A particle is moving along the curve y= 3 \sqrt{3 x + 4}. As the particle passes through the point (4, 12), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

is the curve

y = 3/√(3x + 4) ??

if so, then the particle does not pass through your given point (4,12)

after you establish where your error is,
the method to solve the problem would be:

differentiate your equation with respect to t

your differential equation contains a dy/dt and a dx/dt term.
sub in dx/dt = 4 when x = ? and y = ? from the correct given point.

now the distance from the origin of a general point on the curve is

d^2 = x^2 + y^2
2d(dd/dt) = 2x(dx/dt) + 2y(dy/dt)

sub in all the stuff from above

To find the rate of change of the distance from the particle to the origin, we need to consider the position vector of the particle.

The position vector of the particle at any point P(x, y) is given by r = xi + yj, where i and j are the unit vectors along the x and y-axes respectively.

In this case, the position vector for the particle is r = xi + (3√(3x+4))j.

To find the rate of change of the distance from the particle to the origin, we can use the chain rule from calculus to find the derivative of the distance function.

Let d = √(x^2 + y^2) be the distance from the particle to the origin.

Taking the derivative of d with respect to time (t) gives:
dd/dt = (dd/dx)(dx/dt) + (dd/dy)(dy/dt),

where dx/dt is the rate at which x is changing and dy/dt is the rate at which y is changing.

Since dx/dt = 4 units per second, we can substitute this value into the equation above.

Let's find the derivatives dd/dx and dd/dy first.

dd/dx = (1/2)((x^2 + y^2)^(-1/2))(2x)
= x/(√(x^2 + y^2))

dd/dy = (1/2)((x^2 + y^2)^(-1/2))(2y)
= y/(√(x^2 + y^2))

Now we can substitute dd/dx = x/(√(x^2 + y^2)) and dd/dy = y/(√(x^2 + y^2)), along with dx/dt = 4, into the equation:

dd/dt = (x/(√(x^2 + y^2)))(4) + (y/(√(x^2 + y^2)))(dy/dt).

We can use the given point (4, 12) to find the values of x and y at that point.

Substituting x = 4 and y = 12 into the equation, we get:

dd/dt = (4/(√(4^2 + 12^2)))(4) + (12/(√(4^2 + 12^2)))(dy/dt).

Simplifying this expression, we have:

dd/dt = (4/(√(16 + 144)))(4) + (12/(√(16 + 144)))(dy/dt).

dd/dt = (16/(√160))(4) + (48/(√160))(dy/dt).

Now we need to find dy/dt, the rate at which y is changing.

Since y = 3√(3x+4), we can take the derivative of y with respect to x and multiply it by dx/dt to find dy/dt.

dy/dt = (dy/dx)(dx/dt).

First, let's find dy/dx.

dy/dx = (d/dx)(3√(3x+4))
= (3/2)(1/√(3x+4))(d/dx)(3x+4)
= (3/2)(1/√(3x+4))(3)
= (9/2√(3x+4)).

Substituting dx/dt = 4 into the equation, we have:

dy/dt = (9/2√(3x+4))(4).

Now we can substitute this value for dy/dt into the expression for dd/dt:

dd/dt = (16/(√160))(4) + (48/(√160))((9/2√(3x+4))(4)).

Since we are interested in finding the rate of change of the distance from the particle to the origin when x = 4 and y = 12, we can substitute these values into the equation:

dd/dt = (16/(√160))(4) + (48/(√160))((9/2√(3(4)+4))(4)).

dd/dt = (16/(√160))(4) + (48/(√160))((9/2√(12+4))(4)).

Simplifying this expression will give us the rate of change of the distance from the particle to the origin at the given instant.

To find the rate of change of the distance from the particle to the origin, we need to use the distance formula between two points in the coordinate plane.

The distance formula between two points (x₁, y₁) and (x₂, y₂) is given by:

√((x₂ - x₁)² + (y₂ - y₁)²)

In this case, one of the points is the origin (0, 0), and the other point is the particle's position given by the curve y = 3√(3x + 4). Let's denote the particle's x-coordinate as x and its y-coordinate as y.

So, the distance formula becomes:

√((x - 0)² + (y - 0)²)

Since y = 3√(3x + 4), we can rewrite it as:

√(x² + (3√(3x + 4))²)

To find the rate of change of the distance from the particle to the origin, we can take the derivative with respect to time (t) on both sides of the equation, and then substitute the values given in the question.

Let's differentiate both sides of the equation with respect to t:

d/dt [√(x² + (3√(3x + 4))²)] = d/dt [√(x² + 27(3x + 4))]

The left side of the equation gives us the rate of change of the distance from the particle to the origin, while the right side represents the rate at which the particle's x-coordinate is changing, which is given as 4 units per second.

Therefore, we have:

d/dt [√(x² + 27(3x + 4))] = 4

To simplify the equation, we need to apply the chain rule. Let's denote √(x² + 27(3x + 4)) as u:

d/dt [u] = 4

Using the chain rule, we have:

d/dt [u] = du/dx * dx/dt

To find du/dx, we differentiate u with respect to x:

du/dx = 1/2 * (x² + 27(3x + 4))^(-1/2) * (2x + 27(3))

Now, substituting dx/dt = 4 and du/dx into the original equation, we get:

1/2 * (x² + 27(3x + 4))^(-1/2) * (2x + 81) * 4 = 4

Simplifying further:

(x² + 27(3x + 4))^(-1/2) * (2x + 81) = 1

By substituting the x-coordinate of the particle at the given point (4, 12) into the equation, you can find the rate of change of the distance from the particle to the origin at that instant.