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# Parametric equations Watch

1. Q: the parametric equation of a curve are

x=(t+2)/(t+1)

y=2/(t+3)

Show that dy/dx >0

here is what I've done so far

rearanged y to get t=...

t=(2/y)-3

then Sub into equation x to get

x= (2/y)-1 / (2/y) -2

Then flipped one of the equation and timed them out to get

x = (4/y^2)-6/y +2

then differentiate to get dx/dy then flip the equation to get dy/dx

dx/dy = (6/y^2) - (8/y^3)

there for
dy/dx = (y^2/6)-(y^3/6)

the problem is I dunno how to prove that
dy/dx > 0
q: The parametric equation of a curve are

x=(t+2)/(t+1)

y=2/(t+3)

show that dy/dx >0

here is what i've done so far

rearanged y to get t=...

T=(2/y)-3

then sub into equation x to get

x= (2/y)-1 / (2/y) -2

then flipped one of the equation and timed them out to get

x = (4/y^2)-6/y +2

then differentiate to get dx/dy then flip the equation to get dy/dx

dx/dy = (6/y^2) - (8/y^3)

there for
dy/dx = (y^2/6)-(y^3/6)

the problem is i dunno how to prove that
dy/dx > 0

you need to think about parametric differentiation
Q: the parametric equation of a curve are

x=(t+2)/(t+1)

y=2/(t+3)

Show that dy/dx >0
As TeeEm says, you need parametric differentiation. Do you know how to write dy/dx in terms of dy/dt and dx/dt?
4. Think so. I just have to defferenciate them both to get dx/dt and dy/dt then do I divide the answers to get me dy/dx ?
Think so. I just have to defferenciate them both to get dx/dt and dy/dt then do I divide the answers to get me dy/dx ?
differentiate

And yes - dy/dx = (dy/dt) / (dx/dt)
6. Ty

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Updated: January 31, 2015
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