a particle is moving along the curve whose equation is" y=x^2? At what point on the curve are the x and y coordinates changing at the opposite rate?
If they are changing at opposite rates, the slope is -1
dy/dx = 2x
so when is that -1?
To find the point on the curve where the x and y coordinates are changing at the opposite rate, we need to determine the derivative of both x and y with respect to time (t) and compare their values.
Given the equation of the curve y = x^2, we can differentiate both sides of the equation to find the rates of change of x and y with respect to x.
So, dy/dx = 2x.
Now, to find the rates of change of x and y with respect to time (dt), we need to consider the chain rule. We have the equation x = x(t), which relates how the x-coordinate changes with respect to time.
Differentiating both sides of the equation x = x(t) with respect to t, we get dx/dt = d(x(t))/dt.
By substituting the equation dx/dt = d(x(t))/dt, we have dx/dt = d(x(t))/dt = d(x^2)/dt = 2x*dx/dt.
Now, we can compare the rates of change of x and y with respect to time by setting the equations dy/dt = 2x*dx/dt and solving for x.
Since we know dy/dx is equal to 2x, and we want dy/dt to have the opposite rate of change compared to dx/dt, we can equate dy/dx = -dx/dt.
Substituting the expressions 2x for dy/dx and dx/dt, we have 2x = -dx/dt.
Now, we can solve this equation to find the x-coordinate at which the rates of change of x and y are opposite.
2x = -dx/dt
2x + dx/dt = 0
Integrating both sides,
∫(2x + dx/dt) dt = ∫0 dt
2∫x dt + ∫dx/dt dt = C
Simplifying,
2xt + x = C
So, the equation 2xt + x = C represents the relationship between the x-coordinate and time (t) at the point where the x and y coordinates are changing at the opposite rate.
To find the specific point on the curve, we need more information about the particle's location at a particular time or any other constraints.