Problem 52 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=\sec ^{2} \pi t$$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = 2\pi \sec^2(\pi t)\tan(\pi t) \).

Step 1: Identify the Function and Required Derivative

We are given the function \( y = \sec^{2} \pi t \) and need to find \( \frac{dy}{dt} \). This involves differentiating \( y \) with respect to \( t \).

Step 2: Recognize the Derivative Formula for Secant Squared

Recall that the derivative of \( \sec^2(x) \) with respect to \( x \) is \( 2\sec^2(x)\tan(x) \). We'll apply this formula here, acknowledging that the argument is a function of \( t \).

Step 3: Chain Rule Application

Given \( y = \sec^{2}(\pi t) \), use the chain rule. Let \( u = \pi t \), hence \( y = \sec^2(u) \). The derivative \( \frac{dy}{dt} \) is \( 2\sec^2(u)\tan(u) \cdot \frac{du}{dt} \).

Step 4: Differentiate the Inner Function

Differentiate the inner function \( u = \pi t \) with respect to \( t \), which gives \( \frac{du}{dt} = \pi \).

Step 5: Substitute and Simplify

Substitute \( u = \pi t \) and \( \frac{du}{dt} = \pi \) back into the derivative expression: \( \frac{dy}{dt} = 2\sec^2(\pi t)\tan(\pi t) \cdot \pi \). Simplifying, we get \( \frac{dy}{dt} = 2\pi \sec^2(\pi t)\tan(\pi t) \).

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus that allows us to differentiate composite functions. When you have a function within another function, like in our example of

• \(y = \sec^2(\pi t)\)

, the chain rule is essential.To apply the chain rule, you first identify the outer function and the inner function. For our function:

• The outer function is \(\sec^2(u)\), where \(u = \pi t\).
• The inner function is \(u = \pi t\).

To differentiate, we first find the derivative of the outer function with respect to the inner function, giving us \(2 \sec^2(u) \tan(u)\), as the derivative of \(\sec^2(x)\) is \(2 \sec^2(x) \tan(x)\). Then, we multiply by the derivative of the inner function with respect to \(t\), which is \(\frac{du}{dt} = \pi\). Putting it together,

• \(\frac{dy}{dt} = 2 \sec^2(\pi t) \tan(\pi t) \cdot \pi = 2\pi \sec^2(\pi t) \tan(\pi t)\)

. The chain rule is especially useful in situations where the function you're working with is a composition of simpler functions, allowing us to break down the problem and solve it step by step.

Derivative

In calculus, a derivative represents the rate of change of a function with respect to a variable. Essentially, it tells us how one quantity changes as another quantity changes. When we are finding the derivative \(\frac{dy}{dt}\), we're looking for the rate of change of \(y\) with respect to \(t\).The derivative of fundamental trigonometric functions is particularly important to learn. Knowing the derivative formulas of trigonometric functions, like \(\sec^2(x)\), helps simplify complex calculus problems. Each derivative gives us vital information:

• It tells us how the function \(y\) behaves as \(t\) changes.
• It can also help in approximating functions using linear models.

In our exercise, applying the formula for the derivative of \(\sec^2(x)\) with respect to \(x\), which is \(2\sec^2(x)\tan(x)\), is crucial for finding the derivative \(\frac{dy}{dt}\). This understanding allows us to calculate and predict the behavior of the function more accurately over any given interval.

Trigonometric Functions

Trigonometric functions, like \(\sec(x)\), \(\tan(x)\), \(\sin(x)\), and \(\cos(x)\), are fundamental in calculus and many areas of mathematics. These functions describe relationships between the angles and sides of triangles, and they have periodic properties that make them cyclic in nature.When dealing with derivatives of trigonometric functions, it’s important to remember their characteristic transformations and periodicity. For example, the function \(\sec(x)\) is the reciprocal of \(\cos(x)\), and it has its own unique properties and transformations:

• \(\sec(x) = \frac{1}{\cos(x)}\).
• The function is undefined where \(\cos(x) = 0\).

Its derivative, \(\sec(x)\tan(x)\), demonstrates the function's sensitivity to changes in angle and highlights its oscillating nature.Understanding these functions thoroughly opens up solving deep mathematical problems involving oscillations, waveforms, and circular motion. In our problem, using \(\sec^2(\pi t)\) simplified by trigonometric identities and derivative formulas allowed us to effectively calculate \(\frac{dy}{dt}\), showing how interconnected these concepts are in calculus.