Posted by drew on .
A particle moves in the plane along the curve y=xln(x). At what rate is the distance from the origin increasing at the moment its xcoordinate is 2 cm and its xcoordinate is increasing at a rate of 17 cm/sec?

calculus (related rates) 
bobpursley,
distance=sqrt(y^2 + x^2)
ddistance/dt= 1/2distance *( 2ydy/dt+2xdx/dt)
now examine y dy/dt= x ln(x) (ln(x)+1) dx/dt
ddistance/dt=rateyou want=1/2distance*(2xlnx+1)17cm/sec + x dx/dt)
check that, I typed it in a hurry.