# One more particle question/s

Really struggling with these particle questions, jsut can't grasp the concept

A moving particle at time t ∈ [0, 10] (seconds) has position vector in metres from the origin (0, 0, 0) given by the vector function r(t) = (10 − t)i + (t 2 − 10t)j + sin tk.

i. Describe the path of the particle, as seen from above (the positive k-direction), and
also describe it in three dimensions.

ii. Find the curvature of the path, at t = 2π ≈ 6.28 seconds.

iii. Find the angle between the path (at start and end-points) and the k-direction.

And this one

A particle’s path, in two dimensions, is described by its position vector (in metres and
time t ∈ [1, 2] seconds) relative to point (0, 0, 0) by r(t) = (2t + 1)i + (4 − t 2 )j.

i. Sketch the path of the particle.

ii. Find the value of t ∗ at which the particle has greatest distance from (0, 0, 0).
Hint: optimising squared distance may be the simpler method here.

iii. Show that at position r(t ∗ ), the particles velocity is not perpendicular to r(t∗ ).

From a practice exam, for my final this week

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Dick
Homework Helper
You've posted two problems with a total of six parts. And you haven't given us any attempt to solve any of them. Read the forum rules. You have to start TRYING to solve one of these.

sorry i have tried.

i have so far

i.
x = 10 - t => t = 10 - x (1)

y = t^2 - 10t (2)
Substitute (1) into (2).
which gives y=(10-x)^2 - 10(10-x) which describes the particles path?

ii. and then i need to substitute 2*pi into the equation (not 100% sure which one though) to find the position

iii. and to find the angle of the path i need to use the dot product but i can't work out which other vector to use to get the dot product.

The second question:

The distance between the particle and the origin |r| and its square |r|^2 will be minimum at the same time t*

|r|^2 = (2t+1)^2+(4-t^2)^2
this then needs to be expanded which i think i am doing wrong

once the expansion is right i need to differentiate the equation?

but part iii. i can't work out how to do

sorry i'm a bit new here, i have definitely tried them just thought i might be on the wrong track

Dick
Homework Helper
That's a start. Good. Now you'll hopefully get a lot of help. You have y=(10-x)^2-10(10-x). That's the path of the particle viewed from above, in the xy plane. It's a parabola, do you know why? To describe it in 3d you have to use words creatively, it's not any kind of a standard curve. Finding angles of the path and curvature requires calculus. For angles you need to take the dot product of tangent vectors. Can you find the tangent vector to the curve in the first question?

the tangent vector is that r(t) / ||r(t)|| and then substitute in r(0)?

Dick
Homework Helper
No. The tangent vector is r'(t). The normalized tangent vector is r'(t)/|r'(t)|. Notice the derivatives.

so the derivative of r(t) would be:

r(t) = (10 − t)i + (t 2 − 10t)j + sin tk.

r'(t) = ti + 2tj + (-cos)tk ?

also just to back track a little the 2 * pi gets subbed into y=(10-x)^2 - 10(10-x)?

Dick
Homework Helper
You need a lot of work on derivatives. Start with the x component. What's the derivative of 10-t? On the backtrack you need to substitute 2pi for t, not x, and you need to substitute it into the 3d form. We haven't even started the curvature question, you'll need a second derivative for that. Can you find a formula? For iii), you should get one vector at t=0 and another at t=10. Use them to find the angles with the k direction.

so pi would be substituted into

r(t) = (10 − t)i + (t 2 − 10t)j + sin tk
giving (10-2*pi) + (2*pi^2 - 10(2*pi))+sin(2*pi)

and would the equation for r'(t) be:
i + 2tj + -cosk?

Dick
Homework Helper
Don't even try to substitute 2pi into anything until you have at least the tangent right. And you don't. I asked you what is d/dt of (10-t) and you got that one wrong. Next I'll ask you what is d/dt of t^2-10t. What's your guess? Then if you want to practice while I sleep, what is d/dt sin(t). I'll give you a hint. It's NOT -cos(t).

d/dt of 10-t is 1
and the other equation is
d/dt 2t -10

is that right?

giving i + (2t -10)j + cos(t)?

Dick
Homework Helper
No. Still not. d/dt(10-t) is NOT 1. Try again. The other parts are ok.

is it -1i + (2t-10)j + cos(t)?

Dick
Homework Helper
Yessss. That's the tangent vector for the curve in the first problem. Can you use that to answer part iii)? What's the angle between that vector and k at t=0 and t=10

so i put 2*pi into that equation and then for part iii:

theta= cos^-1 (u.v/ |u||v|)

with r(t) being u and k being v

Dick