In the problem, x and y are differentiable functions of t. Find dx/dt when x = 9, y = 2, and dy/dt= 2.
x^3y^2 = 2916
f(x,y) = x^3y^2
df = (df/dx) dx + (df/dy) dy
df/dt=(df/dx) dx/dt + (df/dy) dy/dt
but f is a constant = 2916
df/dx = 3x^2y^2 = 3*81*4= 972
df/dy = 2x^3y = 2*729*2= 2916
so 0 = 972 (dx/dt) + 2916 (2)
dx/dt = -6
To find dx/dt when x = 9, y = 2, and dy/dt = 2, we can use implicit differentiation. This means we will differentiate both sides of the equation with respect to t and use the chain rule.
First, let's take the derivative of both sides of the equation using the product and power rules of differentiation.
d/dt(x^3y^2) = d/dt(2916)
Now, applying the product rule on the left side, we get:
d/dt(x^3) * y^2 + x^3 * d/dt(y^2) = 0
Using the power rule, we can simplify this to:
3x^2 * dx/dt * y^2 + x^3 * 2y * dy/dt = 0
Now, we can substitute the given values: x = 9, y = 2, and dy/dt = 2, into the equation:
3(9)^2 * dx/dt * (2)^2 + (9)^3 * 2(2) = 0
Simplifying further:
3(81) * dx/dt * 4 + 729 * 4 = 0
243 dx/dt + 2916 = 0
Subtracting 2916 from both sides:
243 dx/dt = -2916
Dividing both sides by 243:
dx/dt = -2916/243
Simplifying this expression gives:
dx/dt = -12
Therefore, when x = 9, y = 2, and dy/dt = 2, dx/dt equals -12.