1. ## word problem, if your up for a challenge

A particle moves that its position vector r at the time t is given by

r = c ( t - 1/3 t^3) i + ct^2 j

Where c is a positive constant.

Find the velocity at time t(> 0) and show that’s its magnitude increase with time.

Determine its direction of motion at time t=0 show that the directions turns through a right angle between t =0 and t=1.

Show also that the component of force acting on the particle in the direction of the vector j is constant throughout the motion.

Given that c=1, find the maximum displacement of the particle in the I direction and draw a rough sketch of the path of the particle for values of t> 0, showing clearly the positions of the particle at times t=0, 1, radical 3, 2.

2. Originally Posted by sderosa518
A particle moves that its position vector r at the time t is given by

r = c ( t - 1/3 t^3) i + ct^2 j

Where c is a positive constant. Find the velocity at time t(> 0) and show that’s its magnitude increase with time. Determine its direction of motion at time t=0 show that the directions turns through a right angle between t =0 and t=1. Show also that the component of force acting on the particle in the direction of the vector j is constant throughout the motion.
Given that c=1, find the maximum displacement of the particle in the I direction and draw a rough sketch of the path of the particle for values of t> 0, showing clearly the positions of the particle at times t=0, 1, radical 3, 2.
Hints for the first couple (if you're up to the challenge of showing all your work and saying where you get stuck):

v = dr/dt.

Consider |v| = |dr/dt| and show that d|v|/dt > 0.

Direction of motion given by direction of dv/dt.

Take the dot product of dv/dt at t = 0 with dv/dt at t = 1.

F = ma and a = dv/dt.

3. Thank you, but what do you mean find the diravative of what. This question is part of my homework for Calculus Class, plus its my major. I am in college. i was hopin if possible if you can do and explained how to get the answer step by step.

Thank you

4. Originally Posted by sderosa518
Thank you, but what do you mean find the diravative of what. This question is part of my homework for Calculus Class, plus its my major. I am in college. i was hopin if possible if you can do and explained how to get the answer step by step.

Thank you
You need to differentiate the given vector function. Have you been taught how to do this? If r = f(t) i + g(t) j + h(t) k then dr/dt = df/dt i + dg/dt j + dh/dt k.

5. Honestly, my professor isnt that easy going. I am 100% that he didnt taught us this, but we can get help if we need it. Isnt vectors in Calculus II. I am in Cal I. He wants us to think, but nothin is comin to mind. If you can, can you write it out and show me and maybe I can do another example that is in the book.

TY

6. Hello sderosa518
Originally Posted by sderosa518
A particle moves that its position vector r at the time t is given by

r = c ( t - 1/3 t^3) i + ct^2 j

Where c is a positive constant.

Find the velocity at time t(> 0) and show that’s its magnitude increase with time.

Determine its direction of motion at time t=0 show that the directions turns through a right angle between t =0 and t=1.

Show also that the component of force acting on the particle in the direction of the vector j is constant throughout the motion.

Given that c=1, find the maximum displacement of the particle in the I direction and draw a rough sketch of the path of the particle for values of t> 0, showing clearly the positions of the particle at times t=0, 1, radical 3, 2.
$\textbf{r} = c(t-\tfrac13t^3)\textbf{i}+ct^2\textbf{j}$

$\Rightarrow \textbf{v}=\frac{d\textbf{r}}{dt}=c(1-t^2)\textbf{i}+2ct\textbf{j}$

The magnitude of the velocity $= \sqrt{c^2(1-t^2)^2+4c^2t^2}$
$=c\sqrt{1-2t^2+t^4+4t^2}$

$=c\sqrt{1+2t^2+t^4}$

$=c\sqrt{(1+t^2)^2}$

$=c(1+t^2)$
which clearly increases as $t$ increases, for $t > 0$, since $c$ is a positive constant.

When $t = 0,\; \textbf{v}= c\textbf{i}$. So the direction of motion when $t=0$ is parallel to the vector $\textbf{i}$ (i.e. parallel to the $x$-axis).

When $t = 1,\; \textbf{v}= 2c\textbf{j}$. So the direction of motion when $t=1$ is parallel to the vector $\textbf{j}$ (i.e. parallel to the $y$-axis), and is therefore perpendicular to its direction when $t = 0$.

Acceleration $= \textbf{a} = \frac{d\textbf{v}}{dt} = -2ct\textbf{i}+2c\textbf{j}$. The component in the $\textbf{j}$ direction is therefore constant. So the component of the external force that causes this acceleration is also constant in this direction.

The maximum displacement in the $\textbf{i}$ direction will occur when the component of the velocity in this direction is zero; i.e. when $c(1-t^2)=0$; i.e. when $t = 1$ (for $t>0$)

So when $t = 1$ and $c = 1$, the maximum displacement in the $\textbf{i}$ direction is $1(1-\tfrac13.1^3) = \tfrac23$.

I'll leave you to work out the sketch of the path of the particle. Simply give $t$ various sensible values (including those given in the question) and work out the position vector $\textbf{r}$ for each value. Plot these points to sketch the path.