Question

If xy^2 = 20, and x is decreasing at the rate of 3 units per second, the rate at which y is changing when y = 2 is nearest to?

Answer 1

xy^2 = 20 => y^2 = 20/x

[y^2]' = [20/x]'

2y [dy/dt] = [-20/x^2] [dx/dt]

dy/dt = [-10/yx^2] [dx/dt]


[dx/dt] = - 3 u/s ....(the negative sign was included to account for decreasing)

y = 2 => x(2)^2 = 20 => x = 20/4 = 5

dy/dt = [-10/(2*5^2)] [- 3]

dy/dt = [-1/5][- 3] = 3/5

Answer: dy/dt = 3/5 units/s