Pure Mathematics | Hyperbolic functions

🚀 FP3 Chapter Notes: Hyperbolic Functions (The Exponential Cousins)

Welcome to the exciting world of Hyperbolic Functions! Don't worry if the name sounds intimidating—these functions are essentially just cleverly disguised combinations of the exponential function, \(e^x\). They are the "cousins" of the standard trigonometric functions (sin, cos, tan), and they pop up everywhere in physics and engineering, especially when dealing with shapes formed by hanging cables or motion in resistive media.

In this chapter, we will master their definitions, explore their unique identities, and learn how to differentiate and integrate them. Let's dive in!

1. Defining the Hyperbolic Functions

1.1 Definitions in Terms of Exponentials

The three main hyperbolic functions are derived directly from the exponential function \(e^x\). They are defined as follows:

• Hyperbolic Sine (sinh x):
  \( \sinh x = \frac{e^x - e^{-x}}{2} \)
• Hyperbolic Cosine (cosh x):
  \( \cosh x = \frac{e^x + e^{-x}}{2} \)
• Hyperbolic Tangent (tanh x): (Defined as \(\frac{\sinh x}{\cosh x}\))
  \( \tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} \)

⭐ Memory Aid:
Think of the 'h' in hyperbolic.

Cosh has the plus sign (like the letter C usually means positive or cosine).
Sinh has the minus sign (Sine often relates to odd functions, and this formula uses subtraction).

1.2 Graphs and Properties

Understanding the graphs is key to knowing their domains and ranges.

i) Cosh x (The Catenary Curve)

• Domain: \(x \in \mathbb{R}\)
• Range: \(\cosh x \geq 1\)
• Shape: Looks similar to a parabola (\(y=x^2\)), but it’s flatter at the bottom. This shape is physically known as a catenary, which is the curve formed by a cable or chain hanging freely between two points (e.g., power lines).
• Symmetry: It is an even function (\(\cosh(-x) = \cosh x\)).

ii) Sinh x

• Domain: \(x \in \mathbb{R}\)
• Range: \(\sinh x \in \mathbb{R}\)
• Shape: Always increasing, passing through the origin \((0, 0)\).
• Symmetry: It is an odd function (\(\sinh(-x) = -\sinh x\)).

2. Fundamental Hyperbolic Identities

2.1 The Master Identity (The Crucial Difference)

In circular trigonometry, the master identity is \(\cos^2 \theta + \sin^2 \theta = 1\).

For hyperbolic functions, the sign changes:

$$ \mathbf{\cosh^2 x - \sinh^2 x = 1} $$

⚠️ Common Mistake Alert: Students often mistakenly write \(\cosh^2 x + \sinh^2 x = 1\). Remember, in the hyperbolic world, the minus sign is mandatory!

Did you know? This identity is why they are called hyperbolic functions! The set of points \((x, y)\) such that \(x^2 - y^2 = 1\) defines a hyperbola.

2.2 Derived Identities

By dividing the master identity \(\cosh^2 x - \sinh^2 x = 1\) by \(\cosh^2 x\) or \(\sinh^2 x\), we get two further identities (just like in circular trig):

i) Dividing by \(\cosh^2 x\): $$ 1 - \tanh^2 x = \operatorname{sech}^2 x $$ (Where \(\operatorname{sech} x = \frac{1}{\cosh x}\))

ii) Dividing by \(\sinh^2 x\): $$ \coth^2 x - 1 = \operatorname{cosech}^2 x $$ (Where \(\coth x = \frac{1}{\tanh x}\) and \(\operatorname{cosech} x = \frac{1}{\sinh x}\))

2.3 Double Angle Formulas

The addition formulas for sinh and cosh are very similar to circular trig, but watch the signs!

$$ \cosh(A+B) = \cosh A \cosh B + \sinh A \sinh B $$ $$ \sinh(A+B) = \sinh A \cosh B + \cosh A \sinh B $$

By setting \(A=B=x\), we get the double angle identities:

• \(\cosh 2x\) (three forms): $$ \cosh 2x = \cosh^2 x + \sinh^2 x $$ $$ \cosh 2x = 2\cosh^2 x - 1 $$ $$ \cosh 2x = 1 + 2\sinh^2 x $$
• \(\sinh 2x\): $$ \sinh 2x = 2 \sinh x \cosh x $$

Remember: All these identities are proven using the basic exponential definitions in Section 1. If you get stuck in an exam, go back to the definitions!

3. Differentiation of Hyperbolic Functions

Differentiating hyperbolic functions is usually easier than differentiating circular functions because you often don't have to worry about the minus signs!

3.1 Standard Derivatives

These results are mandatory for FP3 and can be proved using the exponential definitions.

• $$ \frac{d}{dx}(\sinh x) = \cosh x $$
• $$ \frac{d}{dx}(\cosh x) = \sinh x $$ (Notice: NO minus sign here, unlike \(\frac{d}{dx}(\cos x) = -\sin x\). This is a relief!)
• $$ \frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x $$
• $$ \frac{d}{dx}(\operatorname{coth} x) = -\operatorname{cosech}^2 x $$
• $$ \frac{d}{dx}(\operatorname{sech} x) = -\operatorname{sech} x \tanh x $$
• $$ \frac{d}{dx}(\operatorname{cosech} x) = -\operatorname{cosech} x \coth x $$

3.2 Using the Chain Rule

Just like in standard calculus, if you have a function of a function, use the chain rule: $$ \frac{d}{dx} (f(g(x))) = f'(g(x)) \cdot g'(x) $$

Example: Find the derivative of \(y = \cosh(3x^2 + 1)\).

Step 1: Identify the inner function \(g(x) = 3x^2 + 1\). Thus \(g'(x) = 6x\).
Step 2: Differentiate the outer function: \(\frac{d}{du}(\cosh u) = \sinh u\).
Step 3: Combine: $$ \frac{dy}{dx} = \sinh(3x^2 + 1) \cdot (6x) = 6x \sinh(3x^2 + 1) $$

4. Inverse Hyperbolic Functions (Area Functions)

4.1 Definitions and Notation

The inverse hyperbolic functions are also known as the Area Functions (written as arsinh, arcosh, artanh, etc.).

If \(y = \sinh x\), then \(x = \operatorname{arsinh} y\).

These functions are defined in terms of natural logarithms (\(\ln\)). These logarithmic forms are crucial, as they are used to solve equations and prove differentiation results.

i) Inverse Sinh (Area Hyperbolic Sine): $$ \operatorname{arsinh} x = \ln(x + \sqrt{x^2 + 1}), \quad x \in \mathbb{R} $$

ii) Inverse Cosh (Area Hyperbolic Cosine): $$ \operatorname{arcosh} x = \ln(x + \sqrt{x^2 - 1}), \quad x \geq 1 $$ Note: Because \(\cosh x \geq 1\), the domain of \(\operatorname{arcosh} x\) is restricted to \(x \geq 1\).

iii) Inverse Tanh (Area Hyperbolic Tangent): $$ \operatorname{artanh} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right), \quad |x| < 1 $$ Note: Because \(\tanh x\) is strictly between -1 and 1, the domain of \(\operatorname{artanh} x\) is restricted to \(-1 < x < 1\).

4.2 Solving Equations Using Log Forms (A Step-by-Step Example)

If you need to solve an equation like \(\sinh x = 5\), you can use the log form or go straight back to the exponential definition.

Example: Solve \(\sinh x = 5\).

Method 1: Using the log definition: $$ x = \operatorname{arsinh} 5 = \ln(5 + \sqrt{5^2 + 1}) $$ $$ x = \ln(5 + \sqrt{26}) $$

Method 2: Using the exponential definition (often required for proofs):
$$ \frac{e^x - e^{-x}}{2} = 5 $$ $$ e^x - e^{-x} = 10 $$ Multiply by \(e^x\) and rearrange into a quadratic in \(e^x\). Let \(y = e^x\): $$ y - \frac{1}{y} = 10 \implies y^2 - 1 = 10y $$ $$ y^2 - 10y - 1 = 0 $$ Using the quadratic formula, since \(y = e^x\) must be positive: $$ e^x = 5 + \sqrt{25 - 4(1)(-1)} / 2 = 5 + \sqrt{26} $$ $$ x = \ln(5 + \sqrt{26}) $$

5. Differentiation of Inverse Hyperbolic Functions

The derivatives of inverse hyperbolic functions are essential because they define the standard integration forms you must recognize in the final section.

5.1 Standard Differentiation Results

When you differentiate the logarithmic forms (Section 4.1), you arrive at these simple algebraic fractions:

• $$ \frac{d}{dx}(\operatorname{arsinh} x) = \frac{1}{\sqrt{x^2 + 1}} $$
• $$ \frac{d}{dx}(\operatorname{arcosh} x) = \frac{1}{\sqrt{x^2 - 1}}, \quad x > 1 $$
• $$ \frac{d}{dx}(\operatorname{artanh} x) = \frac{1}{1 - x^2}, \quad |x| < 1 $$

5.2 Proof Sketch (Why are they so simple?)

Let's see why \(\frac{d}{dx}(\operatorname{arsinh} x) = \frac{1}{\sqrt{x^2 + 1}}\).
If \(y = \operatorname{arsinh} x\), then \(x = \sinh y\).
Differentiate implicitly with respect to \(x\): $$ 1 = \cosh y \cdot \frac{dy}{dx} $$ $$ \frac{dy}{dx} = \frac{1}{\cosh y} $$ Using the identity \(\cosh^2 y - \sinh^2 y = 1\), we know that \(\cosh y = \sqrt{1 + \sinh^2 y}\) (we take the positive root since \(\cosh y > 0\)).
Since \(x = \sinh y\), substitute \(x\) back in: $$ \frac{dy}{dx} = \frac{1}{\sqrt{1 + x^2}} $$ See? The identities make the calculus magic happen!

6. Integration

Integration of hyperbolic functions involves reversing the differentiation process.

6.1 Direct Integrals

These are the basic results:

• $$ \int \sinh x \, dx = \cosh x + C $$
• $$ \int \cosh x \, dx = \sinh x + C $$
• $$ \int \operatorname{sech}^2 x \, dx = \tanh x + C $$

Remember: Just like differentiation, the signs are usually positive, making integration straightforward.

6.2 Standard Integrals Leading to Inverse Hyperbolic Functions

This is often the most tested area of the chapter. You must be able to recognize these integral forms, as they relate directly to the derivatives of the inverse functions (Section 5.1).

The following standard results are derived from the differentiation of the logarithmic forms. When integrating these forms, you can often use a substitution or "completion of the square," but ultimately, the integral will yield an inverse hyperbolic function.

i) Forms leading to \(\operatorname{arsinh}\) (or \(\ln\) form): $$ \int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \operatorname{arsinh}\left(\frac{x}{a}\right) + C = \ln\left(x + \sqrt{x^2 + a^2}\right) + C $$ (Note: Since \(a\) is usually 1 in the log form, for a general \(a\), the \(\ln\) form simplifies slightly as the term involving \(\ln a\) is absorbed into the constant \(C\).)

ii) Forms leading to \(\operatorname{arcosh}\) (or \(\ln\) form): $$ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \operatorname{arcosh}\left(\frac{x}{a}\right) + C = \ln\left(x + \sqrt{x^2 - a^2}\right) + C, \quad |x| > a $$

iii) Forms leading to \(\operatorname{artanh}\) (or \(\ln\) form): $$ \int \frac{1}{a^2 - x^2} \, dx = \frac{1}{a}\operatorname{artanh}\left(\frac{x}{a}\right) + C = \frac{1}{2a}\ln\left(\frac{a+x}{a-x}\right) + C, \quad |x| < a $$

iv) Forms leading to \(\operatorname{arcoth}\): $$ \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{a}\operatorname{arcoth}\left(\frac{x}{a}\right) + C = \frac{1}{2a}\ln\left(\frac{x-a}{x+a}\right) + C, \quad |x| > a $$

6.3 Dealing with More Complex Integrands

When the integral doesn't exactly match the standard forms, you often need to use completing the square in the denominator or the root.

Example: Integrate \(\int \frac{1}{\sqrt{x^2 + 6x + 13}} \, dx\).

Step 1: Complete the square on the denominator: $$ x^2 + 6x + 13 = (x+3)^2 - 9 + 13 = (x+3)^2 + 4 $$ Step 2: Rewrite the integral: $$ \int \frac{1}{\sqrt{(x+3)^2 + 2^2}} \, dx $$ Step 3: Use substitution (\(u = x+3\), \(du = dx\)). This matches the \(\int \frac{1}{\sqrt{u^2 + a^2}} \, du\) form, where \(a=2\).
Step 4: Apply the standard result for \(\operatorname{arsinh}\): $$ \ln\left(u + \sqrt{u^2 + a^2}\right) + C $$ Step 5: Substitute back \(u = x+3\) and \(a=2\): $$ \ln\left( (x+3) + \sqrt{(x+3)^2 + 4} \right) + C $$ $$ \ln\left( x+3 + \sqrt{x^2 + 6x + 13} \right) + C $$