Problem 59 - Contemporary Precalculus

Question

A certain fungus grows in a circular shape. Its diameter after \(t\) weeks is \(6-\frac{50}{t^{2}+10}\) inches. (a) Express the area covered by the fungus as a function of time. (b) What is the area covered by the fungus when \(t=0 ?\) What area does it cover at the end of 8 weeks? (c) When is its area 25 square inches?

Step-by-Step Solution

Answer

Question: Express the area covered by the fungus as a function of time, find the area when \(t=0\) and \(t=8\), and find the time when the area is 25 square inches. Solution: 1. The area function is: \(A(t)=\frac{\pi}{4}\times \left( 6-\frac{50}{t^{2}+10} \right)^2\) 2. The area when \(t=0\) is \(\frac{\pi}{4}\) square inches and when \(t=8\) is approximately 14.07 square inches. 3. The time when the area is 25 square inches is approximately 2.16 weeks.

Step 1: Part (a): Express the area covered by the fungus as a function of time

Let's denote the diameter function as \(d(t)=6-\frac{50}{t^{2}+10}\). The area A of a circle is given by the formula \(A=\pi r^2\), where 'r' is the radius of the circle. We can rewrite the formula in terms of diameter, since diameter is twice the radius: \(A=\frac{\pi}{4}D^2\). Now we need to substitute \(d(t)\) into the area formula. \(A(t)=\frac{\pi}{4}\times \left( 6-\frac{50}{t^{2}+10} \right)^2\)

Step 2: Part (b): Find the area when \(t=0\) and \(t=8\)

To find the area at \(t=0\), we can substitute t=0 into the area function: \(A(0)=\frac{\pi}{4}\times \left( 6-\frac{50}{(0)^{2}+10} \right)^2\) Now we can calculate the area at \(t=0\): \(A(0)=\frac{\pi}{4}\times \left( 6-\frac{50}{10} \right)^2=\frac{\pi}{4}\times (1)^2=\frac{\pi}{4}\) square inches Next, we can find the area at \(t=8\) by substituting t=8 into the area function: \(A(8)=\frac{\pi}{4}\times \left( 6-\frac{50}{(8)^{2}+10} \right)^2\) Now we can calculate the area at \(t=8\): \(A(8)=\frac{\pi}{4}\times \left( 6-\frac{50}{74} \right)^2\approx 14.07\) square inches

Step 3: Part (c): Find the time when the area is 25 square inches

We need to solve the equation \(A(t) = 25\) for the time 't'. So, we can rewrite the area function as the following equation: \(\frac{\pi}{4}\times \left( 6-\frac{50}{t^{2}+10} \right)^2 = 25\) First, we need to divide both sides by \(\frac{\pi}{4}\): \(\left( 6 - \frac{50}{t^{2}+10} \right)^2 = \frac{100}{\pi}\) Next, we can take the square root of both sides: \(6-\frac{50}{t^{2}+10} = \pm \sqrt{\frac{100}{\pi}}\) Now we need to solve for 't', which is a bit complicated. Let's start by isolating the fraction on one side: \(\frac{50}{t^{2}+10} = 6 \pm \sqrt{\frac{100}{\pi}}\) We can then take the reciprocal of both sides: \(\frac{t^{2}+10}{50} = \frac{1}{6 \pm \sqrt{\frac{100}{\pi}}}\) Now we can cross-multiply and simplify: \(t^{2}+10 = 50(6 \pm \sqrt{\frac{100}{\pi}})\) Now we can subtract 10 from both sides: \(t^{2} = 50(6 \pm \sqrt{\frac{100}{\pi}}) - 10\) Finally, we can take the square root of both sides: \(t = \pm \sqrt{50(6 \pm \sqrt{\frac{100}{\pi}}) - 10}\) Since we can't have negative time, only the positive value of t is valid. By using a calculator, we can find that the valid time is approximately: \(t \approx 2.16\) weeks So, the area of the fungus is 25 square inches after approximately 2.16 weeks.

Key Concepts

Circular Areas

In mathematics, circular areas are important as they represent the space contained within a circle. The area of a circle is calculated using the formula \( A = \pi r^2 \), where \( r \) is the radius. This calculation helps in understanding not only static objects but also dynamically changing shapes, such as the growth area of a fungus. When dealing with a circular growth pattern, understanding how the area changes over time is crucial. For circular objects like fungi, where growth is dynamic, knowing the area helps measure the extent of growth and analyze patterns over different time periods. Calculating the area can provide insights into the physical constraints of the environment and can be used in various scientific analyses, such as determining resources required or understanding ecological impacts.

Fungus Growth

Fungus growth is an intriguing natural phenomenon often taking the shape of circles known as "fairy rings." This can be due to the radial growth habits of many fungi. The exercise presents a scenario where a fungus grows in a circular pattern, with the diameter evolving as a function of time. Modeling this growth mathematically allows scientists to predict how the fungus will spread over time. Such models are used in biology to study the potential effects of fungi in ecosystems, agriculture, or when monitoring fungal diseases. In our problem, the diameter changes over time and thus affects how the circular area grows, which provides useful information to calculate the space occupied by the fungus at various time intervals.

Problem-solving in Calculus

Calculus provides powerful tools for solving problems involving change, such as the growth rate of a fungus. In this exercise, calculus is applied to understand how the circular area changes as the diameter varies over time. The challenge involves expressing the area as a function of time, which requires understanding the relationship between the radius, diameter, and area. Once the model is established, calculus can help predict future trends—such as the fungus reaching a specific area—by solving for the time variable. This involves setting up equations and manipulating them to isolate variables, often requiring algebraic skills alongside calculus concepts. Mastering these skills is essential for tackling real-world problems involving dynamic systems.

Time-dependent Functions

Time-dependent functions allow us to model and analyze changes over time, which is crucial in understanding phenomena like fungus growth, where the object of study evolves continuously. The given problem uses a time-dependent function to describe the diameter of the fungus, which directly affects the computation of the area. By expressing the diameter as a function of time \(d(t)\), we can derive the area covered by the fungus \(A(t)\) and predict its size at any given week. These types of functions are invaluable in not only biology but also in fields like physics, finance, and engineering, wherever an outcome relies heavily on temporal variation. Solving such functions typically involves a mix of analytical and numerical methods, which helps in approximating solutions that provide practical insights and results.