calculus A particle is moving along the curve y= sqrt(x) As the particle passes through the point (16, 4), i

Question

A particle is moving along the curve y= sqrt(x) As the particle passes through the point (16, 4), its x-coordinate increases at a rate of 3 cm/s. How fast is the distance from the particle to the origin changing at this instant?

Round your answer to the nearest hundredth.

Answer 1

dy = dx/(2sqrt(x)), dx = 3 => dy = 3/8 => v = sqrt (dx^2 + dy^2) = 3sqrt(65)/8 = 3.0233...

Answer 2

D^2= x^2+y^2 so D= sqrt(x^2+y^2) =sqrt(x^2+x)
Using chain rule
dD/dt=dD/dx*dx/dt = 1/2sqrt(x^2+x) *( 2x+1)*3 =3.00cm/s Rounded to the nearest hundred

Answer 3

We have dx/dt=3 at x=16 ,y=4
The distance from origin
S=sqrt(x^2+y^2)=sqrt(x^2+x)
The rate of change for distance
ds/dt=[1/2sqrt(x^2+x)]
*(2x+1)*dx/dt
At x=16
ds/dt=3

Answer 4

that will do; no more help