The volume of water in a container is x(2x²+x+5)cm³ when the depth is Xcm water is added at the rate of 60cm³5¯¹ when the depth is 5cm,at what rate is the level rising.
To find the rate at which the level is rising, we need to differentiate the volume equation with respect to time. However, the given equation is in terms of depth (X) and not time. We need to use a related rate equation to convert it into a time-dependent equation.
Let's assume that t represents time. We have the equation for volume in terms of depth:
V = x(2x² + x + 5) cm³
According to the question, the depth is increasing at a rate of 60 cm³/s when the depth is 5 cm. Let's denote the rate of change of depth with respect to time as dx/dt.
Using the chain rule of differentiation, we can relate the rate of change of volume (dV/dt) with the rate of change of depth (dx/dt).
dV/dt = (dV/dx) * (dx/dt)
Let's find the partial derivatives of V with respect to x:
dV/dx = d/dx (x(2x² + x + 5))
= (d/dx)(2x³ + x² + 5x)
= 6x² + 2x + 5
Now, we have:
dV/dt = (6x² + 2x + 5) * (dx/dt)
Given that dx/dt = 60 cm³/s when x = 5 cm, we can substitute these values into the equation:
dV/dt = (6(5)² + 2(5) + 5) * 60
= (150 + 10 + 5) * 60
= 165 * 60
= 9900 cm³/s
So, the level is rising at a rate of 9900 cm³/s.