The valume of water in a container is x (2x²+x+5)cm³ when the depth is x cm.water is added at a content rate of 60cm³ per second.at what rate in the level of water tisng when the depth is 5cm.
5 answers
Yes
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@This8thGradeGirl0_0 -- not every problem is multiple-choice. Many times they want to see your work, to be sure you know the proper steps to follow.
v = 2x^2 + x + 5
dv/dt = (4x+1) dx/dt
so when x=5, we have
60 = 21 dx/dt
so the level is rising at a rate of 60/21 cm/s
v = 2x^2 + x + 5
dv/dt = (4x+1) dx/dt
so when x=5, we have
60 = 21 dx/dt
so the level is rising at a rate of 60/21 cm/s
V = x(2x²+x+5) <-- Volume
dV/dt = 60cm³/s <-- Differentiation of volume w.r.t. time
dx/dt = ? <-- Differentiation of depth w.r.t. time
x = 5 <-- Depth
V = x(2x²+x+5)
V = 2x³+x²+5x
dV/dt = (6x²+2x+5)(dx/dt)
60 = [6(5)²+2(5)+5](dx/dt)
60 = [6(25)+10+5](dx/dt)
60 = [150+15](dx/dt)
60 = 165(dx/dt)
60/165 = dx/dt
4/11 = dx/dt
Thus, the rate that the water is rising is at 4/11 cm/s
dV/dt = 60cm³/s <-- Differentiation of volume w.r.t. time
dx/dt = ? <-- Differentiation of depth w.r.t. time
x = 5 <-- Depth
V = x(2x²+x+5)
V = 2x³+x²+5x
dV/dt = (6x²+2x+5)(dx/dt)
60 = [6(5)²+2(5)+5](dx/dt)
60 = [6(25)+10+5](dx/dt)
60 = [150+15](dx/dt)
60 = 165(dx/dt)
60/165 = dx/dt
4/11 = dx/dt
Thus, the rate that the water is rising is at 4/11 cm/s
@oobleck I think you forgot the x at the beginning of the volume since volume is not in square units.
3 answers