Question

Show steps please

let x and y be the functions of time t that are related by the equation 2x^2 + 3y^3 - 4 x y = 36. At the same time t =1, the value of x=2, the value of y = -2 and the value of dy/dt is 4. Find the value of dx/dt at time t=1

Answer 1

At time t=1, the value of dx/dt for the equation 2x^2 + 3y^3 - 4xy = 36 with x=2, y=-2, and dy/dt=4 is -7.

Step-by-step explanation:

To find the value of dx/dt at time t=1 for the functions x(t) and y(t) that satisfy the equation 2x^2 + 3y^3 - 4xy = 36, we need to differentiate this equation with respect to t. Given that at t=1, x=2, y=-2, and dy/dt=4, we can plug these values into the differentiated equation to solve for dx/dt.

The differentiation with respect to t is:

• 4x(dx/dt) + 9y^2(dy/dt) - 4(dx/dt)y - 4x(dy/dt) = 0

Plugging the values in:

• 4(2)(dx/dt) + 9(-2)^2(4) - 4(dx/dt)(-2) - 4(2)(4) = 0
• 8(dx/dt) + 144 + 8(dx/dt) - 32 = 0
• 16(dx/dt) + 112 = 0
• dx/dt = -112/16
• dx/dt = -7

Thus, the value of dx/dt at time t=1 is -7.