A point is moving along the curve #y=sqrt(x)# in such a way that its x coordinate id increasing at the rate of 2 units per minute. At what rate is its slope changing (a) when x=1? (b) when x=4?

Question

Luke Alvarez · Accepted Answer

#x=1 => d/dt(dy/dx) = -1/2 color(white)".""min"^-1#

#x=4 => d/dt(dy/dx) = -1/16 color(white)".""min"^-1# Let's call our time variable #t#, in minutes. We know that: #dx/dt = 2 color(white)".""min"^-1# And we're trying to find the rate at which the slope is changing with respect to time: #d/dt(dy/dx)# First, let's evaluate #dy/dx#: #dy/dx = d/dx(sqrt(x)) = 1/(2sqrtx)# Now, we need to evaluate #d/dt(dy/dx)#: #d/dt(1/(2sqrt(x))) = 1/2(d/dt1/sqrtx)# #= 1/2(d/dx1/sqrtx)(dx/dt)# #= 1/2(-1/(2x^(3/2)))(2 color(white)"." "min"^-1)# #= -1/(2x^(3/2)) color(white)".""min"^-1# Finally, all we have to do is plug in #x=1# and #x=4# to answer the two parts to the problem. #x=1 =>-1/(2x^(3/2)) color(white)".""min"^-1= -1/(2)color(white)".""min"^-1# #x=4 => -1/(2x^(3/2)) color(white)".""min"^-1 = -1/(2*8) color(white)".""min"^-1 = -1/16 color(white)".""min"^-1# Final Answer

Axel Alvarez · Answer

undefined (a) When x=1, the slope is changing at a rate of 1/4 units per minute.
(b) When x=4, the slope is changing at a rate of 1/8 units per minute.