Problem 25 - Calculus - AP Edition

Question

A conical water tank with vertex down has a radius of \(10 \mathrm{ft}\) at the top and is \(24 \mathrm{ft}\) high. If water flows into the tank at a rate of \(20 \mathrm{ft}^{3} / \mathrm{min}\), how fast is the depth of the water increasing when the water is 16 ft deep?

Step-by-Step Solution

Answer

\( \frac{dh}{dt} \approx 0.0727 \) ft/min.

Step 1: Understand the problem

We are given a conical tank where water is being filled at a constant rate, and we need to find how fast the water's depth is increasing when its depth is 16 feet.

Step 2: Write down known quantities

The radius of the base of the cone is \( r = 10 \) ft, height \( h = 24 \) ft, and the rate of change of volume \( \frac{dV}{dt} = 20 \) ft³/min. We need to find \( \frac{dh}{dt} \) when the height of the water is \( h = 16 \) ft.

Step 3: Establish the relationship between volume and height

The formula for the volume of a cone is \( V = \frac{1}{3} \pi r^2 h \). To use this, we express the radius \( r \) in terms of height \( h \) for the water, using similar triangles. For the whole tank, \( \frac{r}{h} = \frac{10}{24} \).

Step 4: Express radius in terms of height of water

Using the ratio from the tank's dimensions, we have \( r = \frac{10}{24} h = \frac{5}{12}h \). Substitute this into the volume equation to get \( V \) as a function of \( h \).

Step 5: Substitute and simplify the volume formula

Substitute \( r = \frac{5}{12}h \) into \( V = \frac{1}{3} \pi r^2 h \) to get \( V = \frac{1}{3} \pi \left( \frac{5}{12}h \right)^2 h = \frac{1}{3} \pi \frac{25}{144}h^3 = \frac{25\pi}{432}h^3 \).

Step 6: Differentiate volume with respect to time

Differentiate \( V = \frac{25\pi}{432}h^3 \) with respect to \( t \). Using the chain rule, \( \frac{dV}{dt} = \frac{25\pi}{432} \cdot 3h^2 \cdot \frac{dh}{dt} \). Simplifying gives \( \frac{dV}{dt} = \frac{25\pi}{144}h^2 \cdot \frac{dh}{dt} \).

Step 7: Solve for \( \frac{dh}{dt} \)

Substitute \( \frac{dV}{dt} = 20 \) ft³/min and \( h = 16 \) ft into the equation: \( 20 = \frac{25\pi}{144} (16)^2 \cdot \frac{dh}{dt} \).

Step 8: Simplify and calculate \( \frac{dh}{dt} \)

Calculate \( (16)^2 = 256 \). Then solve \( 20 = \frac{25\pi}{144} \times 256 \times \frac{dh}{dt} \). This simplifies to \( \frac{dh}{dt} = \frac{20 imes 144}{25\pi \times 256} \).

Step 9: Calculate the numerical value

Finally, compute the numerical value to get \( \frac{dh}{dt} \approx 0.0727 \) ft/min.

Key Concepts

Conical Tank

A conical tank is a three-dimensional geometric shape that has a circular base and a pointed top, resembling a cone in geometry. In this exercise, the tank is positioned with the vertex downwards. This orientation means that the widest part of the cone, the circular base, is at the top, and the narrow, pointed end is at the bottom.
Understanding the dimensions of the tank is crucial because they define how the volume changes at different water levels. For our specific problem, the conical tank has a radius of 10 feet at the top and a height of 24 feet. These measurements form a linear relationship essential for solving related rates problems.
Conical tanks can often be found in applications like storage tanks for liquids, where the shape allows for effective drainage or filling. The geometry of the cone aids in establishing relationships that can help us solve practical problems using calculus.

Volume of a Cone

The volume of a cone is calculated using the formula:\[V = \frac{1}{3} \pi r^2 h\]where \(V\) is the volume, \(r\) is the radius of the base, and \(h\) is the height of the cone. In our related rates problem, this formula helps us relate how the volume of water in the tank changes with its height.
However, as the tank fills, the radius of the water surface and the height of the water can change. To find the volume of water in the tank as it fills, we express the variable radius \(r\) in terms of the current water height \(h\).
Once the radius is expressed in terms of height using similar triangles, the volume equation becomes a function purely of height. This adaptation allows us to directly relate changes in volume with changes in height, which simplifies to solving for the rate of change of height.

Similar Triangles

The concept of similar triangles is pivotal in this problem to establish a relationship between the radius and height of the water in the conical tank. Since the conical tank and the water inside it form similar triangles, their corresponding sides are proportional.
For our full conical tank:

• The entire tank has a radius of 10 feet and a height of 24 feet.
• The water forms a similar, smaller cone as it fills the tank.

By applying the principle of proportionality, we derive that:\[\frac{r}{h} = \frac{10}{24}\]This proportion lets us express \(r\) (the radius of the water level) in terms of \(h\) (the water depth):\[r = \frac{5}{12} h\]This step is crucial for transforming the volume equation into one that depends only on \(h\), facilitating easier differentiation.

Differentiation with Respect to Time

Differentiation with respect to time is a core calculus technique used to solve related rates problems. In these problems, variables change with time, and we are interested in understanding how one variable changes as another does. For this conical tank problem, we are given the rate at which the volume changes over time and need to find the rate at which the water depth changes.
Once the volume of the water in the tank is expressed as \( V = \frac{25\pi}{432}h^3 \), we apply differentiation with respect to time \(t\) to link the rates of change:

• The rate of change of volume is given by \( \frac{dV}{dt} = 20 \, \text{ft}^3/\text{min} \).
• We need to find \( \frac{dh}{dt} \), the rate of change of water depth.

Using the chain rule, differentiate the volume function:\[\frac{dV}{dt} = \frac{25\pi}{144}h^2 \cdot \frac{dh}{dt}\]Substituting known values and solving for \(\frac{dh}{dt}\) gives us the rate of change in water depth when the water reaches a certain height. This is how we determine how quickly the water level rises inside the tank.