Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Let \(\mathrm{G}(\mathrm{x})\) be the area of the region bounded by the graphs of \(y=\frac{1}{1+t^{2}}, y=0,\) \(t=1\) and \(t=x .\) a. Compute approximate values for \(y(1.0), y(1.5), y(2.0), y(2.5),\) and \(y(3.0)\). b. Compute approximate values for \(G(1.0), G(1.5), G(2.0), G(2.5),\) and \(G(3.0)\). c. Sketch the graph of \(y=\frac{1}{1+t^{2}}\) on [0,3] . d. Sketch the graph of \(G\) on [1,3] . e. Estimate \(G^{\prime}(2)\).

Step-by-Step Solution

Answer

\( y(x) = 0.5, 0.308, 0.2, 0.138, 0.1;\ G(x) = 0, 0.32175, 0.55357, 0.66818, 0.70483;\ G'(2) = 0.2. \)

Step 1: Understanding the function

The given function is \( y = \frac{1}{1+t^2} \). This function describes a curve where \( y \) is the reciprocal of \( 1 + t^2 \). Our job is to compute \( y \) values for different \( t \) values.

Step 2: Calculating y-values

To find the \( y \)-values for \( t=1.0, 1.5, 2.0, 2.5, \) and \( 3.0 \), substitute each value of \( t \) into the formula.- \( y(1.0) = \frac{1}{1+1^2} = \frac{1}{2} = 0.5 \)- \( y(1.5) = \frac{1}{1+(1.5)^2} = \frac{1}{3.25} \approx 0.308 \)- \( y(2.0) = \frac{1}{1+2^2} = \frac{1}{5} = 0.2 \)- \( y(2.5) = \frac{1}{1+(2.5)^2} = \frac{1}{7.25} \approx 0.138 \)- \( y(3.0) = \frac{1}{1+3^2} = \frac{1}{10} = 0.1 \)

Step 3: Understanding the area function

\( G(x) \) represents the area under the curve \( y = \frac{1}{1+t^2} \) from \( t = 1 \) to \( t = x \). This requires integration of the function from 1 to \( x \).

Step 4: Calculating G-values (approximations)

Compute the approximate area, \( G(x) \), using numerical integration or known results.The integral of \( \frac{1}{1+t^2} \) is \( \tan^{-1}(t) \), so:- \( G(1.0) = \int_1^{1.0} \frac{1}{1+t^2} \ dt = \tan^{-1}(1.0) - \tan^{-1}(1) = 0 \)- \( G(1.5) = \tan^{-1}(1.5) - \tan^{-1}(1) \approx 0.32175 \)- \( G(2.0) = \tan^{-1}(2.0) - \tan^{-1}(1) \approx 0.55357 \)- \( G(2.5) = \tan^{-1}(2.5) - \tan^{-1}(1) \approx 0.66818 \)- \( G(3.0) = \tan^{-1}(3.0) - \tan^{-1}(1) \approx 0.70483 \)

Step 5: Sketching the graph of y on [0,3]

To sketch \( y = \frac{1}{1+t^2} \) on \([0, 3]\), note the curve is symmetrical and decreases as \( t \) increases. Mark key points such as \( (1, 0.5), (1.5, 0.308), (2, 0.2), (2.5, 0.138), \) and \( (3, 0.1) \). Draw a smooth decreasing curve through these points.

Step 6: Sketching the graph of G on [1,3]

For the graph of \( G(x) \) on \([1, 3]\), plot the points \( (1, 0), (1.5, 0.32175), (2, 0.55357), (2.5, 0.66818), (3, 0.70483) \). Note the slope decreases as \( x \) increases because the curve \( y \) is decreasing.

Step 7: Estimating G'(2)

To estimate \( G'(2) \), determine the rate of change at \( x = 2 \). Since \( G'(x) \) by the Fundamental Theorem of Calculus is \( \frac{1}{1+x^2} \), substitute \( x = 2 \) to get \( G'(2) = \frac{1}{5} = 0.2 \).

Key Concepts

Integration

Integration is a fundamental concept in calculus that allows us to find the area under a curve. In this exercise, when we talk about the area of the region bounded by a curve, we are referring to the process of integration. To calculate the area under the curve given by the function \( y = \frac{1}{1+t^2} \), we integrate this function with respect to the variable \( t \).

The specific challenge here is to compute \( G(x) \), which represents the area under the curve from \( t = 1 \) to \( t = x \). Using the formula for this integral, we have:

• The integral of \( \frac{1}{1+t^2} \) is \( \tan^{-1}(t) \).
• Therefore, \( G(x) = \tan^{-1}(x) - \tan^{-1}(1) \) which provides us with the values of \( G(x) \) for any \( x \) in the interval.

Graph Sketching

Sketching graphs is a valuable skill in calculus, as it helps visualize the behavior of functions across specific intervals. For this problem, we have two functions to sketch: \( y = \frac{1}{1+t^2} \) and \( G(x) \).

**For \( y = \frac{1}{1+t^2} \):**
This function describes a curve that decreases as \( t \) increases. The graph is symmetrical around the y-axis, and key points were used to determine its shape. By plotting points like \( (1, 0.5) \), and \( (3, 0.1) \), you'll see a smooth, decreasing curve. This asymmetric but consistent decrease conveys how the function approaches zero, but never actually reaches it as \( t \) goes to infinity.

**For \( G(x) \):**
The function \( G(x) \) describes the accumulated area under the original curve as \( t \) moves from 1 to \( x \). While sketching, you'll note the slope of \( G(x) \) decreases with higher values of \( x \). This is because the rate of change (which is \( y(x) \)) also decreases as \( x \) increases, resulting in a graph that smoothly flattens out over the interval from 1 to 3.

Area Under Curve

Understanding the 'area under the curve' concept is crucial because it represents the accumulation of quantities over an interval. In this exercise, \( G(x) \) signifies the areas under the curve \( y = \frac{1}{1+t^2} \) from \( t = 1 \) to \( t = x \). In mathematical terms, this is expressed as:

• \( G(x) = \int_{1}^{x} \frac{1}{1+t^2} \, dt \)
• The value of \( G(x) \) will change depending on the values of \( x \), showing how much 'total area' has accumulated from 1 to \( x \).

This area gives a concrete meaning to the integral, which measures how much of the curve is "captured" between the two boundaries.

Practically, when you determine \( G(1.5) \), \( G(2.0) \), and so on, you're measuring either physically tangible quantities (like distance) or abstract ones (like total opportunity). It reinforces the concept that integration involves finding these cumulative measures over specified regions.

Derivative Estimation

The estimation of derivatives is essential in understanding the rate at which functions change. In this problem, estimating the derivative \( G'(2) \) gives insight into how fast \( G(x) \) is changing at \( x = 2 \). By the Fundamental Theorem of Calculus:

• \( G'(x) = y(x) = \frac{1}{1+x^2} \)
• Substituting \( x = 2 \), we find \( G'(2) = \frac{1}{5} = 0.2 \).

This tells us that at \( x = 2 \), \( G(x) \) is increasing at a rate of 0.2.

Derivative estimation is particularly useful in predicting function behavior without needing to compute massive data. It helps describe not just where a function is heading, but also how briskly it progresses along its path. Understanding \( G'(x) \) as the slope of \( G(x) \) at each point strengthens your grasp of calculus considerations like velocity and acceleration in dynamic systems.