Asked by Jo
A particle is moving along the curve whose equation is (xy^3)/(1+y^2)= 8/5. Assume the x-coordinate is increasing at the rate of 6 units/second when the particle is at the point (1,2). At what rate is the y-coordinate of the point changing at that instant? Is it rising or falling?
Answers
Answered by
Steve
Use implicit differentiation:
xy^3 / (1+y^2) = 8/5
(y^3 + 3xy^2 y')(1+y^2) - xy^3 (2yy') = 0
It's all over (1+y^2)^2, but that can be ignored, since it's never 0.
y'(3xy^2 + 3xy^4 - 2xy^4) = -y^3(1 + y^2)
y' = -y^3 (1+y^2)/(3xy^2 + xy^4)
= -y/x * (1+y^2)/(3 + y^2)
Take it from there.
xy^3 / (1+y^2) = 8/5
(y^3 + 3xy^2 y')(1+y^2) - xy^3 (2yy') = 0
It's all over (1+y^2)^2, but that can be ignored, since it's never 0.
y'(3xy^2 + 3xy^4 - 2xy^4) = -y^3(1 + y^2)
y' = -y^3 (1+y^2)/(3xy^2 + xy^4)
= -y/x * (1+y^2)/(3 + y^2)
Take it from there.
Answered by
v
-10/7 units per second
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