Post a New Question

calulus

posted by .

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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question