Question

10. The height of a rectangular box is increasing at a rate of 2 meters per second
while the volume is decreasing at a rate of 5 cubic meters per second. If
the base of the box is a square, at what rate is one of the sides of the base
decreasing, at the moment when the base area is 64 square meters and the
height is 8 meters?

11. Sand is pouring out of a tube at 1 cubic meter per second. It forms a pile
which has the shape of a cone. The height of the cone is equal to the radius of
the circle at its base. How fast is the sandpile rising when it is 2 meters high?

12. A water tank is in the shape of a cone with vertical axis and vertex downward.
The tank has radius 3 m and is 5 m high. At first the tank is full of water,
but at time t = 0 (in seconds), a small hole at the vertex is opened and the
water begins to drain. When the height of water in the tank has dropped to
3 m, the water is flowing out at 2 m3/s. At what rate, in meters per second,
is the water level dropping then?

Answer 1

Answer on Question #64601 – Math – Calculus

Question

Q1. The height of a rectangular box is increasing at a rate of 2 meters per second while the volume is decreasing at a rate of 5 cubic meters per second. If the base of the box is a square, at what rate is one of the sides of the base decreasing, at the moment when the base area is 64 square meters and the height is 8 meters?

Solution

The volume of the box with the square base is

$$V = h \cdot b ^ {2}.$$

The rate of change of volume equal to its time derivative

$$\frac {d V}{d t} = V ^ {\prime}.$$

Find the rate of changing of the box volume:

$$V ^ {\prime} = (h \cdot b ^ {2}) ^ {\prime} = h ^ {\prime} \cdot b ^ {2} + 2 h b b ^ {\prime}$$

(here we used product rule

$$\left(f (x) \cdot g (x)\right) = f ^ {\prime} (x) \cdot g (x) + f (x) \cdot g ^ {\prime} (x)$$

and chain rule

$$\left(f (g (x))\right) ^ {\prime} = f ^ {\prime} (g (x)) g ^ {\prime} (x)$$

Find the rate of changing of the base side $b'$

$$V ^ {\prime} = h ^ {\prime} \cdot b ^ {2} + 2 h b b ^ {\prime} => 2 h b b ^ {\prime} = V ^ {\prime} - h ^ {\prime} \cdot b ^ {2} => b ^ {\prime} = (V ^ {\prime} - h ^ {\prime} \cdot b ^ {2}) / 2 h b$$

where $h' = 2m / \text{sis}$ the rate of the box height increasing, $V' = -5m^3 / s$ is the rate of the box volume decreasing (minus sign indicates a decrease in volume)

$b = \sqrt{64} = 8$ . Plug given values into

$$b ^ {\prime} = (V ^ {\prime} - h ^ {\prime} \cdot b ^ {2}) / 2 h b$$

we get

$$b ^ {\prime} = \frac {- 5 - 2 \cdot 6 4}{2 \cdot 8 \cdot 8} = - \frac {5 + 1 2 8}{1 2 8} \approx - 1. 0 4 m / s$$

(minus sign indicates a decrease in base side).

Answer: rate of the base side decreasing is $b' \approx -1.04m / s$ .

Question

Q2. Sand is pouring out of a tube at 1 cubic meter per second. It forms a pile which has the shape of a cone. The height of the cone is equal to the radius of the circle at its base. How fast is the sandpile rising when it is 2 meters high?

Solution

The volume of the cone is equal to

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

where $R$ is the radius of the circle at its base, $h$ is the height of the cone. Since by assumption of task the height of the cone is equal to the radius of the circle, i.e. $R = h$, then

$$V = \frac {1}{3} \pi h ^ {3} = \frac {1}{3} \pi R ^ {3}.$$

The rate of change of volume equal to its time derivative

$$\frac {d V}{d t} = V ^ {\prime}.$$

Find the rate of change of the volume of the sandpile:

$$V ^ {\prime} = \left(\frac {1}{3} \pi h ^ {3}\right) ^ {\prime} = \frac {1}{3} \pi \cdot 3 h ^ {2} h ^ {\prime} = \pi h ^ {2} h ^ {\prime} = \pi h ^ {2} R ^ {\prime}.$$

Then the rate of increase in the height and radius of the cone is

$$h ^ {\prime} = R ^ {\prime} = V ^ {\prime} / \pi h ^ {2}$$

Plug given values $(V' = 1 \, \text{m}^3/\text{sec}$ and $h = 2 \, \text{m})$ into last formula, we get the rate of increase in the height and radius of the cone

$$h ^ {\prime} = R ^ {\prime} = \frac {1}{4 \pi} \approx 0,08 \, \text{m/sec} = 8 \, \text{cm/sec}.$$

Answer: the rate of increase of height and radius of the sandpile when it is 2 meters high, is $h' = R' = 0.08 \, \text{m/sec} = 8 \, \text{cm/sec}$.

Question

Q3. A water tank is in the shape of a cone with vertical axis and vertex downward. The tank has radius $3\,\text{m}$ and is $5\,\text{m}$ high. At first the tank is full of water, but at time $t = 0$ (in seconds), a small hole at the vertex is opened and the water begins to drain. When the height of water in the tank has dropped to $3\,\text{m}$, the water is flowing out at $2\,\text{m}^3/\text{s}$. At what rate, in meters per second, is the water level dropping then?

Solution

Volume of a cone is

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

where $R$ is the radius of the cone, $h$ is the height. We know height of the cone and radius of the cone at the top so we can find the radius for the any height using a proportion of similar triangles. The ratio of the height of the tank to the radius:

$$\frac {R}{h} = \frac {3}{5}$$

When the height of the water is $h$ meters the radius is:

$$R = \frac{3}{5} h$$

Plug this into the $V = \frac{1}{3}\pi R^2 h$, we get

$$V = \frac{1}{3} \pi \left(\frac{3}{5} h\right)^2 h = \frac{3\pi}{25} h^3$$

The rate of change of volume equal to its time derivative $\frac{dV}{dt} = V'$.

Take derivative of each side

$$V' = \frac{9\pi}{25} h^2 h'$$

Plug numbers:

$$V' = \frac{9\pi}{25} 3^2 h' = 2 \, \mathrm{m}^3/\mathrm{sec}$$

Then

$$h' = \frac{2 \cdot 25}{81\pi} \approx 0.2 \, \mathrm{m}/\mathrm{sec}$$

Answer: the rate of decreasing water level is $h' \approx 0.2 \, \mathrm{m}/\mathrm{sec}$.

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