So I have the equations x'(t) = 4*y(t) and y'(t)=-x(t). Going through all the steps, you get(adsbygoogle = window.adsbygoogle || []).push({});

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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Finding the maximum speed for a vector?

**Physics Forums | Science Articles, Homework Help, Discussion**