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

Results 1 to 6 of 6

- September 13th 2009, 12:47 AM #1

- Joined
- May 2009
- Posts
- 15

## 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..

- September 13th 2009, 03:01 AM #2

- Joined
- Apr 2009
- Posts
- 53

- September 13th 2009, 03:17 AM #3

- Joined
- Apr 2005
- Posts
- 17,908
- Thanks
- 2328

In other words, you want to find dy/dt. Differentiating both sides of the given equation and using the chain rule, , so

.

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?

Could you explain me with answers plzz

Thank you..

- September 13th 2009, 03:20 AM #4

- Joined
- Sep 2009
- Posts
- 24

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

The particle is rising as the dy/dt at that point is +ve.**dy/dt = 20**

- September 13th 2009, 03:41 AM #5

- Joined
- Apr 2009
- Posts
- 53

- September 13th 2009, 07:39 AM #6

- Joined
- Sep 2009
- Posts
- 24