A bug walks along the circle x^2 + y^2 = 4. When it is at the point (1, (3)^1/2)), dx/dt = 2. Find dy/dt at this instant.
2x dx/dt +2y dy/dt=0
you have x, y, and dx/dt. Solve for dy/dt
To find dy/dt, we can differentiate both sides of the equation x^2 + y^2 = 4 with respect to time t and then solve for dy/dt.
Differentiating x^2 + y^2 = 4 implicitly with respect to t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we know dx/dt = 2 and the point (1, (3)^1/2) lies on the circle, we can substitute these values into the equation:
2(1)(2) + 2((3)^1/2)(dy/dt) = 0
4 + 2√3(dy/dt) = 0
2√3(dy/dt) = -4
dy/dt = -4 / (2√3)
Simplifying the expression, we have:
dy/dt = -2 / √3
Therefore, dy/dt at the instant when the bug is at the point (1, (3)^1/2) is equal to -2 / √3.