# calulus

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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?

• calulus -

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

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