Finding the maximum speed for a vector?

In summary, the conversation discusses a set of equations and their solutions, as well as the goal of finding the maximum speed of a particle. The conversation also includes a discussion about the validity of certain equations and an explanation for why x must be in a specific interval in order for the square root to be a real number. The conversation ends with the individual expressing their understanding of the topic after receiving help.
  • #1
mathgirl313
22
0
So I have the equations x'(t) = 4*y(t) and y'(t)=-x(t). Going through all the steps, you get
x(t) = c1*2*cos(2t) + c2*2*sin(2t) and
y(t) = -c1*sin(2t) + c2*cos(2t).
And of course, this can just can be written as the vector.
So after all that you are to find the maximum speed of the particle, which I have the whole solution to, I just don't understand parts of it.

If a particle starts at ([itex]\alpha[/itex],[itex]\beta[/itex]) then x(0) -[ [itex]\stackrel{2c1}{c2}[/itex] ] = [ [itex]\stackrel{\alpha}{\beta}[/itex] ]. I don't understand why this holds? (and that's my best attempt at a vector..sorry!)

And anyway c1 = [itex]\alpha[/itex] /2 and c2 = [itex]\beta[/itex] and using other earlier equations (x(t)^2)/4 + y(t)^2 = ([itex]\alpha[/itex] ^2)/4 + [itex]\beta[/itex] ^2. I don't understand

So we have the speed is given by [itex]\sqrt{16*y(t)^2 + x(t)^2}[/itex] = f(x,y), which we want to maximize, subject to x^2/4 + y^2 = ([itex]\alpha[/itex]^2) + ([itex]\beta[/itex])^2.

Rearranging the equation we are constrained by we get: f(x,y) = g(x) = [itex]\sqrt{4*(\alpha)^2 + 16*(\beta)^2 - 3*x^2}[/itex], where x is in (-[itex]\sqrt{(\alpha)^2 + 4*(\beta)^2}[/itex], [itex]\sqrt{(\alpha)^2 + 4*(\beta)^2}[/itex]. Why is x in this interval?

The rest of it from that point I understand. I just don't see why x is in that given interval, or why that original equation for x(0) holds? Thank you so much for any help!
 
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  • #2
mathgirl313 said:
So I have the equations x'(t) = 4*y(t) and y'(t)=-x(t). Going through all the steps, you get
x(t) = c1*2*cos(2t) + c2*2*sin(2t) and
y(t) = -c1*sin(2t) + c2*cos(2t).
And of course, this can just can be written as the vector.
So after all that you are to find the maximum speed of the particle, which I have the whole solution to, I just don't understand parts of it.

If a particle starts at ([itex]\alpha[/itex],[itex]\beta[/itex]) then x(0) -[ [itex]\stackrel{2c1}{c2}[/itex] ] = [ [itex]\stackrel{\alpha}{\beta}[/itex] ]. I don't understand why this holds? (and that's my best attempt at a vector..sorry!)
"Start" here is interpreted as "t= 0". Putting t= 0 in your equations for x and y, you get x(0)= 2c1 and y(0)= c2. Set those equal to [itex]\alpha[/itex] and [itex]\beta[/itex].

And anyway c1 = [itex]\alpha[/itex] /2 and c2 = [itex]\beta[/itex] and using other earlier equations (x(t)^2)/4 + y(t)^2 = ([itex]\alpha[/itex] ^2)/4 + [itex]\beta[/itex] ^2. I don't understand
Perhaps because it's not true. What is true is that since [itex]x(0)= \alpha[/itex] and [itex]y(0)= \beta[/itex] direct substitution gives [itex]x(0)^2/4+ y(0)^2= \alpha^2/4+ \beta^2[/itex]. It is not true for general "t".

So we have the speed is given by [itex]\sqrt{16*y(t)^2 + x(t)^2}[/itex] = f(x,y), which we want to maximize, subject to x^2/4 + y^2 = ([itex]\alpha[/itex]^2) + ([itex]\beta[/itex])^2.

Rearranging the equation we are constrained by we get: f(x,y) = g(x) = [itex]\sqrt{4*(\alpha)^2 + 16*(\beta)^2 - 3*x^2}[/itex], where x is in (-[itex]\sqrt{(\alpha)^2 + 4*(\beta)^2}[/itex], [itex]\sqrt{(\alpha)^2 + 4*(\beta)^2}[/itex]. Why is x in this interval?
In order that that square root be a real number, the quantity in the root must not be negative. If x is not in that integral, it will be negative.

The rest of it from that point I understand. I just don't see why x is in that given interval, or why that original equation for x(0) holds? Thank you so much for any help!
 
  • #3
Thanks for the help! I pretty much get all of it now.

The last part though, I know the square root had to be greater than or equal to zero...but when I tried solving I got a different interval. I had 4/3 and 16/3 as coefficients for alpha and beta...but that was used for something different later in the problem...I'm not sure how this interval came up, I must have solved incorrectly or solved the wrong equation.
 

1. What is the maximum speed for a vector?

The maximum speed for a vector is the highest possible magnitude or numerical value of its velocity. This is typically represented by the vector's length or magnitude.

2. How do you calculate the maximum speed for a vector?

To calculate the maximum speed for a vector, you can use the formula: maximum speed = magnitude of the vector / time. This formula takes into account the magnitude or length of the vector and the time it takes for the vector to reach that magnitude.

3. Can a vector have a negative maximum speed?

No, a vector's maximum speed cannot be negative. The maximum speed refers to the highest possible magnitude of the vector's velocity, which is always a positive value. However, the vector's velocity can have a negative direction, indicating a change in direction or a decrease in magnitude.

4. How does the direction of a vector affect its maximum speed?

The direction of a vector does not affect its maximum speed, as the maximum speed is determined by the magnitude of the vector's velocity. However, the direction of the velocity can affect the vector's overall movement and trajectory.

5. Can the maximum speed of a vector change?

Yes, the maximum speed of a vector can change if the magnitude or direction of its velocity changes. The maximum speed is not a fixed value and can vary depending on the conditions and forces acting on the vector.

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