Problem 50 - Advanced Engineering Mathematics

Question

Use (8) to find the indicated derivative. $$ z=u^{3} v-u v^{4} ; u=e^{-5 t}, v=\sec 5 t ; \frac{d z}{d t} $$

Step-by-Step Solution

Answer

Find \( \frac{dz}{dt} \) using the chain rule and simplify.

Step 1: Identify the Functions

Given the function \( z = u^3 v - uv^4 \), we are also given \( u = e^{-5t} \) and \( v = \sec 5t \). We need to find \( \frac{dz}{dt} \).

Step 2: Understand the Derivative Formula

We're tasked with finding the derivative of \( z \) with respect to \( t \). Given \( z = u^3 v - uv^4 \), notice that both \( u \) and \( v \) are functions of \( t \). This means we must use the chain rule alongside the product rule.

Step 3: Differentiate Each Component

Break it down into parts. The first term is \( u^3 v \).- Its derivative with respect to \( t \) is \( \frac{d}{dt}(u^3 v) = u^3 \frac{dv}{dt} + v \frac{d}{dt}(u^3) \).Next, differentiate \( v \) and \( u^3 \):- \( \frac{dv}{dt} = \frac{d}{dt}(\sec 5t) = 5 \sec 5t \tan 5t \).- \( \frac{d}{dt}(u^3) = 3u^2 \cdot \frac{du}{dt} = 3u^2 (-5e^{-5t}) = -15 u^2 e^{-5t} \).

Step 4: Differentiate the Second Component

For the second term, \( -uv^4 \), we find its derivative:- \( \frac{d}{dt}(uv^4) = u \frac{d}{dt}(v^4) + v^4 \frac{du}{dt} \).Differentiate \( v^4 \) and \( u \):- \( \frac{d}{dt}(v^4) = 4v^3 \cdot \frac{dv}{dt} = 4v^3 \cdot (5 \sec 5t \tan 5t) = 20v^3 \sec 5t \tan 5t \).- \( \frac{du}{dt} = \frac{d}{dt}(e^{-5t}) = -5e^{-5t} \).

Step 5: Combine the Derivatives

Combine these results to get \( \frac{dz}{dt} \):- For \( u^3 v \): - \( u^3 \cdot 5 \sec 5t \tan 5t + v \cdot (-15 u^2 e^{-5t}) \).- For \( -uv^4 \): - Combined derivative is \( - \left( u \cdot 20v^3 \sec 5t \tan 5t + v^4 \cdot (-5e^{-5t}) \right) \).

Step 6: Simplify the Expression

Combining all these gives:\[\frac{dz}{dt} = u^3 \cdot 5 \sec 5t \tan 5t - 15u^2 v e^{-5t} - 20uv^3 \sec 5t \tan 5t + 5v^4 e^{-5t}.\]Substitute back \( u = e^{-5t} \) and \( v = \sec 5t \) to get the final answer.

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used when differentiating composite functions. When you have a function within another function, applying the chain rule allows you to find the derivative of the outer function with respect to the inner one. For example, if you have a function such as \( y = f(g(x)) \), the chain rule states that the derivative, \( \frac{dy}{dx} \), is found by multiplying the derivative of the outer function \( f' \) with respect to \( g \) by the derivative of the inner function \( g' \) with respect to \( x \). This is written as:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

In the given exercise, we need to use the chain rule because the functions \( u \) and \( v \), which depend on \( t \), are part of the composite function \( z = u^3 v - uv^4 \). To differentiate each component like \( u^3 \), it becomes important to remember that \( u \) in itself is a function of \( t \), requiring an application of the chain rule to fully differentiate the expression with respect to \( t \).

• This means, for a term like \( u^3 \), you need to multiply the derivative of \( u^3 \) with \( t \) by the derivative of \( u \) itself, which is another application of the chain rule.

Product Rule

The Product Rule is essential for differentiating products of two functions. When you have two functions, say \( f(x) \) and \( g(x) \), the derivative of their product is not just the product of their derivatives. Instead, it's given as:

• \( \frac{d}{dx}[f(x)g(x)] = f(x)\frac{d}{dx}[g(x)] + g(x)\frac{d}{dx}[f(x)] \)

This rule is crucial in problems where two functions are multiplied together, as in the terms \( u^3 v \) and \( uv^4 \) from the exercise. When applying the product rule:

• The derivative of \( u^3 v \) requires using the product rule to differentiate \( u^3 \cdot v \).
• The same is needed for \( -uv^4 \), where the differentiation has to consider both parts of the product separately.

The usefulness of the product rule is seen in how it ensures that both parts of the product are treated correctly, and it rectifies a common misconception that the derivative of \( ab \) is simply \( a'(b) + b'(a) \), which is not specific enough without the product rule.

Derivative

In calculus, the derivative is a way to measure how a function changes as its input changes. The derivative can give you the slope of the function's graph at any given point, thereby indicating how the function value rises or falls as the input increases. It's a fundamental tool for understanding rates of change.In the given exercise, the derivative \( \frac{dz}{dt} \) represents the rate at which the function \( z = u^3 v - uv^4 \) changes with respect to time \( t \). This requires differentiating each part of the expression thoroughly:

• For \( u^3 v \), the approach is to apply both the chain rule and product rule to account for the derivation of \( u^3 \) and \( v \) with respect to \( t \).
• The same approach goes for \( -uv^4 \) where the differentiation of \( u \) and \( v^4 \) happens similarly.

This technique of finding the derivative highlights how calculus allows us to precisely model dynamical systems and understand their behavior over time. Calculating derivatives leads us to insights into marginal changes, rates of change, and the sensitivity of one variable concerning another.