Limiting Value of a Vector Function

• Sirsh
In summary: 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...Yes, that would be the correct way to do it.
Sirsh

Homework Statement

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 = ..."

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!

Sirsh said:
The answer says "s = ..."
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.

blue_leaf77 said:
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.

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

Sirsh said:
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

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.

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

Dick said:
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.

blue_leaf77 said:
Like Dick says, first calculate ##|\mathbf{v}|##, then find its limit as ##t## approaches infinity.

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.

Sirsh said:
lim t → ∞ √((-e-t-6*sin(2t))2+(2)2+(-e-t+6*cos(2t))2)
In order to compute the limit easier, simplify the above equation by expanding the squares and then employ a trigonometric identity.

blue_leaf77 said:
In order to compute the limit easier, simplify the above equation by expanding the squares and then employ a trigonometric identity.

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?

Sirsh said:
Still not sure if this is the correct way to evaluate this?
Not so correct, you missed some terms as well as problem with signs. Let me ask you, how do you expand ##(a+b)^2##?

blue_leaf77 said:
Not so correct, you missed some terms as well as problem with signs. Let me ask you, how do you expand ##(a+b)^2##?

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?

Sirsh said:
I'm thinking that the terms cos^2 and sin^2 may have another form?
Yes they do, take a look at this.
Sirsh said:
I'm not that familiar with trigonometric identities
Then you should familiarize yourself.

blue_leaf77 said:
Yes they do, take a look at this.

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.

Sirsh said:
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.

Yes, that looks ok.

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

What is the limiting value of a vector function?

The limiting value of a vector function is the value that the function approaches as the input variable approaches a certain value, typically infinity or negative infinity. It can also be referred to as the limit of the function.

Why is the limiting value of a vector function important?

The limiting value of a vector function is important because it helps us understand the behavior of the function as the input variable gets larger or smaller. It can also help us determine if the function is continuous at a certain point.

How is the limiting value of a vector function calculated?

The limiting value of a vector function can be calculated by evaluating the function at values that are very close to the desired input variable. As these values get closer and closer, the limiting value will become more accurate.

Can the limiting value of a vector function be infinite?

Yes, the limiting value of a vector function can be infinite. This can occur when the function approaches infinity as the input variable approaches a certain value.

How does the limiting value of a vector function relate to derivatives and integrals?

The limiting value of a vector function can be used to find the derivative and integral of the function at a certain point. By evaluating the function at values close to the point, we can determine the slope of the tangent line and the area under the curve, respectively.

• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
11
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
375
• Calculus and Beyond Homework Help
Replies
2
Views
714
• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
20
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
949
• Calculus and Beyond Homework Help
Replies
1
Views
824
• Calculus and Beyond Homework Help
Replies
12
Views
1K