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
dy/dt = -x dx/dt/y
= -1(2)/√3
= -2/√3
To find dy/dt, we can use the information given about dx/dt and the relationship between x and y on the circle x^2 + y^2 = 4.
First, let's find dy/dx using implicit differentiation.
Taking the derivative of both sides of the equation x^2 + y^2 = 4 with respect to x, we get:
2x + 2y(dy/dx) = 0
Next, solve for dy/dx:
dy/dx = -(2x) / (2y)
= -x / y
Now, we can substitute the given values dx/dt = 2 and (1, √3) into the derivative dy/dx to find dy/dt at this instant.
dy/dx = -x / y
= -(1) / (√3)
= -1 / (√3)
Since dx/dt = 2, and dy/dx = dy/dt / dx/dt, we can rearrange the equation to solve for dy/dt:
dy/dt = (dy/dx) * (dx/dt)
= (-1 / (√3)) * 2
= -2 / (√3)
Therefore, at the given instant, dy/dt is equal to -2 / (√3).
To find dy/dt at a specific point on the given circle, we can use implicit differentiation. Let's differentiate the equation x^2 + y^2 = 4 with respect to t (time) using the chain rule.
Differentiating both sides of the equation with respect to t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we are interested in finding dy/dt when the bug is at the point (1, √3), we can substitute these values into the equation.
At the point (1, √3), x = 1 and y = √3. We are given dx/dt = 2. Substituting these values, we get:
2(1)(2) + 2(√3)(dy/dt) = 0
4 + 2√3(dy/dt) = 0
Now, let's solve for dy/dt:
2√3(dy/dt) = -4
dy/dt = -4 / (2√3)
To simplify the expression, we can rationalize the denominator by multiplying the numerator and denominator by √3:
dy/dt = (-4 / (2√3)) * (√3 / √3)
dy/dt = -4√3 / (2 * 3)
dy/dt = -2√3 / 3
Therefore, at the instant when the bug is at the point (1, √3), dy/dt is equal to -2√3 / 3.