MH1810 Math

Chapter 5b: Differentiation

Imron Rosyadi

5.4 Problems on Rate of Change

The derivative \(f'(x)\) is the instantaneous rate of change of \(f(x)\) with respect to \(x\). This concept has vast applications in various fields.

Applications of Instantaneous Rates of Change:

• Physics: velocity, acceleration, electricity, etc.
• Chemistry: rate of reaction
• Biology: population growth
• Economics: concepts of marginalism (e.g., marginal cost, marginal revenue)
• Engineering: many diverse applications

5.4 Problems on Rate of Change

Example 5.4.1: Volume of a Spherical Cell

The volume of a growing spherical cell is \(V = \frac{4}{3}\pi r^3\), where \(r\) is radius in micrometers.

1. Find the average rate of change of \(V\) with respect to \(r\) when \(r\) changes from 5 to 8 \(\mu m\). \[ \frac {V (8) - V (5)}{8 - 5} = \frac {\frac{4}{3}\pi (8^3) - \frac{4}{3}\pi (5^3)}{3} = \frac {4 \pi}{3} \cdot \frac {512 - 125}{3} = \frac {4 \pi}{3} \cdot \frac {387}{3} = 4\pi \cdot 129 = 516 \pi \mu m^3/\mu m. \]
2. Find the instantaneous rate of change of \(V\) with respect to \(r\) when \(r = 5\mu m\).

First, find \(V'(r) = \frac{d}{dr}(\frac{4}{3}\pi r^3) = 4\pi r^2\).

Then, \(V'(5) = 4\pi (5^2) = 100\pi \mu m^3/\mu m\).

This section brings together all the differentiation rules and concepts we’ve learned and applies them to real-world problems. The core idea is that the derivative quantifies instantaneous change.

The list of applications highlights the interdisciplinary nature of calculus. Engineers use these concepts constantly to design, analyze, and optimize systems.

Example 5.4.1 illustrates the difference between average rate of change and instantaneous rate of change. The average rate is a simple slope over an interval. The instantaneous rate, however, requires the derivative. For the spherical cell, \(V'(r) = 4\pi r^2\) is the derivative. Evaluating it at \(r=5\) gives us the exact rate at which the volume is increasing with respect to the radius at that specific instant. This rate has units of volume per unit radius.

Example: Motion of a Ball

Example 5.4.2: Ball Rolling Down an Inclined Plane

If a ball is given a push so that it has an initial velocity of \(5m/s\) down an inclined plane, then the distance it has rolled after \(t\) seconds is \(x = 5t + 3t^2\).

1. Find the velocity after 2s.

Velocity is the first derivative of position:

\(x'(t) = \frac{dx}{dt} = \frac{d}{dt}(5t + 3t^2) = 5 + 6t\).

At \(t=2s\), velocity \(x'(2) = 5 + 6(2) = 17 \text{ m/s}\).

2. How long does it take for the velocity to reach \(35m/s\)?

Set velocity equal to \(35\): \(5 + 6t = 35\).

\(6t = 30 \Rightarrow t = 5 \text{ s}\).

3. What is the acceleration after 2s?

Acceleration is the second derivative of position (or first derivative of velocity): \(x''(t) = \frac{d^2x}{dt^2} = \frac{d}{dt}(5 + 6t) = 6\).

The acceleration is constant, \(6 \text{ m/s}^2\). So, after 2s, the acceleration is \(6 \text{ m/s}^2\).

This example directly applies derivatives to kinematics, the study of motion. Position, velocity, and acceleration are intrinsically linked through differentiation.

• The first derivative of the position function \(x(t)\) gives the instantaneous velocity \(x'(t)\). We calculate this and evaluate it at \(t=2\) seconds.
• To find when the velocity reaches a certain value, we set the velocity function equal to that value and solve for \(t\).
• The second derivative of the position function \(x''(t)\) (or the first derivative of the velocity function) gives the instantaneous acceleration. In this case, the acceleration is constant.

This problem clearly demonstrates how derivatives are used to quantify motion and understand how forces affect an object over time.

Example: Related Rates (Air Expansion)

Example 5.4.3: Air Expansion

When air expands without losing or gaining heat, its pressure \(P\) and volume \(V\) satisfy \(PV^{1.4} = C\) (a constant).

At one instant:

• \(V = 400 \mathrm{cm}^3\)
• \(P = 80 \mathrm{kPa}\)
• Pressure is decreasing at \(10 \mathrm{kPa/min}\) (\(\frac{dP}{dt} = -10 \mathrm{kPa/min}\)).

At what rate is the volume increasing at this instant?

Solution:

The equation is \(PV^{1.4} = C\). Since \(P\) and \(V\) both depend on time \(t\), we differentiate implicitly with respect to \(t\): \[ \frac {d}{d t} (P V^{1.4}) = \frac {d}{d t} (C). \] Using the Product Rule on the left side: \[ \frac {d P}{d t} \cdot V ^ {1. 4} + P \cdot (1. 4 V ^ {0. 4}) \cdot \frac {d V}{d t} = 0. \] Now, solve for \(\frac{dV}{dt}\): \[ 1. 4 P V ^ {0. 4} \frac {d V}{d t} = - \frac {d P}{d t} \cdot V ^ {1. 4}. \] \[ \frac {d V}{d t} = - \frac {\frac {d P}{d t} \cdot V ^ {1. 4}}{1. 4 P V ^ {0. 4}} = - \frac {\frac {d P}{d t} \cdot V}{1. 4 P}. \] Substitute the given values: \(V = 400\), \(P = 80\), \(\frac{dP}{dt} = -10\): \[ \frac {d V}{d t} = - \frac {(- 1 0) (4 0 0)}{1. 4 (8 0)} = - \frac {- 4000}{112} \approx 35.71 \mathrm{cm}^3 / \mathrm{min}. \] So, the volume is increasing at approximately \(35.71 \mathrm{cm}^3 / \mathrm{min}\).

This is a classic “related rates” problem, a common application of implicit differentiation. The trick is to identify the relationship between the variables (P and V), recognize that they are both changing with respect to a third variable (time \(t\)), and then differentiate the entire equation implicitly with respect to \(t\).

Notice how the product rule is essential for differentiating \(PV^{1.4}\). Also, remember that whenever you differentiate a variable that is implicitly a function of \(t\) (like \(V\) or \(P\)), you must apply the chain rule, resulting in a \(\frac{dV}{dt}\) or \(\frac{dP}{dt}\) term.

After differentiating, the algebra involves solving for the desired rate (\(\frac{dV}{dt}\) in this case) and then plugging in all the given instantaneous values. The negative sign for \(\frac{dP}{dt}\) indicates that pressure is decreasing. The positive result for \(\frac{dV}{dt}\) means volume is increasing, which makes sense if pressure is decreasing (Boyle’s Law connection).

Example: Related Rates (Motorcyclist & Radar)

Example 5.4.4: Motorcyclist and Radar Gun

A traffic police officer is \(6\mathrm{m}\) above a road. A motorcyclist passes a lamppost.

At that instant:

• Distance from motorcycle to bridge (horizontal, \(x\)) is \(8\mathrm{m}\).
• Distance from police to motorcycle (diagonal, \(y\)) is \(10\mathrm{m}\).
• The distance \(y\) is decreasing at \(52\mathrm{km/h}\) (\(\frac{dy}{dt} = -52 \mathrm{km/h}\)).
• Speed limit: \(60\mathrm{km/h}\).

Can the motorcyclist be fined?

Solution:

Let \(x(t)\) be the horizontal distance and \(y(t)\) the diagonal distance. The height of the bridge is \(h = 6\mathrm{m} = 0.006\mathrm{km}\).

These distances are related by the Pythagorean theorem: \(x(t)^2 + h^2 = y(t)^2\).

Differentiate implicitly with respect to \(t\): \[ 2x(t)x'(t) + 0 = 2y(t)y'(t). \] Solving for \(x'(t)\) (the speed of the motorcycle): \[ x'(t) = \frac{y(t)y'(t)}{x(t)}. \] At the specific instant:

• \(h = 6\mathrm{m} = 0.006\mathrm{km}\).
• \(x(t_0) = 8\mathrm{m} = 0.008\mathrm{km}\).
• \(y(t_0) = 10\mathrm{m} = 0.01\mathrm{km}\).
• \(y'(t_0) = -52\mathrm{km/h}\) (decreasing distance).

Substitute values: \[ x'(t_0) = \frac{(0.01 \mathrm{km})(-52 \mathrm{km/h})}{0.008 \mathrm{km}} = \frac{-0.52}{0.008} \mathrm{km/h} = -65 \mathrm{km/h}. \]

The speed of the motorcycle is \(|x'(t_0)| = 65 \mathrm{km/h}\).

Since \(65 \mathrm{km/h} > 60 \mathrm{km/h}\), the motorcyclist can be fined.

This is another classic related rates problem, involving geometry and physics. The setup is a right triangle, where the horizontal distance from the lamppost to the motorcycle is \(x\), the vertical height of the bridge is \(h\), and the diagonal distance from the police to the motorcycle is \(y\).

The first step is to establish the geometric relationship using the Pythagorean theorem. Then, we differentiate this equation implicitly with respect to time \(t\), because all these distances are changing over time. Remember that \(h\) is a constant, so its derivative is zero.

After differentiating and solving for \(x'(t)\) (the motorcycle’s speed), we plug in the given values at the specific instant. It’s crucial to be consistent with units; converting meters to kilometers is necessary here. The negative sign for \(y'(t_0)\) indicates that the distance \(y\) is decreasing. The resulting negative sign for \(x'(t_0)\) means the motorcycle is moving in a direction that makes \(x\) decrease (i.e., towards the lamppost, from the police’s perspective, or away from a reference point on the road if \(x\) was increasing). The speed is the absolute value of the velocity.

Comparing the calculated speed with the speed limit allows us to answer the question. This problem beautifully illustrates how calculus can be used to solve practical problems encountered in daily life (or by traffic police!).

5.5 Linearization

Linearization approximates a function \(f(x)\) near a point \(x=a\) using a linear function \(L(x)\).

Definition 5.5.1: Linearization of \(f\) at \(a\)

The linearization of \(f\) at \(a\) is the linear function:

\[ L (x) = f (a) + f ^ {\prime} (a) (x - a). \]

Remarks:

1. The equation \(y = L(x)\) is the equation of the tangent line to the curve \(y = f(x)\) at \(x = a\).

2. Linearization provides a local approximation: \(f(x) \approx L(x)\) for \(x\) near \(a\).

3. Higher-order approximations exist, such as Taylor polynomials, which use higher derivatives to create more accurate approximations over larger intervals. \[ P _ {n} (x) = f (a) + (x - a) f ^ {\prime} (a) + \frac {(x - a) ^ {2}}{2 !} f ^ {\prime \prime} (a) + \dots + \frac {(x - a) ^ {n}}{n !} f ^ {(n)} (a). \\ \]

Linearization is a fundamental concept that allows us to approximate complicated functions with simpler linear ones, specifically, the tangent line at a point. This is incredibly useful for computations and understanding local behavior.

The definition shows that the linearization is simply the equation of the tangent line. It’s the best linear approximation of the function near the point of tangency.

The remarks highlight several key points:

• It literally IS the tangent line.
• It’s a “local” approximation, meaning it’s accurate only very close to the point ‘a’. As you move further away, the tangent line might deviate significantly from the curve.
• For more accurate approximations over larger intervals, or for more complex functions, we need to use higher-order polynomials, like Taylor polynomials, which incorporate second, third, and higher derivatives. Linearization is simply the first-degree Taylor polynomial.

Examples of Linearization

Example 5.5.2: Approximate \(\ln(1.01)\)

Consider \(f(x) = \ln(1 + x)\). Use the linearization of \(f\) at \(a = 0\) to approximate \(\ln(1.01)\).

Solution:

1. Find \(f(a)\): \(f(0) = \ln(1 + 0) = \ln 1 = 0\).

2. Find \(f'(x)\): \(f'(x) = \frac{1}{1 + x}\).

3. Find \(f'(a)\): \(f'(0) = \frac{1}{1 + 0} = 1\).

4. Form the linearization \(L(x) = f(a) + f'(a)(x - a)\):

   \(L(x) = 0 + 1(x - 0) = x\).

5. Approximate \(\ln(1.01)\):

   Since \(1.01 = 1 + 0.01\), we use \(x = 0.01\).

   \(\ln(1.01) \approx L(0.01) = 0.01\).

   (Actual value \(\ln(1.01) \approx 0.00995\))

Examples of Linearization

Example 5.5.3: Approximate \(\sqrt{4.001}\)

Use the linearization of \(f(x) = \sqrt{x}\) at \(a = 4\) to approximate \(\sqrt{4.001}\).

Solution:

1. Find \(f(a)\): \(f(4) = \sqrt{4} = 2\).

2. Find \(f'(x)\): \(f'(x) = \frac{1}{2\sqrt{x}}\).

3. Find \(f'(a)\): \(f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{2 \cdot 2} = \frac{1}{4}\).

4. Form the linearization \(L(x) = f(a) + f'(a)(x - a)\):

   \(L(x) = 2 + \frac{1}{4}(x - 4)\).

5. Approximate \(\sqrt{4.001}\):

   Use \(x = 4.001\).

   \(\sqrt{4.001} \approx L(4.001) = 2 + \frac{1}{4}(4.001 - 4) = 2 + \frac{1}{4}(0.001) = 2 + 0.00025 = 2.00025\).

   (Actual value \(\sqrt{4.001} \approx 2.000249968...\))

These examples demonstrate the practical utility of linearization for approximating function values without a calculator.

For \(\ln(1.01)\), choosing \(f(x)=\ln(1+x)\) and \(a=0\) is strategic because \(\ln(1)=0\) and \(f'(0)\) is easy to compute. This leads to the simple approximation \(L(x)=x\). So, \(\ln(1.01) \approx 0.01\). Notice how close this is to the actual value.

For \(\sqrt{4.001}\), we choose \(f(x)=\sqrt{x}\) and \(a=4\), because 4 is a perfect square close to 4.001, and \(f(4)\) and \(f'(4)\) are easy to calculate. We build the tangent line equation and then plug in \(x=4.001\). The approximation \(2.00025\) is very accurate.

These examples show how to choose the function \(f(x)\) and the point \(a\) wisely to make the calculations as simple as possible while maintaining a good approximation. The closer \(x\) is to \(a\), the better the approximation.

Example: Approximating \(\sqrt[3]{7.99}\)

Example 5.5.4: Approximate \(\sqrt[3]{7.99}\) by linearization.

Question: Which function \(f(x)\) and at which point \(a\) would you choose?

Solution:

1. Choose \(f(x)\): The value we want to approximate is a cube root, so let \(f(x) = \sqrt[3]{x} = x^{1/3}\).
2. Choose \(a\): We need a point \(a\) near \(7.99\) for which \(f(a)\) and \(f'(a)\) are easy to calculate. The closest perfect cube to \(7.99\) is \(8\). So, we choose \(a = 8\).

Now, we follow the steps for linearization:

1. Find \(f(a)\): \(f(8) = \sqrt[3]{8} = 2\).

2. Find \(f'(x)\): \(f'(x) = \frac{d}{dx}(x^{1/3}) = \frac{1}{3}x^{(1/3) - 1} = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}}\).

3. Find \(f'(a)\): \(f'(8) = \frac{1}{3(8)^{2/3}} = \frac{1}{3(\sqrt[3]{8})^2} = \frac{1}{3(2)^2} = \frac{1}{3 \cdot 4} = \frac{1}{12}\).

4. Form the linearization \(L(x) = f(a) + f'(a)(x - a)\):

   \(L(x) = 2 + \frac{1}{12}(x - 8)\).

5. Approximate \(\sqrt[3]{7.99}\):

   Use \(x = 7.99\).

   \(\sqrt[3]{7.99} \approx L(7.99) = 2 + \frac{1}{12}(7.99 - 8) = 2 + \frac{1}{12}(-0.01)\).

   \(L(7.99) = 2 - \frac{0.01}{12} = 2 - \frac{1}{1200} \approx 2 - 0.0008333 = 1.9991667\).

   (Actual value \(\sqrt[3]{7.99} \approx 1.9991663...\))

This example further reinforces the strategy for using linearization. The first crucial step is always to identify the function \(f(x)\) and the nearby point \(a\) that will simplify your calculations.

Here, since we want to approximate a cube root, \(f(x) = x^{1/3}\) is the natural choice. The number \(7.99\) is very close to \(8\), which is a perfect cube, making \(a=8\) an ideal point.

Once \(f(x)\) and \(a\) are chosen, the rest is mechanical: find \(f(a)\), find \(f'(x)\), evaluate \(f'(a)\), construct the linearization \(L(x)\), and then plug in the value you want to approximate. The arithmetic can be done by hand, and the result is a very good approximation. This method is a testament to the power of derivatives in practical estimation.

Estimation of Change using Differentials

Differentials provide a way to estimate the change in a function \(\Delta f\) based on a small change in its input \(dx\).

Suppose \(x\) changes from \(x = a\) to \(x = a + dx\) (i.e., \(\Delta x = dx\)).

The actual change in \(f\) is: \[ \Delta f = f (a + d x) - f (a). \]

Differentials (\(df\))

When the change \(dx\) is small, we approximate \(\Delta f\) using the change in the linearization (tangent line).

The differential of \(f\), denoted \(df\), is defined as: \[ d f = f ^ {\prime} (a) d x. \] For small \(dx\), we have the approximation: \[ \Delta f \approx d f, \quad \text{or} \quad \Delta f \approx \frac {d f}{d x} d x. \]

Estimation of Change using Differentials

Remarks on Changes:

Type of Change	Actual Change	Estimated Change
Absolute Change	\(\Delta f = f(a+dx) - f(a)\)	\(df = f'(a)dx\)
Relative Change	\(\frac{\Delta f}{f(a)}\)	\(\frac{df}{f(a)}\)
Percentage Change	\(100 \frac{\Delta f}{f(a)}\)	\(100 \frac{df}{f(a)}\)

Warning

The derivative \(\frac{df}{dx}\) is NOT the quotient of the differential \(df\) and the change \(dx\) in the strict sense, but \(df = \frac{df}{dx}dx\) is a very useful formula for approximations.

Differentials offer a slightly different perspective on linearization, focusing on the change in the function’s output rather than its absolute value.

When \(x\) changes by a small amount \(dx\), the function \(f(x)\) changes by \(\Delta f\). We approximate this actual change \(\Delta f\) using the tangent line. The change along the tangent line is called the differential \(df\).

The formula \(df = f'(a)dx\) is extremely powerful. It essentially says that for a small change in \(x\), the change in \(f\) is approximately the slope of the tangent at \(a\) times the change in \(x\).

The table neatly summarizes how differentials can be used to estimate absolute, relative, and percentage changes, which are common metrics in engineering and science. The warning note is important: while we often write \(\frac{df}{dx}\), the ‘d’ terms are not strictly separate numbers that can be divided like a fraction until we define differentials. However, the operational formula \(df = f'(x)dx\) works perfectly for approximations.

Example: Error Estimation with Differentials

Example 5.5.5: Error in Circular Disk Area

The radius of a circular disk is \(24~\mathrm{cm}\) with a maximum error in measurement of \(0.2~\mathrm{cm}\).

1. Use differentials to estimate the maximum error in the calculated area of the disk.
2. What is the relative error? What is the percentage error?

Solution:

Let the radius be \(r\). The area is \(A(r) = \pi r^2\).

The derivative is \(A'(r) = 2\pi r\).

The differential of \(A\) is \(dA = A'(r) dr\).

Given values: \(r = 24~\mathrm{cm}\) and \(dr = \pm 0.2~\mathrm{cm}\) (maximum error).

1. Maximum error in calculated area (\(dA\)):

   \(dA = A'(24)dr = 2\pi (24)(0.2) = 48\pi (0.2) = 9.6\pi~\mathrm{cm}^2\).

   This is approximately \(9.6 \times 3.14159 \approx 30.16 \mathrm{cm}^2\).

2. Relative error and Percentage error:

   First, calculate the area at \(r=24\): \(A(24) = \pi (24)^2 = 576\pi~\mathrm{cm}^2\).

   Relative error: \(\frac{dA}{A(r)} = \frac{9.6\pi}{576\pi} = \frac{9.6}{576} = \frac{1}{60} \approx 0.01667\).

   Percentage error: \((0.01667) \times 100\% \approx 1.667\%\).

This example demonstrates a practical application of differentials in error analysis. In engineering and scientific measurements, it’s common to have some uncertainty (error) in a measured quantity and need to estimate how that uncertainty propagates to a calculated quantity.

Here, the measured quantity is the radius \(r\), and the calculated quantity is the area \(A\).

• We first express the calculated quantity \(A\) as a function of the measured quantity \(r\).
• Then, we find the derivative \(A'(r)\), which gives us the instantaneous rate of change of area with respect to radius.
• The maximum error in measurement of \(r\) is \(dr\).
• Using the differential \(dA = A'(r)dr\), we estimate the maximum absolute error in the area.

Beyond the absolute error, relative and percentage errors provide a more normalized view of the error, making it easier to compare across different scales. The calculations show that even a small error in radius can lead to a noticeable error in area, but the percentage error helps contextualize it. This is a fundamental concept in experimental science and quality control.

5.6 Newton’s Method

Newton’s Method (also Newton-Raphson method) is a numerical technique to approximate solutions to an equation \(f(x) = 0\).

It’s an iterative method that uses the tangent line to refine an initial guess.

Idea of Newton’s Method:

1. Start with an initial guess \(x_0\) close to the root.
2. Draw the tangent line to \(f(x)\) at \((x_0, f(x_0))\).
3. The \(x\)-intercept of this tangent line becomes the next approximation, \(x_1\).
4. Repeat the process.

Figure: Illustration of Newton’s Method, showing how tangent lines are used to iteratively approximate a root of a function.

5.6 Newton’s Method

Iteration Formula:

The equation of the tangent at \((x_n, f(x_n))\) is \(y - f(x_n) = f'(x_n)(x - x_n)\).

To find the \(x\)-intercept \(x_{n+1}\), set \(y=0\):

\(0 - f(x_n) = f'(x_n)(x_{n+1} - x_n)\).

Solving for \(x_{n+1}\): \[ x _ {n + 1} = x _ {n} - \frac {f (x _ {n})}{f ^ {\prime} (x _ {n})}. \]

Newton’s Method is a powerful numerical technique for finding roots of functions (\(f(x)=0\)). It’s an iterative process that refines an initial guess by repeatedly drawing tangent lines.

The core idea is geometric: if you have an initial guess \(x_0\), you find the tangent line to the function at that point. The point where this tangent line crosses the x-axis, \(x_1\), is usually a better approximation of the root than \(x_0\). You then repeat this process: find the tangent at \(x_1\), find its x-intercept \(x_2\), and so on.

The iteration formula is derived directly from the equation of the tangent line. By setting \(y=0\) (because we’re looking for the x-intercept), we can algebraically solve for \(x_{n+1}\) in terms of \(x_n\), \(f(x_n)\), and \(f'(x_n)\). This formula is the heart of Newton’s Method. It requires the function to be differentiable and its derivative not to be zero at the current approximation.

Example: Applying Newton’s Method

Example 5.6.1: Use Newton’s method to solve \(x^{3} - x + 1 = 0\).

Solution:

Let \(f(x) = x^3 - x + 1\).

Then \(f'(x) = 3x^2 - 1\).

First, find an initial guess:

\(f(-1) = (-1)^3 - (-1) + 1 = -1 + 1 + 1 = 1\).

\(f(-2) = (-2)^3 - (-2) + 1 = -8 + 2 + 1 = -5\).

Since \(f(-1)\) is positive and \(f(-2)\) is negative, a root lies between -2 and -1 (by IVT). Let’s choose \(x_0 = -1\).

Newton’s iteration formula: \[ x _ {n + 1} = x _ {n} - \frac {x _ {n} ^ {3} - x _ {n} + 1}{3 x _ {n} ^ {2} - 1}. \]

Example: Applying Newton’s Method (cont.)

Let’s compute the first few iterates:

• \(x_0 = -1\)
• \(x_1 = -1 - \frac{(-1)^3 - (-1) + 1}{3(-1)^2 - 1} = -1 - \frac{1}{2} = -1.5\).
• \(x_2 = -1.5 - \frac{(-1.5)^3 - (-1.5) + 1}{3(-1.5)^2 - 1} = -1.5 - \frac{-0.875}{5.75} \approx -1.5 + 0.15217 = -1.34783\).
• \(x_3 = -1.34783 - \frac{f(-1.34783)}{f'(-1.34783)} \approx -1.34783 - \frac{-0.10068}{4.449905} \approx -1.32520\).
• \(x_4 = -1.32520 - \frac{f(-1.32520)}{f'(-1.32520)} \approx -1.32520 - \frac{-0.002058}{4.264635} \approx -1.324718\).
• \(x_5 = -1.324718 - \frac{f(-1.324718)}{f'(-1.324718)} \approx -1.324718 - \frac{-9.2 \times 10^{-7}}{4.264633} \approx -1.324718\).

Approximated root (to five decimal places): \(x \approx -1.32472\).

Here we apply Newton’s Method step-by-step to find a root of a cubic equation.

First, we define \(f(x)\) and find its derivative \(f'(x)\). An initial guess is crucial; using the Intermediate Value Theorem, we can determine an interval where a root exists and pick a point within or near that interval. Here, \(x_0 = -1\) is a reasonable start.

Then, we iteratively apply the formula. Notice how quickly the approximations converge to the actual root. In just a few iterations, we go from an initial guess of -1 to a highly accurate approximation. This rapid convergence is one of the strengths of Newton’s Method, provided the initial guess is sufficiently close to the root and \(f'(x)\) is not zero or close to zero near the root. The table clearly illustrates this convergence.

Other Uses of Newton’s Method

Newton’s method isn’t just for polynomial roots; it’s used for various numerical approximations.

Example 5.6.2: Approximating Reciprocals (\(\frac{1}{\alpha}\))

Given \(\alpha \neq 0\), estimate \(x = \frac{1}{\alpha}\).

Solution:

The equation is \(x = \frac{1}{\alpha}\), which can be rewritten as \(\frac{1}{x} - \alpha = 0\).

Let \(f(x) = \frac{1}{x} - \alpha\).

Then \(f'(x) = -\frac{1}{x^2}\).

Apply Newton’s iteration formula: \[ \begin{array}{l} x _ {n + 1} = x _ {n} - \frac {f (x _ {n})}{f ^ {\prime} (x _ {n})} = x _ {n} - \frac {\frac {1}{x _ {n}} - \alpha}{- \frac {1}{x _ {n} ^ {2}}} \\ = x _ {n} + x _ {n} ^ {2} \left(\frac {1}{x _ {n}} - \alpha\right) \\ = x _ {n} + x _ {n} - \alpha x _ {n} ^ {2} = 2x_n - \alpha x_n^2 \\ = x _ {n} \left(2 - \alpha x _ {n}\right). \\ \end{array} \] This iterative formula only involves multiplication and subtraction, which were computationally faster than division on early computers.

Example 5.6.3: Approximating Square Roots (\(\sqrt{\alpha}\))

Estimate \(x = \sqrt{\alpha}\) for \(\alpha > 0\).

Solution:

The equation is \(x = \sqrt{\alpha}\), which can be rewritten as \(x^2 = \alpha\), or \(x^2 - \alpha = 0\).

Let \(f(x) = x^2 - \alpha\).

Then \(f'(x) = 2x\).

Apply Newton’s iteration formula:

\[ \begin{array}{l} x _ {n + 1} = x _ {n} - \frac {f (x _ {n})}{f ^ {\prime} (x _ {n})} = x _ {n} - \frac {x _ {n} ^ {2} - \alpha}{2 x _ {n}} \\ = \frac {2 x _ {n} ^ {2} - (x _ {n} ^ {2} - \alpha)}{2 x _ {n}} = \frac {x _ {n} ^ {2} + \alpha}{2 x _ {n}} \\ = \frac {x _ {n}}{2} + \frac {\alpha}{2 x _ {n}} = \frac{1}{2}\left(x_n + \frac{\alpha}{x_n}\right). \\ \end{array} \] This is also known as the Babylonian method for square roots.

Newton’s Method is incredibly versatile. It’s not limited to finding roots of abstract equations; it can be tailored to perform specific arithmetic operations that were historically complex for computers.

For approximating reciprocals, we set up \(f(x) = 1/x - \alpha = 0\). Applying the Newton’s formula results in a simplified iterative expression, \(x_{n+1} = x_n(2 - \alpha x_n)\). The beauty here is that this formula avoids division by \(\alpha\), which could be computationally expensive or lead to precision issues on older hardware.

Similarly, for approximating square roots, we set up \(f(x) = x^2 - \alpha = 0\). The resulting iteration formula, \(x_{n+1} = \frac{1}{2}(x_n + \frac{\alpha}{x_n})\), is a famous algorithm known as the Babylonian method. This formula converges very quickly and is still the basis for how many calculators and computer programs compute square roots. These examples show the ingenuity of using calculus for efficient numerical computation.

Example: Approximating \(\sqrt{3}\)

Example 5.6.4: Approximate the value of \(\sqrt{3}\).

Solution:

We use the iteration formula for square roots: \(x _ {n + 1} = \frac {1}{2} \left(x _ {n} + \frac {\alpha}{x _ {n}}\right)\).

Let \(\alpha = 3\). We need an initial guess \(x_0\).

Since \(1^2 = 1\) and \(2^2 = 4\), \(\sqrt{3}\) is between 1 and 2. Let’s choose \(x_0 = 2\).

Iterates:

• \(x_0 = 2\).
• \(x_1 = \frac{1}{2}\left(2 + \frac{3}{2}\right) = \frac{1}{2}\left(\frac{4+3}{2}\right) = \frac{1}{2}\left(\frac{7}{2}\right) = \frac{7}{4} = 1.75\).
• \(x_2 = \frac{1}{2}\left(\frac{7}{4} + \frac{3}{7/4}\right) = \frac{1}{2}\left(\frac{7}{4} + \frac{12}{7}\right) = \frac{1}{2}\left(\frac{49+48}{28}\right) = \frac{97}{56} \approx 1.732142857\).
• \(x_3 = \frac{1}{2}\left(\frac{97}{56} + \frac{3}{97/56}\right) = \frac{1}{2}\left(\frac{97}{56} + \frac{168}{97}\right) = \frac{1}{2}\left(\frac{97^2 + 56 \cdot 168}{56 \cdot 97}\right) = \frac{9409 + 9408}{10832} = \frac{18817}{10832} \approx 1.73205081\).

(Actual value \(\sqrt{3} \approx 1.73205081\))

In just 3 iterations, we achieved high accuracy!

This example demonstrates the rapid convergence of Newton’s Method, specifically the Babylonian method for square roots.

We apply the iterative formula \(x_{n+1} = \frac{1}{2}(x_n + \frac{\alpha}{x_n})\) with \(\alpha = 3\). An initial guess of \(x_0=2\) is chosen, as it’s a nearby integer whose square is close to 3.

Observe how quickly the approximations improve. After just one iteration, \(x_1=1.75\) is already a decent approximation. By the third iteration, \(x_3\) is accurate to many decimal places, matching the true value of \(\sqrt{3}\) quite closely. This high efficiency is why Newton’s method (and its specialized forms like the Babylonian method) is so widely used in numerical computation.

5.7 Closed Interval Method for Optimization

Optimization problems involve finding the best possible solution(s), often by maximizing or minimizing a function.

Critical Points:

Critical points are points \(c\) where:

1. \(f'(c) = 0\) (these are called stationary points).
2. \(f'(c)\) fails to exist (these are called singular points).

Extreme Value Theorem Review:

A continuous function \(f\) on a closed and bounded interval \([a, b]\) always attains its global maximum and global minimum values within that interval.

Closed Interval Method (3-Step Procedure):

To find the global maximum and absolute minimum of a continuous function \(f\) on \([a, b]\):

1. Find Critical Points: Determine all critical points of \(f\) in the open interval \((a, b)\) and find the corresponding \(f\)-values.
2. Evaluate Endpoints: Compute \(f(a)\) and \(f(b)\).
3. Compare Values: The largest value of \(f\) from Steps 1 and 2 is the global maximum. The smallest value is the global minimum.

Optimization is a core application of differentiation in all branches of engineering and science. It’s about finding the “best” outcome, whether that’s minimizing cost, maximizing efficiency, or finding the strongest design.

The concept of “critical points” is central to optimization. These are the candidate points where a function’s slope is either zero (a stationary point, like the peak of a hill or the bottom of a valley) or undefined (a singular point, like the sharp corner we saw with \(|x|\)).

For functions defined on a closed and bounded interval, the Extreme Value Theorem guarantees that a global maximum and minimum exist. The Closed Interval Method provides a systematic, three-step procedure to find these global extrema:

1. Identify all critical points within the interval.
2. Evaluate the function at these critical points AND at the endpoints of the interval.
3. Compare all these function values to find the absolute largest and smallest. This method ensures you find the true global maximum and minimum, not just local ones.

Example: Optimization on a Closed Interval

Example 5.7.1: Find the global maximum and global minimum of \(f(t) = \sqrt[3]{t} (8 - t)\) on \([-1, 8]\).

Solution:

The function \(f(t) = t^{1/3}(8-t) = 8t^{1/3} - t^{4/3}\) is continuous on \([-1, 8]\).

Thus, by the Extreme Value Theorem, global extrema exist.

1. Find Critical Points in \((-1, 8)\):

   First, find \(f'(t)\):

   \(f'(t) = \frac{d}{dt}(8t^{1/3} - t^{4/3}) = 8 \cdot \frac{1}{3}t^{-2/3} - \frac{4}{3}t^{1/3}\)

   \(f'(t) = \frac{8}{3t^{2/3}} - \frac{4t^{1/3}}{3} = \frac{8 - 4t}{3t^{2/3}} = \frac{8 - 4t}{3(\sqrt[3]{t})^2}\).

   • Singular Points: \(f'(t)\) is undefined when the denominator is zero, so \(3t^{2/3} = 0 \Rightarrow t = 0\). \(t=0\) is in \((-1, 8)\). This is a critical point.
   • Stationary Points: \(f'(t) = 0\) when the numerator is zero, so \(8 - 4t = 0 \Rightarrow 4t = 8 \Rightarrow t = 2\). \(t=2\) is in \((-1, 8)\). This is a critical point.

Example: Optimization on a Closed Interval (cont.)

2. Evaluate Endpoints:

   • \(f(-1) = \sqrt[3]{-1}(8 - (-1)) = (-1)(9) = -9\).
   • \(f(8) = \sqrt[3]{8}(8 - 8) = (2)(0) = 0\).

3. Compare Values:

   • \(f(0) = \sqrt[3]{0}(8 - 0) = 0\).
   • \(f(2) = \sqrt[3]{2}(8 - 2) = 6\sqrt[3]{2} \approx 6 \times 1.26 = 7.56\).
   • \(f(-1) = -9\).
   • \(f(8) = 0\).

Conclusion:

• Global Maximum: \(f(2) = 6\sqrt[3]{2}\).
• Global Minimum: \(f(-1) = -9\).

Let’s apply the Closed Interval Method to find the global extrema of \(f(t) = \sqrt[3]{t} (8 - t)\) on \([-1, 8]\).

First, we confirm the function is continuous on the closed interval, so extrema are guaranteed. Next, we find the critical points. This involves differentiating the function. It’s often helpful to rewrite the cube root as \(t^{1/3}\) to use the power rule. When finding \(f'(t)\), we encounter both a singular point (where \(f'(t)\) is undefined, here at \(t=0\)) and a stationary point (where \(f'(t)=0\), here at \(t=2\)). Both are important candidates for extrema.

Then, we evaluate the function at these critical points and at the endpoints of the interval (\(t=-1\) and \(t=8\)). This gives us a list of potential maximum and minimum values.

Finally, we compare all these function values. The largest value is the global maximum, and the smallest is the global minimum. Notice that the maximum occurs at a stationary point, and the minimum occurs at an endpoint in this example. This highlights why checking both critical points and endpoints is crucial.

Example: More Optimization on a Closed Interval

Example 5.7.2: Let \(f(x) = (x^{2} - 1)^{2 / 3}\). Find the global maximum and global minimum values of \(f\) on the interval \([-3, 3]\).

Solution:

The function \(f(x) = ((x^2 - 1)^{1/3})^2\) is continuous on \([-3, 3]\).

By the Extreme Value Theorem, global extrema exist.

1. Find Critical Points:

   First, find \(f'(x)\) using the Chain Rule:

   \(f'(x) = \frac{2}{3}(x^2 - 1)^{-1/3} \cdot (2x) = \frac{4x}{3(x^2 - 1)^{1/3}}\).

   • Stationary Point: \(f'(x) = 0\) when the numerator is zero.

     \(4x = 0 \Rightarrow x = 0\).

     \(x=0\) is in \((-3, 3)\).

   • Singular Points: \(f'(x)\) fails to exist when the denominator is zero.

     \(3(x^2 - 1)^{1/3} = 0 \Rightarrow x^2 - 1 = 0 \Rightarrow (x - 1)(x + 1) = 0\).

     So, \(x = 1\) and \(x = -1\).

     Both \(x=1\) and \(x=-1\) are in \((-3, 3)\).

2. Evaluate Endpoints:

   • \(f(-3) = ((-3)^2 - 1)^{2/3} = (9 - 1)^{2/3} = (8)^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4\).
   • \(f(3) = (3^2 - 1)^{2/3} = (9 - 1)^{2/3} = (8)^{2/3} = 2^2 = 4\).

3. Compare Values:

   • \(f(0) = (0^2 - 1)^{2/3} = (-1)^{2/3} = (\sqrt[3]{-1})^2 = (-1)^2 = 1\).
   • \(f(-1) = ((-1)^2 - 1)^{2/3} = (1 - 1)^{2/3} = 0^{2/3} = 0\).
   • \(f(1) = (1^2 - 1)^{2/3} = (1 - 1)^{2/3} = 0^{2/3} = 0\).
   • \(f(-3) = 4\).
   • \(f(3) = 4\).

Conclusion:

• Global Maximum: The largest value is \(4\), which occurs at \(x = -3\) and \(x = 3\).
• Global Minimum: The smallest value is \(0\), which occurs at \(x = -1\) and \(x = 1\).

This example further reinforces the application of the Closed Interval Method, with a function involving fractional exponents.

Again, continuity on the closed interval guarantees extrema. We calculate the derivative using the chain rule, which is essential for \((x^2-1)^{2/3}\). This derivative reveals both stationary points (where \(f'(x)=0\)) and singular points (where \(f'(x)\) is undefined due to the denominator becoming zero). In this case, we have three critical points: \(0, 1, -1\).

Next, we evaluate the function at these critical points and at the endpoints of the interval. A common mistake is to forget the endpoints!

Finally, by comparing all the \(f\)-values, we identify the global maximum and minimum. Notice that in this example, the global maximum occurs at the endpoints, and the global minimum occurs at the singular points. This highlights the importance of considering all types of critical points and endpoints.