Posted by =/ on Wednesday, October 5, 2011 at 12:38am.
A particle is moving along the curve y=5sqrt(3x+1). As the particle passes through the point (5,20) its xcoordinate 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  drwls, Wednesday, October 5, 2011 at 1:01am
dx/dt = 3 units/s at the time of interest
dy/dt = dy/dx * dx/dt = 3 dy/dx
dy/dx = (5/2)/sqrt(3x+1) = (5/2)/4 = 5/8
Compute dy/dt and then use
dR/dt = sqrt[(dy/dt)^2 + (dx/dt)^2] ]
sqrt[3^2 + (15/8)^2] = 3.538 
calculus  =/, Wednesday, October 5, 2011 at 1:38am
I just plugged it in and it says it isn't right. =/

calculus  Reiny, Wednesday, October 5, 2011 at 8:23am
drwls left out the factor of 3 in the derivative, it should have been
dy/dx = (5/2)(3) / √(3x+1)
so when x=5
dy/dx = (15/2) / √16 = 15/8
take it from there 
calculus  drwls, Wednesday, October 5, 2011 at 10:35am
Thanks for the correction, Reiny :)
I am getting sloppier in my old age 
calculus  Anonymous, Monday, October 17, 2011 at 10:53pm
at the point .