Final answer:
The rate of change of x with respect to t (dx/dt) given the conditions x²y² = 100 and dy/dt = 4 when y = 6 is found to be dx/dt = -10/9 by using implicit differentiation.
Step-by-step explanation:
The question involves finding the rate of change of x with respect to t (dx/dt) given that x²y² = 100 and dy/dt = 4 when y = 6.
To solve this, we can differentiate the equation x²y² = 100 with respect to t implicitly. Here are the steps:
- Differentiate both sides of the equation with respect to t: 2xy(dx/dt) + 2x²(dy/dt) = 0 (using the product rule for differentiation).
- Plug in the values y = 6 and dy/dt = 4: 2x(6)(dx/dt) + 2x²(4) = 0.
- Solve for dx/dt: 12x(dx/dt) + 8x² = 0, so dx/dt = -8x² / 12x = -2x/3.
- Find the value of x when y = 6 using the original equation: x²(6)² = 100, which gives x = ±√(100/36), or x = ±√(25/9), so x = ±5/3.
- Choose the positive value of x since rate of change is typically associated with positive values, thus x = 5/3.
- Now calculate dx/dt: dx/dt = -2(5/3)/3 = -10/9.
Therefore, the rate of change of x with respect to t when y = 6 is dx/dt = -10/9.