A particle is moving along the curve whose equation is 8/5=(xy^3)/(1+y^2)85=xy31+y2. 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
Jun 6, 2018

- 60/7607units sec^-1

Explanation:

8/5=[xy^3]/[1+y^2]85=xy31+y2 Cross multiply, 5xy^3=8[1+y^2]5xy3=8[1+y2]........[1][1]

Differentiating....[1] [1] wrtxx using the product rule and the general power rule, implicitly. i.e, d[uv]=vdu+udvd[uv]=vdu+udv [where uu and vv are both functions of xx]

d/dx[5xy^3]=d/dx[8+8y^2]ddx[5xy3]=ddx[8+8y2] therefore, 5y^3+5x[3y^2]dy/dx=16ydy/dx5y3+5x[3y2]dydx=16ydydx, so , [5y^3+15xy^2dy/dx=16ydy/dx][5y3+15xy2dydx=16ydydx]

Solving for dy/dxdydx, ... dy/dx=[[5y^3]/[16y-15xy^2]]dydx=[5y316y15xy2]

We are told that the rate of change of xx wrt t [ time]= 66 units sec^-1sec1 ,i.e, dx/dt=6dxdt=6.

#dx/dt.dy/dx= dy/dt# thus, dy/dt = [6[5y^3]]/[16y-15xy^2]dydt=6[5y3]16y15xy2

so dy/dtdydt [rate of change of yy wrt t ] = [30y^3]/[16y-15xy^2]30y316y15xy2 and at the point [1,2][1,2], dy/dt = [30][2^3]/[16[2]-15[1][2^2]dydt=[30]2316[2]15[1][22] = -60/7607 units sec^-1unitssec1

Hope this was helpful.