the particle's speed.)

However, the particle is moving along the parabolic curve, so the distance it has traveled from the origin (where it was at $ \ t \ = \ 0 \ ) \ $ is measured as the arc-length along that curve, given by $$ s(t) \ \ = \ \ \int_0^t \ \sqrt{ \left(\frac{dx}{dt} \right)^2 \ + \ \left(\frac{dy}{dt} \right)^2 } \ \ dt \ \ = \ \ \int_0^t \ \sqrt{ \left(4 \right)^2 \ + \ \left(10t \right)^2 } \ \ dt $$ $$ = \ \ \int_0^t \ \sqrt{ 16 \ + \ 100t^2 } \ \ dt \ \ = \ \ t·\sqrt{4 + 25t^2} \ + \ \frac45·\ln \left( \frac{5t \ + \ \sqrt{4 + 25t^2}}{2} \right) \ \ . $$ We can apply this arc-length function to find that the particle has traveled $ \ s(2) \ \approx \ 22.25 \ \ $ in $ \ 2 \ $ seconds.

Now the problem asks for the speed of the particle at $ \ t \ = \ 2 \ \ . \ $ We could differentiate this arc-length function, which looks extremely unappetizing until we recall that Leibniz tells us that the derivative of $ \ s(t) \ $ will just be the integrand function $ \ \sqrt{ \left(\frac{dx}{dt} \right)^2 \ + \ \left(\frac{dy}{dt} \right)^2 } \ = \ \sqrt{ 16 \ + \ 100t^2 } \ \ . \ $ So the speed $ \ v(2) \ $ of the particle is the value of this function at $ \ t \ = \ 2 \ \ . $

But we don't need to go through all of this since we have the position vector of the particle's motion, $ \ \vec{s}(t) \ = \ \langle \ 4t \ , \ 5t^2 \ \rangle \ \ . \ $ We do want to take a derivative: we differentiate the components of the position vector with respect to time to obtain $$ \vec{v}(t) \ \ = \ \ \langle \ \frac{dx}{dt} \ , \ \frac{dy}{dt} \ \rangle \ \ = \ \ \langle \ 4 \ , \ 10t \ \rangle \ \ . $$ This gives us the tangent vector [red arrow] to the parabolic curve, which lies along the tangent line to the curve [in red]; this tangent vector is the velocity vector $ \ \vec{v}(t) \ \ $ and the speed $ \ v(t) \ $ is then given by the magnitude of this vector $ \ | \vec{v}(t) | \ \ . \ $ We note that this magnitude is given by the same expression as the integrand function in $ \ s(t) \ $ that we found earlier.