Experts Please help. Thank you.

Question

A particle is moving along the curve whose equation is xy^3/1+y^2 = 8/5.

Assume that the x-coordinate is increasing at the rate of 6 units/s when the particle is at the point (1,2). At what rate is the y-coordinate of the point changing at that instant?

Enter the exact answer.

a. dy/dt = ?

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

Answer 1

x y³ = 8/5 + 8y²/5

d/ d t [ x y³ ] = d/d t[ 8/5 + 8y²/5 ]

y³ d x/ d t + 3x y² d y/ d t = (16y / 5) d y/d t

y³ d x/ d t = (16y / 5) d y/d t - 3x y² d y/ d t

[ (16y / 5) - 3x y² ] d y/ d t = y³ d x/ d t and since y ≠ 0

d y/ d t = { y² / [ (16 / 5) - 3x y ] } d x/ d t

d y/ d t = {5 y² / [ (16 - 15 x y ] } d x/ d t

d y/ d t = {5 ·4 / [ (16 - 15 ·1· 2 ] } ·6

d y/ d t = -60/7 units / sec and therefore the particle is falling at this instant