A particle is moving along the curve below. y = x As the particle passes through the point (4, 2), its x-coordinate increases at a rate of 4 cm/s. How fast is the distance from the particle to the origin changing at this instant? (Round your answer to three decimal places.)

The correct question is;

A particle is moving along the curve

y=√x. As the particle passes through the point (4, 2), its x-coordinate increases at a rate of 4 cm/s. How fast is the distance from the particle to the origin changing at this instant? (Round your answer to three decimal places.)

dS/dt ≈ 4.025 cm/s

Step-by-step explanation:

Let the distance from the origin be S and thus, the formula for that is

S = √x² + y²

The given function is y =√x

Let's plug in the S function for y to get;

S = √x² + (√x)²

S = √(x² + x)

Now, let's find the derivative with respect to time;

So;

dS/dt= [(1/2)/√(x² + x)] (2x+1)dx/dt

Let's plug in the value of x = 4 and dX/dt as 4;

dS/dt= [[(1/2)/√(4² + 4)] x [((2(4)+1)]] x 4

dS/dt= [(36/4)/√5]

dS/dt = 9/√5

dS/dt = 4.025 cm/s

answer: you need the equation in the question so i can answer it

step-by-step explanation:

step-by-step explanation: