# Finding the maximum speed for a vector?

1. Apr 25, 2012

### mathgirl313

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

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

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

Rearranging the equation we are constrained by we get: f(x,y) = g(x) = $\sqrt{4*(\alpha)^2 + 16*(\beta)^2 - 3*x^2}$, where x is in (-$\sqrt{(\alpha)^2 + 4*(\beta)^2}$, $\sqrt{(\alpha)^2 + 4*(\beta)^2}$. 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!

2. Apr 25, 2012

### HallsofIvy

Staff Emeritus
"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 $\alpha$ and $\beta$.

Perhaps because it's not true. What is true is that since $x(0)= \alpha$ and $y(0)= \beta$ direct substitution gives $x(0)^2/4+ y(0)^2= \alpha^2/4+ \beta^2$. It is not true for general "t".

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.

3. Apr 26, 2012

### mathgirl313

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.