Question

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.

Answer 1

Yes

Answer 2

What are your answer choices?

Answer 3

@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

Answer 4

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

Answer 5

@oobleck I think you forgot the x at the beginning of the volume since volume is not in square units.

Answer 6

[AI-generated answer — Explain Bot]

To find the rate at which the water level is rising when the depth is 5 cm, we need to find the derivative of the volume with respect to time.

Let's break down the given information:
- The volume of water in the container is given by V = x(2x² + x + 5) cm³.
- The rate at which water is being added is 60 cm³ per second.

We need to find the rate at which the water level is rising, which can be represented by dV/dt, where V is the volume and t is time.

To find dV/dt, we will differentiate the volume equation with respect to time. However, since the volume is given in terms of x (depth), we need to use the chain rule to differentiate with respect to x and then multiply by the rate at which x is changing with respect to t.

Step 1: Differentiating with respect to x
dV/dx = 2x² + x + 5

Step 2: Differentiating x with respect to t
dx/dt = ? (We need to find this rate from the given information)

Step 3: Combining the equations
dV/dt = dV/dx * dx/dt

Step 4: Evaluating at x = 5 cm
We now need to substitute x = 5 cm into the derivative equation to find the rate at which the water level is rising.

Once we know dx/dt (rate at which x is changing with respect to t), we can substitute it into the equation to find the rate at which the water level is rising when the depth is 5 cm.