Problem 24 - Calculus

Question

Compute \(\sinh (\ln 5)\) and \(\tanh (2 \ln 4)\)

Step-by-Step Solution

Answer

\( \sinh(\ln 5) = 2.4 \) and \( \tanh(2 \ln 4) = \frac{255}{257} \).

Step 1: Understanding Hyperbolic Sine

The hyperbolic sine function is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \). We need to compute \( \sinh(\ln 5) \).

Step 2: Simplifying \( \sinh(\ln 5) \)

Substitute \( x = \ln 5 \) into the hyperbolic sine definition: \[ \sinh(\ln 5) = \frac{e^{\ln 5} - e^{-\ln 5}}{2} = \frac{5 - \frac{1}{5}}{2} \]. This simplifies to \( \frac{25 - 1}{10} = \frac{24}{10} = 2.4 \).

Step 3: Understanding Hyperbolic Tangent

The hyperbolic tangent function is defined as \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \), where \( \cosh(x) = \frac{e^x + e^{-x}}{2} \). We need to compute \( \tanh(2 \ln 4) \).

Step 4: Simplifying \( \tanh(2 \ln 4) \)

First, compute \( \sinh(2 \ln 4) = \frac{e^{2 \ln 4} - e^{-2 \ln 4}}{2} \). Since \( e^{2 \ln 4} = (e^{\ln 4})^2 = 16 \) and \( e^{-2 \ln 4} = \frac{1}{16} \), this gives \( \frac{16 - \frac{1}{16}}{2} = \frac{256/16 - 1/16}{2} = \frac{255/16}{2} = \frac{255}{32} \).

Step 5: Compute \( \cosh(2 \ln 4) \)

Next, compute \( \cosh(2 \ln 4) = \frac{e^{2 \ln 4} + e^{-2 \ln 4}}{2} = \frac{16 + \frac{1}{16}}{2} = \frac{256/16 + 1/16}{2} = \frac{257/16}{2} = \frac{257}{32} \).

Step 6: Calculating \( \tanh(2 \ln 4) \)

Using the formula for hyperbolic tangent, \( \tanh(2 \ln 4) = \frac{\sinh(2 \ln 4)}{\cosh(2 \ln 4)} = \frac{255/32}{257/32} = \frac{255}{257} \).

Key Concepts

Hyperbolic Sine

The hyperbolic sine is a mathematical function often denoted as \( \sinh(x) \). It's quite an interesting function that arises when dealing with hyperbolic geometry and certain types of differential equations. Here's the key formula to remember: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).
This formula might look intimidating at first glance, but it's simpler than it seems. The letter \( e \) represents Euler’s number, a fundamental constant in mathematics approximately equal to 2.71828.
If we want to compute \( \sinh(\ln 5) \), we simply substitute \( \ln 5 \) into our hyperbolic sine formula. Remember, \( \ln 5 \) is the natural logarithm of 5, and it simplifies nicely:

• Find \( e^{\ln 5} \) which equals 5 due to the property of logarithms where \( e^{\ln a} = a \).
• Calculate \( e^{-\ln 5} \), which equals \( \frac{1}{5} \), again due to the same property.

Now plug these into the formula:\[ \sinh(\ln 5) = \frac{5 - \frac{1}{5}}{2} = \frac{24}{10} = 2.4 \].
It's a straightforward process as long as you're comfortable with exponential and logarithmic calculations.

Hyperbolic Tangent

The hyperbolic tangent function, noted as \( \tanh(x) \), might sound complex, but it's just another useful tool in the study of calculus and physics. It is defined as the ratio of the hyperbolic sine and hyperbolic cosine: \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \).
The hyperbolic cosine function, \( \cosh(x) \), is similar to \( \sinh(x) \) but uses addition instead of subtraction: \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
To calculate \( \tanh(2 \ln 4) \), break it down into smaller steps:

• Find \( \sinh(2 \ln 4) \) using the hyperbolic sine formula:
• Since \( e^{2 \ln 4} = 16 \) and \( e^{-2 \ln 4} = \frac{1}{16} \), substitute these back to get:
• \( \sinh(2 \ln 4) = \frac{16 - \frac{1}{16}}{2} = \frac{255}{32} \).

Next, calculate \( \cosh(2 \ln 4) \):

• Using \( \cosh(x) = \frac{e^x + e^{-x}}{2} \) and the values found, \( \cosh(2 \ln 4) = \frac{257}{32} \).

Finally, compute \( \tanh(2 \ln 4) \):

• \( \tanh(2 \ln 4) = \frac{255/32}{257/32} = \frac{255}{257} \).

The hyperbolic tangent then becomes a simple division once you have both \( \sinh(x) \) and \( \cosh(x) \).

Natural Logarithms

Natural logarithms are a fundamental concept in mathematics, denoted as \( \ln \) and based on the number \( e \). The natural logarithm of a number \( x \) is the power to which \( e \) must be raised to obtain \( x \).
A few important properties of natural logarithms include:

• \( \ln(e) = 1 \), because \( e^1 = e \).
• \( \ln(1) = 0 \), since \( e^0 = 1 \).
• \( \ln(a^b) = b \cdot \ln(a) \), showing the power rule for logarithms.

These properties make simplifying expressions involving \( \ln \) much easier. In the context of hyperbolic functions, natural logarithms often appear when computing arguments, such as \( \ln 5 \) or \( 2 \ln 4 \).
Taking the logarithm of numbers like 5 in \( \ln 5 \) allows us to relate back to the exponential function, such as converting \( e^{\ln 5} \) to 5.
Natural logarithms help bridge the gap between exponential growth processes and linear scaling, which is why they're so essential in many areas of science and mathematics.