Problem 8 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Valentina is sipping from a straw she has stuck in a conical cup of lemonade. The cup is an inverted circular cone of height \(12 \mathrm{~cm}\) and radius \(6 \mathrm{~cm}\). When the height of the lemonade in the cup is \(10 \mathrm{~cm}\), Valentina is sipping lemonade at a rate of \(2 \mathrm{~cm}^{3} / \mathrm{second}\). At this moment, how fast is the height of the lemonade in the cup decreasing?

Step-by-Step Solution

Answer

The rate at which the height of the lemonade in the cup is decreasing is approximately -0.0127 cm/s.

Step 1: Determine the volume of the lemonade in the cup

Given that the cup is a cone and Valentina is sipping at a rate of 2 cm^3s, the volume \(V\) can be given by the formula for volume of a cone as follows: \(V = \frac{1}{3}πr^2h\). Since the cup has a constant radius and height ratio (that is, \(\frac{r}{h} = \frac{1}{2}\)) given by the problems' constraints, we can replace the radius \(r\) in the equation with \(\frac{h}{2}\), so that \(V = \frac{1}{3}π*\left(\frac{h}{2}\right)^2*h = \frac{1}{12}πh^3\). This simplification will let us proceed with a function in terms of \(h\) alone.

Step 2: Find the derivative of the volume equation

Taking the derivative of both sides with respect to time \(t\), we get \(\frac{dV}{dt} = \frac{1}{12}π*3h^2*\frac{dh}{dt}\). Simplifying this equation, we get \(\frac{dV}{dt} = \frac{1}{4}πh^2*\frac{dh}{dt}\). This sets up a relationship between the change in volume and change in height over time.

Step 3: Substitute and solve for the unknown

Substitute \(-2 \mathrm{~cm}^{3}/\mathrm{second}\) for \(\frac{dV}{dt}\) and 10 cm for \(h\) into the equation from Step 2. Now, solve for \(\frac{dh}{dt}\), the rate at which the height of the lemonade in the cup is decreasing. This will give us \(\frac{dh}{dt} = -\frac{4}{π*100} = -0.0127 \mathrm{cm/s}.\)

Key Concepts

Volume of a Cone

The concept of the volume of a cone is crucial and is used when dealing with problems relating to the capacity and space within a cone-shaped object. The formula for the volume of a cone is given by \[ V = \frac{1}{3} \pi r^2 h \] where:

• \(V\) is the volume,
• \(r\) is the radius of the base,
• \(h\) is the height of the cone.

In our lemonade problem, the cup can be viewed as an upside-down cone since it is broader at the top and narrower at the bottom. By substituting the given radius and height into the formula, you can find the volume of lemonade in the cup. Additionally, knowing the relationship between the radius and height helps simplify equations to refer solely to the height when solving related rates problems.

Derivatives

Derivatives are a fundamental concept in calculus which involve the rate at which a function is changing at any given point. In the context of our lemonade problem, the derivative is used to understand how quickly a particular measurement, like the volume or height of lemonade, is changing over time. The derivative of the volume \[ \frac{dV}{dt} \] with respect to time \(t\) provides vital information about the rate of change of volume. Using calculus, we can take the derivative of the cone’s volume formula to find out how fast the volume is decreasing as Valentina sips lemonade. Knowing this rate at a specific height allows us to solve related rates problems effectively by connecting changes in multiple dimensions or properties to understand overall dynamics.

Chain Rule

The chain rule in calculus is a principle used to take derivatives of composite functions, which are functions made by combining two or more functions. In our scenario with the lemonade, we need to find the rate at which height changes over time. Since the volume of lemonade depends on both the radius and height, and each of these variables might change over time, the chain rule comes into play. By applying the chain rule, we can express the derivative of a function that is dependent on another variable which also changes over time. Specifically, \( \frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt} \).This expression shows how the rate of change of height \(\frac{dh}{dt}\) can be calculated by knowing both the current rate of change in volume and the relationship between volume and height.

Calculus

Calculus is the branch of mathematics that studies continuous change. It introduces essential tools such as differentiation and integration. When solving the problem of how fast the height of lemonade is decreasing, calculus provides the tools for analyzing changes in the lemonade's volume and height. This is known as a "related rates problem," as we find the relationship between different rates of change. Calculus allows us to handle complex problems involving varying quantities seamlessly by:

• Using derivatives to express rates of change,
• Applying formulas to understand real-world changes,
• Combining information from different dimensions (height, radius, volume) to form comprehensive solutions and insights.

With calculus, students can acquire a deeper understanding of both the techniques and their applications, equipping them to tackle a wide array of mathematical tasks.