A particle is moving along the curve y= 3sqrt3x+1. As the particle passes through the point (5,12), its x-coordinate increases at a rate of 3 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Let P(x,y) be any point on the curve

then we could write P as (x,3√(3x+1))
and if D is the distance to the origin,
D^2 = x^2 + (3√(3x+1))^2
= x^2 + 27x + 9
2D(dD/dt) = 2x(dx/dt) + 27(dx/dt)

at (5,12) D^2 = 25+135+9 = 169
so D = 13

so dD/dt = (2(5)(3) + 27(3))/(2(13))
= 4.26

check my work please

your answer is good

To find the rate of change of the distance from the particle to the origin, we need to find the equation of the distance between the particle and the origin.

The distance formula between two points, (x1, y1) and (x2, y2), is given by:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

In this case, we want to find the distance between the particle and the origin, so (x1, y1) is (0, 0), and (x2, y2) is (x, y) where y = 3√3x + 1.

Therefore, the equation of the distance between the particle and the origin is:
Distance = sqrt((x - 0)^2 + (3√3x + 1 - 0)^2)
= sqrt(x^2 + (3√3x + 1)^2)

To find the rate of change of the distance, we need to differentiate the distance equation with respect to time, t, and then substitute the given values.

Differentiating the distance equation with respect to t, we get:
d(Distance)/dt = d(sqrt(x^2 + (3√3x + 1)^2))/dt

Using the chain rule and differentiating each term, we have:
d(Distance)/dt = (1/2)*(x^2 + (3√3x + 1)^2)^(-1/2)*(2x*dx/dt) + (1/2)*(x^2 + (3√3x + 1)^2)^(-1/2)*2(3√3x + 1)*(9√3*dx/dt)

Simplifying, we have:
d(Distance)/dt = (x*dx/dt + (3√3x + 1)*(9√3*dx/dt))/(sqrt(x^2 + (3√3x + 1)^2))

Substituting the given values, x = 5 and dx/dt = 3, we get:
d(Distance)/dt = (5*3 + (3√3*5 + 1)*(9√3*3))/(sqrt(5^2 + (3√3*5 + 1)^2))
= (15 + (45√3 + 1)*27√3)/(sqrt(25 + (45√3 + 1)^2))

Calculating the values, we have:
d(Distance)/dt ≈ 56.57 units per second

Therefore, the rate of change of the distance from the particle to the origin at this instant is approximately 56.57 units per second.

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

The distance from the particle to the origin can be found using the distance formula, which is D = √(x^2 + y^2). In this case, we can substitute the given equation of the curve, y = 3√3x + 1, to find y in terms of x.

Since the particle passes through the point (5, 12), we can substitute x = 5 and y = 12 into the equation to get the specific values for D.

D = √(5^2 + (3√3(5) + 1)^2)
= √(25 + (15√3 + 1)^2)
= √(25 + 225 + 30√3 + 3 + 2√3 + 9)
= √(262 + 32√3)

Now, to find the rate of change of the distance to the origin, we need the derivative of D with respect to time. Let's call this rate of change dD/dt.

dD/dt = (dD/dx) * (dx/dt)

We can find dD/dx by differentiating the expression for D with respect to x:

dD/dx = (1/2) * (262 + 32√3)^(-1/2) * (0 + 16√3)
= 8√3 / √(262 + 32√3)

Given that the x-coordinate of the particle is increasing at a rate of 3 units per second, dx/dt = 3.

Therefore, the rate of change of the distance from the particle to the origin at this instant is:

dD/dt = (8√3 / √(262 + 32√3)) * 3
= 24√3 / √(262 + 32√3)

So, the rate of change of the distance from the particle to the origin at this instant is 24√3 / √(262 + 32√3) units per second.