Given the relationship x^3 + 2y^2 =10, with y > 0 and dy/dt = 4 units/min., find the value of dx/dt at the instant y = 1 unit.

Question

a. 8/3  units/min

 

b. 2/3  units/min

 

 

c. -1/3  units/min

 

d. -4/3 units/min

Answer 1

x³ + 2y² = 10   Since y = 1, x³ + 2 = 10

 

                                              x³ = 8    So, x = 2

 

Differentiating the given equation implicitly with respect to t, we have 

 

   3x²(dx/dt) + 4y(dy/dt) = 0

 

                         dx/dt = -4y(dy/dt)/(3x²)

 

                                 = -4(1)(4)/(3(2²)) = -16/12 = -4/3

 

 

Answer 2

OK. I'm going to find the rate of change of

 

x³ + 2y² =10

 

by taking the derivative.

 

3x²(dx/dt) + 4y(dy/dt) = 0

 

3x²(dx/dt) + 16 = 0

 

There's still a term in x there.

 

dx/dt = -16/3x²

 

That's as close as you can get to a solution unless x is known.