A particle moves along the parabola y equals x squared in the first quadrant in such a way that its x-coordinate (measured in meters) increases at a steady 10 StartFraction m Over sec EndFraction . How fast is the angle of inclination theta of the line joining the particle to the origin changing when x equals 1 m question mark

(dθ/dt) = 1 rad / s = 57.3° /s

Step-by-step explanation:

- A particle moves along curve with function:

                                  y = x^2

- The rate of change of x-coordinate is given by dx/dt = 10 m/s

Find:

How fast is the angle of inclination theta of the line joining the particle to the origin changing when x equals 1 m

Solution:

- The gradient of the line from origin to particle at position ( x , y ) is given by:

                                tan ( θ ) = y / x

Where, θ is the angle between x-axis and line from origin

            x & y are coordinate of the point on given graph.

- To develop a rate of change expression we will derivate the above expression by time t:

                              d / dt (tan ( θ )) = d/dt (y / x )

                              (dθ/dt) / cos^2(θ) = (dy/dt) / (dx/dt)

                              (dθ/dt) = cos^2(θ) * (dy/dt) / (dx/dt)

- The rate of change of angle (dθ/dt) is given by above expression.  

- We will apply the following chain rule to evaluate (dy/dt):

                              (dy/dt) = (dy/dx) * (dx/dt)

                              (dy/dt) = 2x * (10)

                              (dy/dt) = 20*x

@ x = 1,                   (dy/dt) = 20 m/s

@ x = 1,                    y = (1)^2 = 1

                              tan (θ) = 1

                              θ = 45°

- Now use the derived rate of change of angle expression we get:

                            (dθ/dt) = cos^2(45) * 20 / 10

                            (dθ/dt) = 0.5 * 20 / 10

                            (dθ/dt) = 1 rad / s = 57.3° /s