Problem 51 - Advanced Engineering Mathematics

Question

Use (8) to find the indicated derivative. $$ w=\cos (3 u+4 v) ; u=2 t+\frac{\pi}{2}, v=-t-\frac{\pi}{4},\left.\frac{d w}{d t}\right|_{t=\pi} $$

Step-by-Step Solution

Answer

The derivative \( \frac{dw}{dt} \) at \( t = \pi \) is 2.

Step 1: Identify Functions and Apply Chain Rule

First, recognize that \( w = \cos(3u + 4v) \), where \( u = 2t + \frac{\pi}{2} \) and \( v = -t - \frac{\pi}{4} \). We'll use the chain rule to find \( \frac{dw}{dt} \). The formula for the derivative using the chain rule is: \( \frac{dw}{dt} = \frac{dw}{du} \cdot \frac{du}{dt} + \frac{dw}{dv} \cdot \frac{dv}{dt} \).

Step 2: Compute \( \frac{dw}{du} \) and \( \frac{dw}{dv} \)

The derivative of \( w \) with respect to \( u \) is \( \frac{dw}{du} = -\sin(3u + 4v) \cdot 3 \). Similarly, \( \frac{dw}{dv} = -\sin(3u + 4v) \cdot 4 \).

Step 3: Derive \( \frac{du}{dt} \) and \( \frac{dv}{dt} \)

For \( u = 2t + \frac{\pi}{2} \), the derivative is \( \frac{du}{dt} = 2 \). For \( v = -t - \frac{\pi}{4} \), the derivative is \( \frac{dv}{dt} = -1 \).

Step 4: Substitute into Chain Rule Formula

Substitute the results into the chain rule formula: \[ \frac{dw}{dt} = \left(-\sin(3u + 4v) \cdot 3\right) \cdot 2 + \left(-\sin(3u + 4v) \cdot 4\right) \cdot (-1) \] to get \[ \frac{dw}{dt} = -6\sin(3u + 4v) + 4\sin(3u + 4v) = -2\sin(3u + 4v) \].

Step 5: Evaluate at \( t = \pi \)

At \( t = \pi \), substitute into the expressions for \( u \) and \( v \): \( u = 2\pi + \frac{\pi}{2} = \frac{5\pi}{2} \) and \( v = -\pi - \frac{\pi}{4} = -\frac{5\pi}{4} \). Hence, \( 3u + 4v = \frac{15\pi}{2} - 5\pi = \frac{5\pi}{2} \). Substitute this back into \(-2\sin(3u + 4v)\): \( \sin\left(\frac{5\pi}{2}\right) = -1 \).

Step 6: Final Calculation

Calculate the final result: \( \frac{dw}{dt} = -2(-1) = 2 \).

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus, specifically when dealing with composite functions. A composite function is one where a function is applied within another function. In our problem, we have several functions combined:

• Outer function: cosine function \(w = \cos(3u + 4v)\)
• Inner functions for \(u\) and \(v\): \(u = 2t + \frac{\pi}{2}\) and \(v = -t - \frac{\pi}{4}\)

To find the derivative \(\frac{dw}{dt}\), we use the chain rule formula for functions of multiple variables:\[\frac{dw}{dt} = \frac{dw}{du} \cdot \frac{du}{dt} + \frac{dw}{dv} \cdot \frac{dv}{dt}\]This formula allows us to differentiate \(w\) with respect to \(t\) by considering both paths through the chain of differentiation, i.e., via \(u\) and \(v\). Each term corresponds to how changes in \(t\) affect \(w\) through each variable.

Trigonometric Functions

Trigonometric functions like sine and cosine are crucial in this exercise. These functions have well-defined derivatives that are useful in various calculations. In our case:

• The derivative of cosine: \(\frac{d}{dx} \cos(x) = -\sin(x)\)
• The derivative of sine, for future reference: \(\frac{d}{dx} \sin(x) = \cos(x)\)

For the function \(w = \cos(3u + 4v)\), finding \(\frac{dw}{du}\) involves differentiating the cosine function. This gives:\[\frac{dw}{du} = -\sin(3u + 4v) \cdot 3\]and similarly for \(\frac{dw}{dv}\):\[\frac{dw}{dv} = -\sin(3u + 4v) \cdot 4\]Understanding these derivatives is key in applying the chain rule effectively, allowing transformations that involve trigonometric manipulations often seen in calculus.

Calculus of Functions of Multiple Variables

When dealing with functions of multiple variables, we must consider how changing one variable affects the entire function. For instance, if we look at the partial derivatives:

• \(\frac{du}{dt} = 2\) affects \(w\) through \(u\)
• \(\frac{dv}{dt} = -1\) affects \(w\) through \(v\)

In more simple terms, we think of \(w\) as a surface bending and changing in response to \(t\) through both \(u\) and \(v\). Once we have the impacts through partial derivatives:

• Substitute \(du/dt\) and \(dv/dt\) into the chain rule
• Account for changes in both variables in the same calculation

This approximation through partial derivatives, when put into practice, gives us the overall derivative \(\frac{dw}{dt}\) for specific values, like \(t = \pi\), helping us understand how \(w\) reacts to alterations in \(t\). This blending of changes in partial derivatives is integral in solving complex calculus problems.