a particle is moving along the cure y=sqrt x. as the particle passes through the point (4,2), its x-coordinate increase at a rate of 3 cm/s. how fast is the distancefrom the particle to the origin changing at this instant?

distance to origin=sqrt(x+ x^2)


d distance/dt= 1/(2sqrt(x+x^2)*(dx/dt+ 2xdx/dt)

you have x, dx/dt solve for ddistance/dt

To find how fast the distance from the particle to the origin is changing at a specific instant, we need to use the concept of related rates.

Let's denote the distance from the particle to the origin as "d" and the x-coordinate of the particle as "x." We know that the particle is moving along the curve y = √x, and we want to find how fast d is changing with respect to time (dt) when the particle is at the point (4,2) and dx/dt = 3 cm/s.

Since we have the equation y = √x, we can substitute the coordinates of the particle when it is at the point (4,2) into this equation to find the corresponding value of x:

2 = √4
2 = 2

Therefore, when the particle is at (4,2), the x-coordinate is x = 4.

To find the relationship between x and d, we can use the Pythagorean theorem:

d² = x² + y²

Substituting the value of x as 4:

d² = 4² + y²
d² = 16 + y²

Since the curve is y = √x, we substitute this into the equation above:

d² = 16 + (√x)²
d² = 16 + x

Differentiating both sides of the equation with respect to time (dt):

2ddt = 0 + dxdt

Since dx/dt = 3 cm/s, we have:

2ddt = 3

Solving for ddt, we find:

ddt = 3/2 cm/s

Therefore, the distance from the particle to the origin is changing at a rate of 3/2 cm/s when the particle is at the point (4,2).