in b) you were right to find
t = 1/4
but remember the period of 1 - sin(2pi*t)
is 2pi/2pi = 1
so another solution would be 1/4 + 1 or
t = 5/4
so the times between 0 and 2 when the object is at rest is
t = 1/4 and t = 5/4
for c) if
v(t) = 1 - sin(2pi*t)
then x(t) = t + (1/2pi)cos(2pi*t) + c
but you were given x(0) = 0
0 = 0 + (1/2pi)cos(0) + c
0 = )1/2pi) + c
c = -1/(2pi)
so x(t) = t + (1/2pi)cos(2pi*t) - 1/(2pi)
omg i can't believe i just forgot c in part c. it's even called part c!
and thank you for the help with part b. i completely forgot about that!
Let R be the region bounded by the y-axis and the graph of y=xcubed divided by 1+xsquared and y=4-2x,as shown inthe figure above.
find the area of R
find the volume of the solid generated when R is revolved about the x-axis
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