A conical tank (with vertex down) is 28 ft across the top and 7 ft ft
deep. If water is flowing into the tank at the rate of , find
the rate of change of the depth of the water at the instant when it is 2
deep.

Ask by Boone Munoz. in Puerto Rico

Nov 01,2024

Question

A conical tank (with vertex down) is 28 ft across the top and 7 ft ft
deep. If water is flowing into the tank at the rate of , find
the rate of change of the depth of the water at the instant when it is 2
deep.

Ask by Boone Munoz. in Puerto Rico

Nov 01,2024

UpStudy AI · Accepted Answer

To solve this problem, we need to relate the volume of water in the conical tank to the depth of the water.

### Step 1: Determine the dimensions of the cone

The tank is a cone with a diameter of 28 ft, which gives a radius of:

$$
r = \frac{28}{2} = 14 \text{ ft}
$$

The height of the cone is 7 ft.

### Step 2: Set up the relationship between the volume and the depth

The volume $$ V $$ of a cone is given by the formula:

$$
V = \frac{1}{3} \pi r^2 h
$$

where $$ r $$ is the radius of the base and $$ h $$ is the height of the cone.

### Step 3: Relate the radius to the height

Since the tank is conical, the radius $$ r $$ of the water surface at a height $$ h $$ is proportional to $$ h $$. The ratio of the radius to the height of the cone is constant:

$$
\frac{r}{h} = \frac{14}{7} = 2 \implies r = 2h
$$

### Step 4: Substitute $$ r $$ in the volume formula

Now we can express the volume $$ V $$ in terms of $$ h $$:

$$
V = \frac{1}{3} \pi (2h)^2 h = \frac{1}{3} \pi (4h^2) h = \frac{4}{3} \pi h^3
$$

### Step 5: Differentiate the volume with respect to time

We need to find the rate of change of the depth of the water $$ \frac{dh}{dt} $$ when the depth $$ h = 2 $$ ft. We know that water is flowing into the tank at a rate of $$ \frac{dV}{dt} = 9 \text{ ft}^3/\text{sec} $$.

Using the chain rule, we differentiate the volume with respect to time:

$$
\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}
$$

First, we find $$ \frac{dV}{dh} $$:

$$
\frac{dV}{dh} = \frac{d}{dh} \left( \frac{4}{3} \pi h^3 \right) = 4 \pi h^2
$$

### Step 6: Substitute $$ h = 2 $$ ft into $$ \frac{dV}{dh} $$

Now we evaluate $$ \frac{dV}{dh} $$ at $$ h = 2 $$:

$$
\frac{dV}{dh} \bigg|_{h=2} = 4 \pi (2^2) = 4 \pi (4) = 16 \pi
$$

### Step 7: Solve for $$ \frac{dh}{dt} $$

Now we can substitute into the equation for $$ \frac{dV}{dt} $$:

$$
9 = 16 \pi \cdot \frac{dh}{dt}
$$

Solving for $$ \frac{dh}{dt} $$:

$$
\frac{dh}{dt} = \frac{9}{16 \pi}
$$

### Conclusion

Thus, the rate of change of the depth of the water when it is 2 ft deep is:

$$
\frac{dh}{dt} = \frac{9}{16 \pi} \text{ ft/sec}
$$

This is the final answer.
The rate of change of the depth of the water when it is 2 ft deep is $$ \frac{9}{16 \pi} \text{ ft/sec} $$.

A conical tank (with vertex down) is 28 ft across the top and 7 ft ft
deep. If water is flowing into the tank at the rate of , find
the rate of change of the depth of the water at the instant when it is 2
deep.

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A conical tank (with vertex down) is 28 ft across the top and 7 ft ft deep. If water is flowing into the tank at the rate of , find the rate of change of the depth of the water at the instant when it is 2 deep.

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Latest Calculus Questions

A conical tank (with vertex down) is 28 ft across the top and 7 ft ft
deep. If water is flowing into the tank at the rate of , find
the rate of change of the depth of the water at the instant when it is 2
deep.