# Thread: can anyone help me with this problem !

1. ## can anyone help me with this problem !

Question is:

A particle is moving along the curve whose equation is:

xy^3 = 8(1+y^2)/5

and the x-coordinate is increasing at a rate of 6 units/s when the particle is at the point (1, 2).

1. At what rate is the y-coordinate changing at that instant

2. Is the particle rising or falling at that instant?

Could you explain me with answers plzz

Thank you..

2. Did you try to cross-multiply from the initial equation? Rather, 5(xy^3) = 8(1+y^2) ?

3. Originally Posted by mookie
Question is:

A particle is moving along the curve whose equation is:

xy^3 = 8(1+y^2)/5

and the x-coordinate is increasing at a rate of 6 units/s when the particle is at the point (1, 2).

1. At what rate is the y-coordinate changing at that instant
In other words, you want to find dy/dt. Differentiating both sides of the given equation and using the chain rule, $d(xy^3)/dt= (8/5)d(1+y^2)/dt$, so

$y^3 dx/dt+ 3xy^2 dy/dt= (8/5)(2y)dy/dt$.

Now, you are told "x-coordinate is increasing at a rate of 6 units/s when the particle is at the point (1, 2)." so x= 1, y= 2 and dx/dt= 6. Put those into the equation and solve for dy/dt.

2. Is the particle rising or falling at that instant?
Well, you know, I hope, that the particle is "rising" if dy/dt> 0 and "falling" if dy/dt< 0.

Could you explain me with answers plzz

Thank you..

4. differentiate both side with respect to 't'
=> x.(y^2).dy/dt + (y^3).dx/dt = 8/5.2y.dy/dt

Use the given value dx/dt(1,2) = 6 in the above equation

1.4.dy/dt + 8.6 = (32/5)dy/dt
dy/dt(32/5-4) = 48
dy/dt(12/5) = 48
dy/dt = 20

The particle is rising as the dy/dt at that point is +ve.

5. Does dy/dt definitely equal 20?

6. I just missed one coefficient.. thanks for pointing that out...

Please recalculate by replacing
x.(y^2).dy/dt
with
3.x.(y^2).dy/dt

And if dy/dt > 0 ... it is rising ...
if dy/dt < 0 ... it is decending...