Expert: Paul Klarreich - 3/27/2008

Question

It's has been awhile since I worked with derivatives and antiderivative. Can you check if I did these two problems right?

A moving particle has position (x(t),y(t)) at time t. The position of the particle at time t=1 is (2,6) and the velocity vector at any time t>0 is given by (1-(1/t^2)), 2+(1/t^2)).

a, Find the acceleration vector at time t=3.

<x"(t),y"(t)> = (2/27, -2/27)

b, Find the position of the particle at time t=3.
<x(t),y(t)> = (10/3, 17/3)

and can you please help me with these questions?

c, For what t>0 does the line tangent to the path of the particle at (x(t),y(t)) have a slope of 8?

I think I'm suppose to set dy/dx = 8. but I'm not sure what i'm suppose to do after that.

d, The particle approaches a line as t-> infinity. Find the slope of this line. Show the work that leads to your conclusion.

I have no idea how to do part D at all. I guessing I have to graph it..? maybe?

Thank you so much for your help.

Answer
Questioner:   Ann
Category:  Calculus
 
Subject:  parametric?
It's has been awhile since I worked with derivatives and antiderivative. Can you check if I did these two problems right?

A moving particle has position (x(t),y(t)) at time t. The position of the particle at time t=1 is (2,6) and the velocity vector at any time t>0 is given by (1-(1/t^2)), 2+(1/t^2)).

a, Find the acceleration vector at time t=3.

<x"(t),y"(t)> = (2/27, -2/27)

b, Find the position of the particle at time t=3.
<x(t),y(t)> = (10/3, 17/3)

and can you please help me with these questions?

c, For what t>0 does the line tangent to the path of the particle at (x(t),y(t)) have a slope of 8?

I think I'm suppose to set dy/dx = 8. but I'm not sure what i'm suppose to do after that.

d, The particle approaches a line as t-> infinity. Find the slope of this line. Show the work that leads to your conclusion.

I have no idea how to do part D at all. I guessing I have to graph it..? maybe?

Thank you so much for your help.
......................................
Hi, Ann,

vx = 1- 1/t^2,
ax = 2/t^3
ax(3) = 2/27

vy = 2 + 1/t^2
ay = -2/t^3
ay(3) = -2/27

GOOD SO FAR.

To get x,y you integrate:

x = t + 1/t + C0
x(1) = 2 = 1 + 1/1 + C0.
C0 = 0
x = t + 1/t
x(3) = 3 + 1/3 = 10/3

Good.

y = 2t - 1/t + C1
y(1) = 6 = 2 - 1/1 + C1
C1 = 5
y(3) = 2(3) - 1/3 + 5 = 32/3

Hmmm....... which one of us blew something here?
.............................
c, For what t>0 does the line tangent to the path of the particle at (x(t),y(t)) have a slope of 8?

Translate that to: "for what t is  dy/dx = 8?"

dy   dy/dt   2 + 1/t^2
-- = ----- = --------- = 8
dx   dx/dt   1- 1/t^2


 2t^2 + 1
 --------- = 8
  t^2 - 1

2t^2 + 1 = 8t^2 - 8
9 = 6t^2
3 = 2t^2
t = sqrt(3/2)
................................
d, The particle approaches a line as t-> infinity. Find the slope of this line. Show the work that leads to your conclusion.

Translate that to: "What is lim[t->inf] of dy/dx?

Lim     2 + 1/t^2   2 + 0
       --------- = ----- = 2
t->inf  1 - 1/t^2   1 + 0