Problem 49 - Advanced Engineering Mathematics

Question

Use (8) to find the indicated derivative. $$ z=\ln \left(u^{2}+v^{2}\right) ; u=t^{2}, v=t^{-2} ; \frac{d z}{d t} $$

Step-by-Step Solution

Answer

\( \frac{dz}{dt} = \frac{4t^3 - 4t^{-5}}{t^4 + t^{-4}} \).

Step 1: Identify the Function and Substitutions

We have the function \( z = \ln(u^2 + v^2) \) with substitutions \( u = t^2 \) and \( v = t^{-2} \). Our goal is to find the derivative \( \frac{dz}{dt} \).

Step 2: Substitute u and v into the Function

Substitute the expressions for \( u \) and \( v \) into \( z \). This gives \( z = \ln((t^2)^2 + (t^{-2})^2) = \ln(t^4 + t^{-4}) \).

Step 3: Differentiate the Function with Respect to t

To find \( \frac{dz}{dt} \), apply the chain rule. The derivative of \( \ln(f(t)) \) is \( \frac{1}{f(t)} \cdot \frac{df(t)}{dt} \), where \( f(t) = t^4 + t^{-4} \).

Step 4: Compute the Derivative of f(t)

Differentiate \( f(t) = t^4 + t^{-4} \) with respect to \( t \). The derivative is \( \frac{d}{dt}(t^4 + t^{-4}) = 4t^3 - 4t^{-5} \).

Step 5: Combine Using the Chain Rule

Using the chain rule, substitute back: \( \frac{dz}{dt} = \frac{1}{t^4 + t^{-4}} \cdot (4t^3 - 4t^{-5}) \).

Step 6: Simplify the Expression

Simplify the expression \( \frac{dz}{dt} = \frac{4t^3 - 4t^{-5}}{t^4 + t^{-4}} \) to its simplest form, if possible.

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. It allows us to find the derivative of a function that has another function inside it. If a function can be expressed as a composition of two functions, such as \( z = g(f(t)) \), the derivative \( \frac{dz}{dt} \) is found by multiplying the derivative of the outer function \( g \) evaluated at \( f(t) \) by the derivative of the inner function \( f(t) \).

For example, in the provided exercise, we have \( z = \ln(u^2 + v^2) \) where \( u=t^2 \) and \( v=t^{-2} \). To find \( \frac{dz}{dt} \), we substitute the expressions for \( u \) and \( v \) into \( f(t) \) to get \( f(t) = t^4 + t^{-4} \). The chain rule is then applied by first differentiating \( \ln(f(t)) \), which results in \( \frac{1}{f(t)} \) times the derivative of \( f(t) \) with respect to \( t \). This method breaks down complex functions into manageable steps and simplifies the differentiation process.

• Differentiate the outer function.
• Multiply by the derivative of the inner function.
• Simplify the resulting expression if possible.

Natural Logarithm Differentiation

Differentiating a function that involves a natural logarithm is straightforward once you understand the basic rule: the derivative of \( \ln(x) \) is \( \frac{1}{x} \). When dealing with a compound function like \( \ln(u^2 + v^2) \), where \( u \) and \( v \) are functions of another variable \( t \), the process involves both the natural logarithm rule and the chain rule.

In this exercise, the function inside the logarithm is \( u^2 + v^2 \), substituted with \( t^4 + t^{-4} \) in terms of \( t \). Hence, the derivative of \( z = \ln(t^4 + t^{-4}) \) gives \( \frac{1}{t^4 + t^{-4}} \). The differentiation not only requires applying the logarithmic derivative rule but also calls for the subsequent differentiation of \( t^4 + t^{-4} \) itself.

This process highlights how these rules combine to tackle more complex expressions:

• Apply the logarithm differentiation rule.
• Use the result to multiply by the derivative of the internal function.
• Simplify where possible for clarity.

Mathematical Substitution

Mathematical substitution is a technique often used to simplify differentiation tasks by expressing a given function in terms of new variables. This approach enables a clearer pathway to applying differentiation rules, especially in more complex multi-variable scenarios.

In the exercise, substitutions like \( u = t^2 \) and \( v = t^{-2} \) replace the variables in the function \( z = \ln(u^2 + v^2) \). By doing so, we convert the function into a single variable function \( f(t) = t^4 + t^{-4} \), which is easier to differentiate. Substitution reduces the complexity by focusing on a straightforward variable representation.

Here’s a simplified approach to perform substitutions effectively:

• Identify substitute expressions for complicated parts of the function.
• Re-express the function using these new terms.
• Differentiation becomes manageable and often, standard rule applications follow.
• Substitute back only if necessary to interpret results.