blankman013
New member
- Joined
- Jan 24, 2008
- Messages
- 2
I'm not sure if this is in the right section or not, but i'm out of ideas and i figure it couldn't hurt.
General equation of a parabola is x(y)=ay^2+by+c, i need x(0)=10 and x(5)=0.
So C must equal 10, and b=-2-5a. So the equation is now x(y)=ay^2-(2+5a)y+10.
Now, from here i must somehow derive the differential equation
(d^2y/dt^2)+(g/(1+(2ay-(2+5a))^2))=0 with y(0)=5 and y'(0)=0.
g=10m/s^2
I know the denominator is the same as 1+(x'^2), but apparently i have no idea how to derive differential equations. So thank you for any help!! I have more information on the problem but i'm not sure what else would be needed for the derivation.
General equation of a parabola is x(y)=ay^2+by+c, i need x(0)=10 and x(5)=0.
So C must equal 10, and b=-2-5a. So the equation is now x(y)=ay^2-(2+5a)y+10.
Now, from here i must somehow derive the differential equation
(d^2y/dt^2)+(g/(1+(2ay-(2+5a))^2))=0 with y(0)=5 and y'(0)=0.
g=10m/s^2
I know the denominator is the same as 1+(x'^2), but apparently i have no idea how to derive differential equations. So thank you for any help!! I have more information on the problem but i'm not sure what else would be needed for the derivation.