sketch it.
length= L = 1m
1= x^2+y^2
0= 2x dx/dt + 2y dy/dt
dy/dt= -(x/y)dx/dt
but x=1/2 sin PIt/4
dx/dt= 1/2* cos(PIt/4)*PI/4
Now when x=1/16, then y= sqrt (L^2-x^2)
at x=1/16, then y= sqrt(1-1/256)
y= sqrt (255/256)
dx/dt= PI/8 * cos (PIt/4)
and x/y= (1/16)/(255/256)=16/255
dy/dt=-(x/y)dx/dt= you do it.
check my work, I am tired tonite.
The endpoints of a movable rod of length 1 m have coordinates (x, 0) and (0, y) (see figure). The position of the end on the x-axis is given in the equation below, where t is the time in seconds.
x(t) = 1/8 sin(πt/4)
(c) Find the speed of the y-axis endpoint when the x-axis endpoint is (1/16, 0). (Round your answer to four decimal places.)
1 answer
1 answer