Problem 7 - 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 \(x .\) $$ \begin{array}{l}{y=\tan x ; \frac{d x}{d t}=3 \text { feet per second }} \\\ {\begin{array}{llll}{\text { (a) } x=-\frac{\pi}{3}} & {\text { (b) } x=-\frac{\pi}{4}} & {\text { (c) } x=0}\end{array}}\end{array} $$

Step-by-Step Solution

Answer

The rate of change of \(y\) with respect to time \(dy/dt\) at different values of \(x\) are: (a) 12 feet per second, (b) \(6\sqrt{2}\) feet per second, and (c) 3 feet per second.

Step 1: Differentiate \(y = tan(x)\)

Using the standard differentiation formula, we know that the derivative of \(\tan(x)\) is \( \sec^2(x)\). Hence, \(dy/dx = \sec^2(x)\).

Step 2: Apply Chain Rule and Substitute \(dx/dt = 3\)

According to the chain rule, \(dy/dt = dy/dx \cdot dx/dt\), which equals \(\sec^2(x) \cdot 3\).

Step 3: Substitute Given Values of \(x\) to Compute \(dy/dt\)

Now substitute the given values of \(x\) in this function, and compute \(dy/dt\) for each of them. (a) With \(x = -\pi/3\), \(\sec^2(-\pi/3) \cdot 3 = 12\) feet per second; (b) with \(x = -\pi/4\), \(\sec^2(-\pi/4) \cdot 3 = 6\sqrt{2}\) feet per second; and (c) with \(x = 0\), \(\sec^2(0) \cdot 3 = 3\) feet per second.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that helps us differentiate composite functions. Suppose you have a function defined as a composition of two functions, say \( y = f(g(x)) \). The chain rule states that to find the derivative of this composition, you must take the derivative of \( f \) with respect to \( g(x) \) and multiply it by the derivative of \( g(x) \) with respect to \( x \).Essentially, the chain rule can be written as:

• \( \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \)

In our exercise, though the function is not composite, we treat the differentiation with respect to time, \( t \), by using the chain rule. We differentiate the function \( y = \tan(x) \) to get \( dy/dx = \sec^2(x) \), and then apply the chain rule:

• \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \)

Here, \( \frac{dx}{dt} = 3 \) feet per second, which represents the rate at which \( x \) changes with respect to time, \( t \). Using this approach allows us to solve for \( \frac{dy}{dt} \), or how \( y \) changes over time.

Trigonometric Functions

Trigonometric functions are functions that relate the angles of a triangle to the lengths of its sides. Among these functions, tangent (tan) and secant (sec) are particularly important here. When differentiating trigonometric functions, each has its own unique derivative:

• The derivative of \( \tan(x) \) is \( \sec^2(x) \).
• \( \sec(x) \) is a reciprocal trigonometric function which is \( 1/\cos(x) \).

In the context of our problem, we're dealing with \( y = \tan(x) \). This choice leads to the derivative \( dy/dx = \sec^2(x) \) due to the specific differentiation rules of trigonometric functions. This derivative is crucial for finding how the output of the function changes as \( x \) changes, especially when applying the chain rule.

Rate of Change

The rate of change is an important concept in calculus, especially in real-world applications involving motion and growth. It describes how a quantity changes over time. For the problem at hand, we're particularly interested in how both \( x \) and \( y = \tan(x) \) change over time, \( t \).In this exercise:

• \( \frac{dx}{dt} = 3 \) feet per second is given, representing how quickly \( x \) is changing with respect to time.
• \( \frac{dy}{dt} \) is what we're trying to find — the rate at which \( y \), or \( \tan(x) \), changes as \( x \) changes with respect to time.

To calculate \( \frac{dy}{dt} \), we use the chain rule, which combines these rates of change:

• \( \frac{dy}{dt} = \sec^2(x) \cdot \frac{dx}{dt} \)

By substituting different values of \( x \) — \(-\pi/3\), \(-\pi/4\), and \(0\) — we find different rates of change for \( y \) at these points. These results show how the speed of tan function's change can vary greatly based on the input \( x \).