A particle is moving along the curve y= 4 \sqrt{3 x + 1}. As the particle passes through the point (1, 8), its x-coordinate increases at a rate of 2 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Question

A particle is moving along the curve y= 4 \sqrt{3 x + 1}. As the particle passes through the point (1, 8), its x-coordinate increases at a rate of 2 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Answer 1

To find the rate of change of the distance from the particle to the origin, we need to find the rate of change of the distance function.

The distance from a point (x, y) to the origin (0, 0) can be found using the distance formula:
distance = √(x^2 + y^2)

In this case, the x-coordinate of the particle is changing at a rate of 2 units per second. We need to find the y-coordinate of the particle to determine the distance function.

Given that the particle is moving along the curve y = 4 √(3x + 1), we can differentiate this equation to find the rate of change of the y-coordinate with respect to the x-coordinate.

Differentiating both sides of the equation with respect to x, we get:
dy/dx = (d/dx)(4√(3x + 1))

To differentiate the square root, we can use the chain rule:
dy/dx = 4 * (1/2)(3x + 1)^(-1/2)(d/dx)(3x + 1)
= 2(3x + 1)^(-1/2)(d/dx)(3x + 1)
= 2 / √(3x + 1)

Now, we know that the x-coordinate of the particle is increasing at 2 units per second, so dx/dt = 2.

To find dy/dt (the rate of change of the y-coordinate), we multiply dx/dt by dy/dx:
dy/dt = (dy/dx) * (dx/dt)
= 2 / √(3x + 1) * 2
= 4 / √(3x + 1)

Now, we can find the rate of change of the distance from the particle to the origin by finding the derivative of the distance formula with respect to time:

distance = √(x^2 + y^2)

Differentiating both sides with respect to t, we get:
d(distance)/dt = (d/dt)√(x^2 + y^2)

To differentiate the square root, we can use the chain rule:
d(distance)/dt = (1/2)(x^2 + y^2)^(-1/2)(d/dt)(x^2 + y^2)

Now, we substitute the values of x, y, dx/dt, and dy/dt:
d(distance)/dt = (1/2)(x^2 + y^2)^(-1/2)(d/dx)(x^2 + y^2) * (dx/dt) + (d/dy)(x^2 + y^2) * (dy/dt)
= (1/2)(x^2 + y^2)^(-1/2)(2x(dx/dt) + 2y(dy/dt))
= (1/2)(x^2 + y^2)^(-1/2)(2x * 2 + 2y * (4 / √(3x + 1)))

Now, we substitute the values of x and y at the given point (1, 8):
d(distance)/dt = (1/2)(1^2 + 8^2)^(-1/2)(2 * 1 * 2 + 2 * 8 * (4 / √(3 * 1 + 1)))
= (1/2)(1 + 64)^(-1/2)(4 + 8 * (4 / √4))
= (1/2)(65)^(-1/2)(4 + 8 * 2)
= (1/2)(65)^(-1/2)(4 + 16)
= (1/2)(65)^(-1/2)(20)
= 10 / √65

Therefore, the rate of change of the distance from the particle to the origin at the instant when it passes through the point (1, 8) is 10 / √65 units per second.