Homework Help Forum: calculus

Posted by bre on Thursday, October 29, 2009 at 6:56pm.

A particle is moving along the curve y= 3sqrt3x+1. As the particle passes through the point (5,12), its x-coordinate increases at a rate of 3 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

• calculus - Reiny, Thursday, October 29, 2009 at 7:18pm

  Let P(x,y) be any point on the curve then we could write P as (x,3√(3x+1)) and if D is the distance to the origin, D^2 = x^2 + (3√(3x+1))^2 = x^2 + 27x + 9 2D(dD/dt) = 2x(dx/dt) + 27(dx/dt) at (5,12) D^2 = 25+135+9 = 169 so D = 13 so dD/dt = (2(5)(3) + 27(3))/(2(13)) = 4.26 check my work please

  
  

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