Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Find (approximately) equations of the lines tangent to the graphs of a. \(y=1.5^{t}\) at the points \((-1,2 / 3), \quad(0,1),\) and \((1,3 / 2)\) b. \(y=2^{t} \quad\) at the point \(\quad(-1,1 / 2), \quad(0,1),\) and (1,2) c. \(y=3^{t} \quad\) at the point \(\quad(-1,1 / 3), \quad(0,1),\) and (1,3) d. \(y=5^{t} \quad\) at the point \(\quad(-1,1 / 5), \quad(0,1), \quad\) and \(\quad(1,5)\)

Step-by-Step Solution

Answer

The tangent line equations are derived using the derivative formula \(y' = a^t\ln(a)\) and evaluating at each point. Each point has a different slope that leads to an equation for the tangent line.

Step 1: Understand the Problem

We need to find the tangent lines to four different exponential curves at specific points. The equation of the tangent line can be derived using the derivative of each function, evaluated at the given points.

Step 2: Differentiate the Functions

For any exponential function of the form \(y = a^t\), the derivative is \(y' = a^t \ln(a)\). This derivative represents the slope of the tangent line at any point \(t\).

Step 3: Part a: Find the Tangent Equations for \(y=1.5^t\)

1. Compute the derivative \(y' = 1.5^t \ln(1.5)\).2. Evaluate \(y'\) at \(t = -1, 0, \text{and} 1\) to get the slopes.3. Slope at \((-1, \frac{2}{3})\): \( \frac{1}{1.5} \ln(1.5) \approx 0.405\).4. Slope at \((0, 1)\): \(1 \ln(1.5) \approx 0.405\).5. Slope at \((1, \frac{3}{2})\): \(1.5 \ln(1.5) \approx 0.608\).6. Use point-slope form \(y - y_1 = m(x - x_1)\) to get equations: - For \((-1, \frac{2}{3})\): \(y = 0.405(t + 1) + \frac{2}{3}\) - For \((0,1)\): \(y = 0.405(t) + 1\) - For \((1, \frac{3}{2})\): \(y = 0.608(t - 1) + \frac{3}{2}\)

Step 4: Part b: Find the Tangent Equations for \(y=2^t\)

1. Compute the derivative \(y' = 2^t \ln(2)\).2. Evaluate \(y'\) at \(t = -1, 0, \text{and} 1\) to get the slopes.3. Slope at \((-1, \frac{1}{2})\): \(\frac{1}{2} \ln(2)\approx 0.346\).4. Slope at \((0, 1)\): \(1 \ln(2) \approx 0.693\).5. Slope at \((1, 2)\): \(2 \ln(2) \approx 1.386\).6. Use point-slope form to get equations: - For \((-1, \frac{1}{2})\): \(y = 0.346(t + 1) + \frac{1}{2}\) - For \((0, 1)\): \(y = 0.693(t) + 1\) - For \((1, 2)\): \(y = 1.386(t - 1) + 2\)

Step 5: Part c: Find the Tangent Equations for \(y=3^t\)

1. Compute the derivative \(y' = 3^t \ln(3)\).2. Evaluate \(y'\) at \(t = -1, 0, \text{and} 1\) to get the slopes.3. Slope at \((-1, \frac{1}{3})\): \(\frac{1}{3} \ln(3) \approx 0.366\).4. Slope at \((0, 1)\): \(1 \ln(3) \approx 1.099\).5. Slope at \((1, 3)\): \(3 \ln(3) \approx 3.297\).6. Use point-slope form to get equations: - For \((-1, \frac{1}{3})\): \(y = 0.366(t + 1) + \frac{1}{3}\) - For \((0, 1)\): \(y = 1.099(t) + 1\) - For \((1, 3)\): \(y = 3.297(t - 1) + 3\)

Step 6: Part d: Find the Tangent Equations for \(y=5^t\)

1. Compute the derivative \(y' = 5^t \ln(5)\).2. Evaluate \(y'\) at \(t = -1, 0, \text{and} 1\) to get the slopes.3. Slope at \((-1, \frac{1}{5})\): \(\frac{1}{5} \ln(5) \approx 0.322\).4. Slope at \((0, 1)\): \(1 \ln(5) \approx 1.609\).5. Slope at \((1, 5)\): \(5 \ln(5) \approx 8.045\).6. Use point-slope form to get equations: - For \((-1, \frac{1}{5})\): \(y = 0.322(t + 1) + \frac{1}{5}\) - For \((0, 1)\): \(y = 1.609(t) + 1\) - For \((1, 5)\): \(y = 8.045(t - 1) + 5\)

Key Concepts

Derivatives

Derivatives are fundamental concepts in calculus and provide a way to measure how a function changes. They give us the slope of the tangent line to a curve at any given point. This is incredibly useful because it allows us to understand the rate at which the function is changing. For exponential functions such as \(y = a^t\), the derivative is calculated as \(y' = a^t \ln(a)\). Here, the derivative involves the original exponential function multiplied by the natural logarithm of its base \(a\).

• The process of differentiation helps us determine slopes of tangent lines.
• These slopes represent the function's instantaneous rate of change.

These concepts might seem abstract, but understanding derivatives is key to mastering calculus and the changes that occur within functions.

Tangent Lines

Tangent lines are straight lines that just touch a curve at a single point without crossing it. Imagine a car driving along a curvy road: the tangent line at any point represents the direction and slope of the road exactly at that point. By calculating the derivative of a function at a certain point \((t, y)\), we get the slope \(m\) of this tangent line. Alongside the coordinates of the point itself, we can form the equation of the tangent line using the point-slope form.

• Tangent lines provide a linear approximation of the curve at specific points.
• They are particularly useful in predicting and analyzing local behavior of curves.

Grasping how to find these lines helps you better understand the nuances and details in the behavior of various functions.

Exponential Functions

Exponential functions are a central part of many real-world applications and analytical problems. These functions are of the form \(y = a^t\), where \(a\) is a constant base and \(t\) is the exponent. Their most striking feature is their rapid rate of increase or decrease, particularly evident as the exponent \(t\) becomes large.

• Exponential growth occurs when the rate of increase of a quantity is proportional to the current amount.
• These functions define processes like population growth, radioactive decay, and interest calculation.

Understanding exponential behavior helps predict and interpret significant natural and economic phenomena, emphasizing their importance in both mathematics and everyday applications.

Point-Slope Form

The point-slope form of a line equation is a powerful tool to derive equations of tangent lines from given derivatives. It is represented by the formula \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) are the coordinates of a point on the line, and \(m\) represents the line's slope. This form is commonly used to find the equation of a line when its slope and a point on the line are known.

• Point-slope form is intuitive and easy to use for calculating tangent lines.
• It straightforwardly incorporates both the point and slope information into one equation.

By mastering this form, you can quickly and accurately determine the specific linear equations corresponding to tangent lines, facilitating deeper understanding and analysis of function behaviors.