Problem 8 - Calculus of a Single Variable

Question

Moving Point In Exercises \(5-8,\) a point is moving along the graph of the given function at the rate \(d x / d t .\) Find \(d y / d t\) for the given values of \(x .\) $$ \begin{array}{l}{y=\cos x ; \frac{d x}{d t}=4 \text { centimeters per second }} \\ {\begin{array}{llll}{\text { (a) } x=\frac{\pi}{6}} & {\text { (b) } x=\frac{\pi}{4}} & {\text { (c) } x=\frac{\pi}{3}}\end{array}}\end{array} $$

Step-by-Step Solution

Answer

The derivative \(d y / d t\) for the given values of \(x\) are: (a) for \(x=\pi/6\) : \(d y/d t=-2 cm/s \)(b) for \(x=\pi/4\) : \(d y/d t=-\sqrt{2} cm/s \)(c) for \(x=\pi/3\) : \(d y/d t=-2\sqrt{3} cm/s\)

Step 1: Define the Function and the Rate

The function is defined as \(y=\cos x \), and the rate of change of \(x\) is given as \( \frac{d x}{d t}=4 cm/s \) .

Step 2: Differentiate the Function

It is known that the derivative of the cosine function is negative of the sine function, \(d y/d x=-\sin x\). This is the rate of change of \(y\) with respect to \(x\).

Step 3: Apply Chain Rule

To find \(d y/d t\), the chain rule is used. According to this rule, \(d y/d t= (\frac{d y}{d x}) \times (\frac{d x}{d t})\). Substituting the given and calculated values, we get \(d y/d t=-\sin x \times 4 cm/s\).

Step 4: Calculate \(d y/d t\) for \(x=\pi/6, \pi/4, \pi/3\)

Substitute the given \(x\) values into the formula to get the respective \(d y/d t\) values as follows:(a) for \(x=\pi/6\) : \(d y/d t=-\sin (\pi/6) \times 4 = -4/2 = -2 cm/s\)(b) for \(x=\pi/4\) : \(d y/d t=-\sin (\pi/4) \times 4 = -\sqrt{2} cm/s\)(c) for \(x=\pi/3\) : \(d y/d t=-\sin (\pi/3) \times 4 = -2\sqrt{3} cm/s\)

Key Concepts

Derivative of Cosine

When we talk about the derivative of a function, we're exploring how the function changes as its input changes. For the cosine function, defined as \(y = \cos x\), its derivative explains how much \(y\) changes with a small change in \(x\). Specifically, the derivative of the cosine is the negative sine function, denoted as \(\frac{d y}{d x} = -\sin x\). This means that as \(x\) increases, if you picture the wave of the cosine graph, \(y\) decreases in the steepness of its drop, following the sine function in a negative direction.
Understanding the derivative of cosine is key when dealing with related rates or changes involving trigonometric functions. This concept helps us calculate how one variable, such as \(y\), reacts when another, \(x\), changes over time. This is essential in physics and engineering where trigonometric functions are often used to model waveforms, rotational motion, and oscillations, giving us insight into their rate of change.

Chain Rule

The chain rule is a vital tool in calculus for finding the derivative of a composite function. It allows us to differentiate functions that are composed of multiple layers. In the context of our problem, we have a function of the form \(y = f(g(x))\), where \(y = \cos(x)\).
The chain rule states that \( \frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t} \). This becomes useful when both \(x\) and \(y\) change over time. Here's how it helps:

• First, differentiate \(y\) with respect to \(x\), which we found to be \(-\sin x\).
• Then, multiply this with the rate of change of \(x\) with respect to time \(t\), \(\frac{d x}{d t}\), which is 4 in our problem.

Applying the chain rule turns the concept of a static derivative into a dynamic one, reflecting changes over time. This method is essential when working on problems where one variable affects another indirectly, such as in physics and related rates problems.

Trigonometric Derivatives

Trigonometric derivatives are derivatives of trigonometric functions, which include sine, cosine, and tangent, among others. Understanding these derivatives allows you to handle changes in trigonometric functions effectively.
Focusing on our case:

• The derivative of \(\cos x\) is \(-\sin x\), a fundamental identity in calculus.
• For \(\sin x\), the derivative is \(\cos x\).
• These identities help in solving problems where trigonometric functions are involved, especially when they change over time, like oscillations or waves.

Trigonometric derivatives are not just mathematical tools but are crucial in solving practical problems in various fields like engineering, physics, and even computer science, where wave patterns, rotational motion, and signal processing are studied. Understanding these derivatives and how they relate to each other makes solving complex systems involving trigonometry more manageable.