find the rate of change of the distance between the origin and a moving point on the graph of y=x^2+2 if ds/dt=5 centimeters per second.
To find the rate of change of the distance between the origin and a moving point on the graph of y=x^2+2, we start by considering the equation of the curve, which is y = x^2 + 2.
The distance between two points in a two-dimensional plane can be calculated using the Pythagorean theorem. In this case, the two points are the origin (0, 0) and the moving point (x, y).
The distance d between these two points is given by:
d = sqrt(x^2 + y^2)
To find the rate of change of the distance d, we need to differentiate both sides of the equation with respect to time t:
d/dt = d(sqrt(x^2 + y^2))/dt
Now, let's differentiate √(x^2 + y^2) with respect to t using the chain rule:
d/dt (√(x^2 + y^2)) = (∂(√(x^2 + y^2))/∂x)*(dx/dt) + (∂(√(x^2 + y^2))/∂y)*(dy/dt)
We are given that ds/dt = 5 centimeters per second, which means that dx/dt = 5 because the point is moving horizontally along the x-axis.
To find (∂(√(x^2 + y^2))/∂x), we differentiate √(x^2 + y^2) with respect to x:
∂(√(x^2 + y^2))/∂x = (1/2)*(x^2 + y^2)^(-1/2)*(2x) = x/(√(x^2 + y^2))
Since the graph is y = x^2 + 2, we can substitute y with x^2 + 2:
∂(√(x^2 + y^2))/∂x = x/(√(x^2 + (x^2 + 2)^2))
Now we can plug in the values into our equation:
d/dt (√(x^2 + y^2)) = (x/(√(x^2 + (x^2 + 2)^2)))*(dx/dt) + (∂(√(x^2 + y^2))/∂y)*(dy/dt)
d/dt (√(x^2 + y^2)) = (x/(√(x^2 + (x^2 + 2)^2)))*(5) + (∂(√(x^2 + y^2))/∂y)*(dy/dt)
Now all that's left is to find (∂(√(x^2 + y^2))/∂y)*(dy/dt). However, we have not been provided with values for y or dy/dt in the problem statement. Without these values, it is not possible to fully calculate the rate of change of the distance.