particle movement, partial differentiation

Can someone explain how to do this question?

**"A particle moves such that its position at time t is given by r=(t,t**^{2},t^{3}).

Find the rate of change of the distance of the particle from the origin."

I don't understand the way r is described, does it mean the x,y and z coordinates of r? If so, how do you use the chain rule to differentiate this?

Thanks

Laura

Re: particle movement, partial differentiation

Re: particle movement, partial differentiation

Oh ok, that makes a bit more sense. Still struggling with how to differentiate it though.

Re: particle movement, partial differentiation

First there is just the one independent variable, t, so this is NOT "partial differentiation", just ordinary differentiation.

This is a **vector** equation: $\displaystyle r= <x, y, z>= <t, t^2, t^3>$. Do you know how to differentiate t with respect t? What about $\displaystyle t^2$ or $\displaystyle t^3$.

Re: particle movement, partial differentiation

For a vector valued function $\displaystyle \displaystyle \begin{align*} \mathbf{r}\,(t) = \left( x(t), y(t), z(t) \right) \end{align*}$, its rate of change is $\displaystyle \displaystyle \begin{align*} \mathbf{r}'(t) = \left( x'(t), y'(t), z'(t) \right) \end{align*}$.

Re: particle movement, partial differentiation

I get an answer of 1+2t+3t^2, but that is not the correct answer

Re: particle movement, partial differentiation

That's because your answer is also supposed to be a vector. Read my last response.

Re: particle movement, partial differentiation

I am still obviously not fully understanding this and need more of an explanation.

What I have understood so far is that r is a vector with coordinates (x(t),y(t),z(t)). Where x(t)=t, y(t)=t^{2} and z(t)=t^{3}.

In order to find the rate of change of r I must find the derivatives x'(t), y'(t) and z'(t) which are 1, 2t and 3t^{2}, respectively. I also know you can write the coordinates in terms of unit vectors: r= t(**i**)+t^{2}(**j**)+t^{3}(**k**). Is this all correct? What happens to the unit vectors if that is the case?

Re: particle movement, partial differentiation

Quote:

Originally Posted by

**lauramorrison93** I am still obviously not fully understanding this and need more of an explanation.

What I have understood so far is that r is a vector with coordinates (x(t),y(t),z(t)). Where x(t)=t, y(t)=t^{2} and z(t)=t^{3}.

In order to find the rate of change of r I must find the derivatives x'(t), y'(t) and z'(t) which are 1, 2t and 3t^{2}, respectively. I also know you can write the coordinates in terms of unit vectors: r= t(**i**)+t^{2}(**j**)+t^{3}(**k**). Is this all correct? What happens to the unit vectors if that is the case?

They stay there.