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Deathslayer7
2010-02-16, 06:57 PM
the problem is this:

given y^2=4x with a constant velocity v=4 m/s and at x=5

determine the x and y components of velocity.

determine the x and y components of acceleration.

Mando Knight
2010-02-16, 07:11 PM
y²=4x
x=5
x'²+y'²=16
Correct?

dx/dy = y/2
dx/dt = dx/dy * dy/dt
x'=y'*y/2
y'²+(y'*y/2)²=16
y=sqrt(20)

2y''y'+(y'*y/2)(y''*y+y'²)=0 -> 2y''y'+y''*y'*y²/2+y'³*y=0
2x''x'+2y''y'=0

There. Now solve.

Deathslayer7
2010-02-16, 07:35 PM
y²=4x
x=5
x'²+y'²=4
Correct?

dx/dy = y/2
dx/dt = dx/dy * dy/dt
x'=y'*y/2
y'²+y'*y/2=4
y=sqrt(20)

2y''y'+y'²/2+y''*y/2=0
2x''x'+2y''y'=0

There. Now solve.

how'd you get that line?

Astrella
2010-02-16, 07:38 PM
how'd you get that line?

Orthonormal system, so you can apply Pythagoras to the coordinate axes.

Mando Knight
2010-02-16, 07:40 PM
how'd you get that line?

Constant velocity magnitude...

Oops. Should be x'²+y'²=16. Fixing!

Deathslayer7
2010-02-16, 07:57 PM
ok then where does the dx/dy=y/2 come from?

and just to note. I need to learn how to do this for a test tomorrow. This isnt a hw problem or anything. So i need to understand it rather then just "do it."

littlebottom
2010-02-16, 08:06 PM
ok then where does the dx/dy=y/2 come from?

i believe it is the differential? its usually shown as dy over dx (dy/dx), but in this case as we know X and not Y, it is the other way around (dx/dy) you know how to differentiate?

dont hold me to that though.

Mando Knight
2010-02-16, 08:38 PM
It is indeed the derivative of x with respect to y, just like it says (dx/dy). dx/dt and dy/dt are the derivatives with respect to time. It's simple basic calculus stuff that I've forgotten not everyone has memorized.