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Chapter 3: Problem 29

Find \(d p / d q\) $$ p=\frac{\sin q+\cos q}{\cos q} $$

Short Answer

Expert verified
\( \frac{d p}{d q} = \sec^2 q \).

Step by step solution

01

Apply Trigonometric Identity

First, rewrite the expression using the trigonometric identity for tangent and splitting the fraction. The expression given is \( p = \frac{\sin q + \cos q}{\cos q} \). Split it as two separate fractions: \( p = \frac{\sin q}{\cos q} + \frac{\cos q}{\cos q} \). This simplifies to \( p = \tan q + 1 \).
02

Differentiate the Expression

Now, differentiate the expression \( p = \tan q + 1 \) with respect to \( q \). The derivative of \( \tan q \) with respect to \( q \) is \( \sec^2 q \), and the derivative of a constant \( 1 \) is \( 0 \). Thus, \( \frac{d p}{d q} = \sec^2 q \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Identity
Trigonometric identities are formulas that relate the angles and side ratios of triangles.
This exercise uses a basic trigonometric identity to simplify the expression for easier differentiation.
When faced with a fraction like \(\frac{\sin q + \cos q}{\cos q}\), you can split it into two parts for simplification.
  • The expression \(\frac{\sin q}{\cos q}\) is recognized as \(\tan q\), which is a fundamental trigonometric identity.
  • The expression \(\frac{\cos q}{\cos q}\) simplifies to \(1\) because any value divided by itself is \(1\).
By applying these identities, the expression \(p = \tan q + 1\) becomes clearer and easier to work with.
This transformation is a crucial step to make further calculations straightforward.
Differentiation
Differentiation is a technique in calculus to find the rate at which a function is changing at any point.
Once the expression is simplified to \(p = \tan q + 1\), finding its derivative becomes much easier.To differentiate, you need to understand how different functions behave when derived:
  • The derivative of \(\tan q\) is \(\sec^2 q\), a standard result from differentiation rules for trigonometric functions.
  • The derivative of a constant, such as \(1\), is \(0\).
Combining these results gives us \(\frac{d p}{d q} = \sec^2 q\).
This means that the rate of change of \(p\) with respect to \(q\) is determined by \(\sec^2 q\), indicating the function's sensitivity to changes in \(q\).
Tangent Function
The tangent function, denoted as \(\tan q\), is one of the six fundamental trigonometric functions.
It is defined as the ratio of the sine and cosine of an angle: \(\tan q = \frac{\sin q}{\cos q}\).Some things to know about the tangent function:
  • It is periodic with a period of \(\pi\) radians or \(180\) degrees.
  • It has asymptotes, or points where the function tends towards infinity, at odd multiples of \(\frac{\pi}{2}\).
  • The derivative of the tangent function is \(\sec^2 q\), which is the square of the secant function.
This makes the tangent function interesting since it can rapidly increase or decrease.
Understanding these properties helps when you are differentiating expressions involving the tangent function,
as shown in the exercise with the calculation of \(\frac{d p}{d q}\).

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