# a particle is moving along a curve whose equation is:   `(xy^3)/(1+y^2)=8/5 ` assume the x-coordinate is increasing at a rate of 6 units/sec when the particle is at the point (1,2) a) at what rate...

a particle is moving along a curve whose equation is:

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

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

a) at what rate is the y-coordinate of the point changing at that instant?

b) is the particle rising or falling at that instant?

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

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

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

Differentiate both sides with respect to time t.

`5(x*3y^2y'+y^3x') = 8xx2yxxy' ---(1)`

It is given that at (1,2) the rate of increasing of x is 6 units per second.

`x = 1`

`y = 2`

`x' = 6`

Applying the above values in (1) will give you;

`5(x*3y^2y'+y^3x') = 8xx2yxxy'`

`5(1xx3xx2^2xxy'+2^3xx6) = 8xx2xx2xxy'`

`15y'+60 = 8y'`

`y' = -60/7`

So the y coordinate is changing at a rate of `60/7` units per second. Since we have a negative value for y' by the usual notation that means y is decreasing or falling.