Problem 9 - The Calculus with Analytic Geometry

Question

A water tank in the form of an inverted cone is being emptied at the rate of \(6 \mathrm{ft}^{3} / \mathrm{min}\). The altitude of the cone is \(24 \mathrm{ft}\), and the base radius is \(12 \mathrm{ft}\). Find how fast the water level is lowering when the water is \(10 \mathrm{ft}\) deep.

Step-by-Step Solution

Answer

-0.305 ft/min

Step 1: Understand the problem

We need to find how fast the water level is lowering, given the rate at which the volume is decreasing. The tank is an inverted cone with given dimensions.

Step 2: Determine the relationship between the variables

We have a cone with a height of 24 ft and a base radius of 12 ft. As the water level (height) changes, the radius of the water surface also changes. The relationship between the radius and height is proportional: \[ \frac{r}{h} = \frac{12}{24} = \frac{1}{2} \] Therefore, \[ r = \frac{1}{2}h \]

Step 3: Express the volume of the cone

The volume of a cone is given by \[ V = \frac{1}{3}\pi r^2h \] Substituting \[ r = \frac{1}{2}h \] into the volume formula, we get: \[ V = \frac{1}{3}\pi \left( \frac{1}{2}h \right)^2 h = \frac{1}{3}\pi \frac{1}{4} h^3 = \frac{1}{12}\pi h^3 \]

Step 4: Differentiate the volume with respect to time

We need to find \( \frac{dh}{dt} \). Using the chain rule: \[ \frac{dV}{dt} = \frac{d}{dt} \( \frac{1}{12}\pi h^3 \) = \frac{1}{12}\pi \(3h^2 \frac{dh}{dt} \) = \frac{1}{4}\pi h^2 \frac{dh}{dt} \]

Step 5: Plug in the known values

We know \( \frac{dV}{dt} = -6 \mathrm{ft}^3 / \mathrm{min} \) (the volume is decreasing, hence the negative sign), and at the instant we are looking at, \( h = 10 \mathrm{ft} \). Plug these into the differentiated equation: \[ -6 = \frac{1}{4}\pi (10)^2 \frac{dh}{dt} \]

Step 6: Solve for \( \frac{dh}{dt} \)

Isolate \( \frac{dh}{dt} \) to find its value: \[ -6 = \frac{1}{4}\pi \times 100 \frac{dh}{dt} \Rightarrow -6 = 25\frac{\pi}{4} \frac{dh}{dt} \] \[ \frac{dh}{dt} = \frac{-6\times 4}{25\times \pi} = \frac{-24}{25\times \pi} \Rightarrow \frac{dh}{dt} \approx -0.305 \mathrm{ft}/\mathrm{min} \]

Key Concepts

Rate of Change

The rate of change in calculus is essentially the speed at which one variable changes with respect to another variable. In our problem, the rate of change we're interested in is how fast the water level is dropping in the inverted cone.
Specifically, this is represented by \( \frac{dh}{dt} \), which denotes the rate of change of the height of water, with respect to time (t). Here are a few simple points to better understand this concept:

• Rate of change is generally expressed as a derivative in calculus.
• Common notations for rate of change include \( \frac{dy}{dx} \) or \( \frac{dX}{dt} \), where Y and X are variables.
• It measures how a quantity, like height or time, changes over a particular interval.

Now, using this concept helps us find how fast the water depth decreases given the rate at which the volume of water decreases. In simple terms, we're looking at how quickly the depth of water reduces while the water is being drained from the tank.

Related Rates

Related rates problems revolve around finding the rate at which one quantity changes by relating that change to another quantity. In this exercise, the relationship between the volume of water and the water level inside the cone describes a perfect example of related rates.
For the situation at hand, the volume of the cone and both its radius and height are variables that change with time. Since these quantities are interlinked, changes in one quantity affect the others. Here’s what you need to keep in mind:

• Identify all quantities that vary with respect to time.
• Establish the relationship between the variables (e.g., using similar triangles to relate height and radius).
• Differentiate the relationship with respect to time using implicit differentiation.

By understanding the concept of related rates, you see how varying one aspect, such as the rate of volume decrease, influences another aspect, like the rate at which the height changes (\( \frac{dh}{dt} \)). This approach serves to connect different parts of the problem logically, leading to a solution.

Cone Volume Calculation

A crucial part of the solution involves understanding how to calculate the volume of a cone and manipulate that formula given dynamic constraints. The volume of a cone is determined using the formula:
\[ V = \frac{1}{3}\pi r^2 h \]
However, in our problem, the radius (r) and the height (h) are interrelated, decreasing as the water level drops. By recognizing and employing the relationship \( r = \frac{1}{2} h \), we adapt the formula to focus on height alone:

• Substitute \( r = \frac{1}{2} h \) back into the volume formula.
• Refine the formula to \( V = \frac{1}{12}\pi h^3 \), a function based purely on the height.
• Differentiate this function with respect to time to connect the volume change rate with the height change rate.

Understanding this part is key because it simplifies handling the variables and makes it easier to solve the problem. Once simplified, differentiating the volume function concerning time provides the link between the volume rate and the height rate, bringing us closer to the desired solution.