A particle is moving along the curve whose equation is #8/5=(xy^3)/(1+y^2)#. Assume that the x-coordinate is increasing at the rate of 6 units/sec when the particle is at the point (1,2). At what rate is y-coordinate of the point changing at that instant?

1 Answer
Barry H.
Jun 6, 2018

# - 60/7#units sec^-1

Explanation:

#8/5=[xy^3]/[1+y^2]# Cross multiply, #5xy^3=8[1+y^2]#........#[1]#

Differentiating....#[1] # wrt#x# using the product rule and the general power rule, implicitly. i.e, #d[uv]=vdu+udv# [where #u# and #v# are both functions of #x#]

#d/dx[5xy^3]=d/dx[8+8y^2]# therefore, #5y^3+5x[3y^2]dy/dx=16ydy/dx#, so , #[5y^3+15xy^2dy/dx=16ydy/dx]#

Solving for #dy/dx#, ... #dy/dx=[[5y^3]/[16y-15xy^2]]#

We are told that the rate of change of # x# wrt t [ time]= #6# units # sec^-1# ,i.e, #dx/dt=6#.

#dx/dt.dy/dx= dy/dt# thus, #dy/dt = [6[5y^3]]/[16y-15xy^2]#

so #dy/dt# [rate of change of #y# wrt t ] = #[30y^3]/[16y-15xy^2]# and at the point# [1,2]#, #dy/dt = [30][2^3]/[16[2]-15[1][2^2]# = -#60/7# # units sec^-1#

Hope this was helpful.

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