Question

Suppose that x=x(t) and y=y(t) are both functions of t. If y² x=4, and dy/dt=-5 when x=3 and y=1, what is dx/dt?

Answer 1

Using implicit differentiation on the equation y² x = 4 and plugging in given values, it is determined that dx/dt is 30 when x = 3 and y = 1.

Step-by-step explanation:

The question provided requires us to use implicit differentiation to find dx/dt when given the equation y² x = 4, dy/dt at a certain point, and the corresponding values of x and y. Given that y²x = 4, we can differentiate both sides of the equation with respect to t to find dx/dt. Using the product rule and chain rule, we find:

d/dt(y² x) = d/dt(4) \
2y(dy/dt)x + y²(dx/dt) = 0

Plugging in the values x = 3, y = 1, and dy/dt = -5, we get:

(2)(1)(-5)(3) + (1)²(dx/dt) = 0 \
-30 + dx/dt = 0 \
dx/dt = 30

Therefore, the rate of change of x with respect to t, or dx/dt, is 30 when x = 3 and y = 1.