Finding the velocity and the speed for a particle

Question

Sorry for the question not being in the correct format. It's my first time using the site.

For the particle with position given by

x(t) = (1/(t+1))(Rcos(t)i+Rsin(t)j)

for t ∈ (0, ∞), where R > 0 is a constant, calculate the velocity and the speed. Sketch the particle path and indicate the direction of motion with an arrow.

Can someone please help me on how to calculate the velocity and the speed of this vector function. I'm having difficulty computing the first derivative. Do I need to use the quotient rule? I have tried this and the answer doesn't seem correct. For the velocity I get,

((-R/(t+1)^2)((t+1)sin(t)+cos(t))i +(R/(t+1)^2)((t+1)cos(t)-sin(t)).

And for the speed I get,

Sqrt((R^2)/((t+1)^4)[((t+1)^2)+1].

When I also try to eliminate t from the parametric equations I end up with x^2=-y^2 which also doesn't seem correct.

Can some please point me in the right direction. Thank you.

Answer 1

Since you are working in two dimensions, I think a nice way to look at your problem is to consider the position of your particle at time $t$ as a complex number $x(t)$ with real part $\frac{R}{t+1}\cos(t)$, and imaginary part $\frac{R}{t+1}\sin(t)i$, where I am denoting by $i$ the imaginary unit. Then $x(t)=\frac{R}{t+1}e^{i t}$.

You can obtain the velocity simply as $\frac{d}{dt} x(t)= -\frac{R}{(t+1)^2}e^{i t} + i\frac{R}{t+1}e^{i t}$. You can extract the real and imaginary parts of the two terms to obtain the components of the velocity vector, and add their squares and take the square root to obtain its modulus, the speed.

But there's a way that might provide more insight. Note that $i\frac{R}{t+1}e^{i t}$ equals to the original position vector rotated by $\frac{\pi}{2}$, and $-\frac{R}{(t+1)^2}e^{i t}$ equals to the original position vector rotated by $\pi$ and shrunk by a factor $\frac{1}{t+1}$. So your velocity vector equals the original position vector, stretched by a factor $\sqrt{1+(\frac{1}{1+t})^2}$ and rotated by an angle $\frac{\pi}{2}+\arctan(\frac{1}{t+1})$, i.e. $\frac{d}{dt} x(t)=\frac{R}{t+1}\sqrt{1+(\frac{1}{t+1})^2}\cdot e^{i\left(\frac{\pi}{2}+\arctan(\frac{1}{t+1})+t\right)}$. The speed then is equal to its modulus $s(t)=\frac{R}{t+1}\sqrt{1+(\frac{1}{t+1})^2}$, and the two components along the two axes are $s(t)\cos(\theta(t))$ and $s(t)\sin(\theta(t))$ where $\theta(t)=\frac{\pi}{2}+\arctan(\frac{1}{t+1})+t$.