A particle is moving along the graph of x^-(1/3). When x=8, the component of its position is increasing at the rate of 1 cm/sec. How fast is the distance from the origin changing at this moment? How fast is the angle of inclination changing at this moment?
To find how fast the distance from the origin is changing, we can use the Pythagorean theorem:
r^2 = x^2 + y^2
where r is the distance from the origin, x is the x-coordinate, and y is the y-coordinate of the particle.
Differentiating both sides of the equation with respect to time (t), we get:
2r * dr/dt = 2x * dx/dt + 2y * dy/dt
Since our particle is moving along the curve x^-(1/3), we know that x = 8, and we are given dx/dt = 1 cm/sec. Let's find y in terms of x:
y = f(x) = x^-(1/3)
Differentiating y with respect to x using the chain rule:
dy/dx = -(1/3) * x^(-(1/3) + 1)
dy/dx = -(1/3) * x^-(4/3)
Substituting x = 8, we can find dy/dx:
dy/dx = -(1/3) * 8^-(4/3)
Now, we can find the value of dy/dt by multiplying dy/dx by dx/dt:
dy/dt = dy/dx * dx/dt
dy/dt = -(1/3) * 8^-(4/3) * 1 cm/sec
To find how fast the angle of inclination is changing, we can use the arctangent function:
θ = arctan(y / x)
Differentiating both sides of the equation with respect to time (t), we get:
dθ/dt = (1 / (1 + (y/x)^2)) * (dy/dt * x - y * dx/dt)
Substituting the values of x = 8, y = 8^-(1/3), and the values of dy/dt and dx/dt we found earlier, we can solve for dθ/dt.