A Particle Is Moving Along The Curve Y = √x. As...

Anonymous

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A particle is moving along the curve y = √x. As the particle passes through the point (3, 9) its x-coordinate isincreasing at 9cm/sec. How fast is the distance from the particle to the origin changingat this instant? Thanks for any help in advance

 

Answer No.1

PhdKnowledge7

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The point (3, 9) is NOT on the curve. Is the point (9, 3)? y = √x, the distance fromthe particle to the origin = L = √(x2 +y2) = √(x2 + x) dL/dt = (2x + 1) dx/dt/[2√(x2 + x)] plug x = 9, dx/dt = 9 cm/s, we get dL/dt = 9.01 cm/s

Answer No.2

Anonymous

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You're right, the point is 9,3. Is 9.01 the answer to the problem - How fast is the distance from the particle to the origin changing at this instant?

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