- #26

- 365

- 1

[tex]u_x=\frac{d\,x(t)}{d\,t},\,u_y=\frac{d\,y(t)}{d\,t}[/tex] and [tex]u=C[/tex], i.e.

[tex]\frac{d\,y(t)}{d\,t}=2\,x(t)\,\frac{d\,x(t)}{d\,t}\Rightarrow u_y=2\,x\,u_x[/tex]

dropping the [itex]t[/itex] dependence.

Similary taking the derivative once more, in order to find the accerelation

[tex]\frac{d\,u_y(t)}{d\,t}=2\frac{d\,x(t)}{d\,t}\,u_x+2\,x\,\frac{d\,u_x(t)}{d\,t}\Rightarrow a_y=2\,u_x^2+2\,x\,a_x[/tex]

Thus the particle's velocity would be

[tex]u^2=u_x^2+u_y^2 \Rightarrow u^2=u_x^2\,(4\,x^2+1) \Rightarrow u_x=\frac{C}{\sqrt(4\,x^2+1)}[/tex]

and the accerelation

[tex]a^2=a_x^2+a_y^2\Rightarrow a^2=\dots[/tex]

From the last equation you will find [itex]a[/itex] with respect to [itex]x[/itex] thus you are ready to find the maximum value.