# Homework Help: Limiting Value of a Vector Function

1. Mar 17, 2016

### Sirsh

1. The problem statement, all variables and given/known data
The position vector of a particle is given in the terms of t, by,
s = (e-t+3*cos(2t)i+2tj+(e-t+3*sin(2t)k)

Find the limiting value of speed when t approaches positive infinity. The answer says "s = ..."

3. The attempt at a solution
I have evaluated the limits of all the components of this vector function individually:
e-t = 0
3*cos(2t) = -3 to 3
2t = positive infinity
3*sin(2t) = -3 to 3.

I know that the magnitude of the velocity vector is equal to the speed of the particle, so I'm not sure if i should have done the limits on the velocity vector that i found by differentiating the position vector? but the answer is in the form of "s = ..."

Any help would be appreciated, Thanks!

2. Mar 17, 2016

### blue_leaf77

That can be a typo because if the magnitude of the particle's position vector has a limit then it means the particle is gradually coming to a stop (which is clearly not judging from $\mathbf{s}(t)$). You should instead calculate the magnitude of the velocity vector.

3. Mar 17, 2016

### Sirsh

This is the view of the question that I've got:

I'm not sure how I would go about calculating the magnitude of the velocity vector, given all I have is t approaching infinity. Given that this is the form of the velocity vector function:

v = ds/dt = (-e-t-6*sin(2t)i+2j+(-e-t+6*cos(2t)k)

If I were to do the limits of each component and then do the magnitude of that it'd make sense, but the limits of the cosine and sine fluctuate

4. Mar 17, 2016

### Dick

The velocity vector does not have a limit, as you say. The SPEED (which is NOT a vector) does. Write down an expression for the speed and find it's limit.

5. Mar 17, 2016

### blue_leaf77

Like Dick says, first calculate $|\mathbf{v}|$, then find its limit as $t$ approaches infinity.

6. Mar 17, 2016

### Sirsh

Here is my attempt at interpreting what you guys have told me thus far:

s = (e-t+3*cos(2t)i+2tj+(e-t+3*sin(2t)k)
v = ds/dt = (-e-t-6*sin(2t)i+2j-(e-t+6*cos(2t)k)

Speed = length/magnitude of velocity vector, v:

|v| = √((-e-t-6*sin(2t))2+(2)2+(-e-t+6*cos(2t))2)

lim t → ∞ √((-e-t-6*sin(2t))2+(2)2+(-e-t+6*cos(2t))2)

-e^-t = 0
-6*sin(2t) = -6 to 6
2 = 2
6*cos(2t) = -6 to 6

I get to here but am unsure how to interpret the '-6 to 6' limits for the trig functions? I'm assuming if I can evaluate each part of the limit of |v| then I'll be able to get a scalar number from it.

7. Mar 17, 2016

### blue_leaf77

In order to compute the limit easier, simplify the above equation by expanding the squares and then employ a trigonometric identity.

8. Mar 17, 2016

### Sirsh

So by expanding the squares I have:

|v| = (-e^-2t-36*sin^2(2t))+(2^2)+(-e^-2t+36*cos^2(2t))^(0.5)

Limits for t approaching infinity:
-e^-2t = 0
-36*sin^2(2t) = 0 to 36
2^2 = 4
36*cos^2(2t) = 0 to 36

Still not sure if this is the correct way to evaluate this?

9. Mar 17, 2016

### blue_leaf77

Not so correct, you missed some terms as well as problem with signs. Let me ask you, how do you expand $(a+b)^2$?

10. Mar 17, 2016

### Sirsh

Ah! My mistake, trying to reply as I woke up lol.

|v| = (-e^-t-6*sin(2t))^2+(2)^2+(-e^-t+6*cos(2t))^2)^(0.5)
Doing the expansion
(-e^-t-6*sin(2t))(-e^-t-6*sin(2t)+4+(-e^-t+6*cos(2t))(-e^-t+6*cos(2t))
=> (e^-2t+6*e^-t*sin(2t)+6*e^-t*sin(2t)+36*sin^2(2t)+4+(e^-2t-6*e^-t*cos(2t)-6*e^-t*cos(2t)+36*cos^2(2t))
=> e^-2t+12*e^-t*sin(2t)+36*sin^2(2t)+4+e^-2t-12*e^-t*cos(2t)+36*cos^2(2t))

Would this be the correct way of expanding the equation? I'm not that familiar with trigonometric identities but I'm thinking that the terms cos^2 and sin^2 may have another form?

11. Mar 17, 2016

### blue_leaf77

Yes they do, take a look at this.
Then you should familiarize yourself.

12. Mar 17, 2016

### Sirsh

e^-2t+12*e^-t*sin(2t)+36*sin^2(2t)+4+e^-2t-12*e^-t*cos(2t)+36*cos^2(2t))

=> 2e^-2t+12e^-t*sin(2t)+4-12*e^-t*cos(2t)+36*(sin^2(2t)+cos^2(2t))
=> 2e^-2t+12e^-t*(sin(2t)-cos(2t))+4+36

Lim t->infinity of 2e^-2t+12e^-t*(sin(2t)-cos(2t))+4+36

limit of any e^-2t function is 0
Therefore,
the limit of the expansion is equal to 4+36 = 40

Hence, |v| = √(40).

Does this seem like the correct usage of the trigonometric identities? I'm not sure if there would be any use (if it is possible) to simplify the sin(2t)-cos(2t) given that the limit of e^-t is zero anyway.

13. Mar 17, 2016

### Dick

Yes, that looks ok.

14. Mar 18, 2016

### blue_leaf77

No, that's fine. As you know, there is no point simplifying these terms as they go to zero.