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

- 60/7units sec^-1

Explanation:

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

Differentiating....[1] wrtx 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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