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Chapter 3: Problem 23

In Exercises \(21-24,\) find \(d p / d q\) $$ p=\frac{\sin q+\cos q}{\cos q} $$

Short Answer

Expert verified
The derivative \( \frac{dp}{dq} \) is \( \sec^2 q \).

Step by step solution

01

Understand the Problem

We need to find the derivative \( \frac{dp}{dq} \), where the function \( p = \frac{\sin q + \cos q}{\cos q} \). This equation is asking us to differentiate the function with respect to \( q \).
02

Simplify the Function

Simplify the function \( p \) before differentiating. We can rewrite it as:\[ p = \frac{\sin q}{\cos q} + \frac{\cos q}{\cos q} = \tan q + 1 \]
03

Differentiate Each Term

Differentiate the simplified expression term by term. We use the derivative of \( \tan q \), which is \( \sec^2 q \), and the derivative of a constant, which is 0.\[ \frac{dp}{dq} = \frac{d}{dq}(\tan q) + \frac{d}{dq}(1) = \sec^2 q + 0 \]
04

Write the Final Result

Combine the derivative results, which gives:\[ \frac{dp}{dq} = \sec^2 q \]

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Key Concepts

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

Trigonometric Functions
Trigonometric functions are the building blocks of mathematics when you deal with angles and triangles. In this problem, we deal with the trigonometric functions sine, cosine, and tangent. These functions are essential for solving problems involving periodic phenomena and are our keys to unlocking the angles.
  • The sine function, written as \( \sin q \), refers to the ratio of the length of the opposite side to the hypotenuse in a right triangle.
  • The cosine function, \( \cos q \), is the ratio of the length of the adjacent side to the hypotenuse.
  • The tangent function, \( \tan q \), defined as \( \frac{\sin q}{\cos q} \), represents the ratio of the sine to the cosine.
Understanding these relationships helps us break down complex expressions into simpler, familiar terms. In our exercise, we transformed the original equation using these trigonometric relationships to make differentiation easier.
Simplifying Expressions
Simplifying expressions is an important skill in mathematics. By simplifying, we not only make our calculations easier but also make our results more insightful.
In the given exercise, we started by simplifying \( p = \frac{\sin q + \cos q}{\cos q} \). This was done via breaking the expression down:
  • \( \frac{\sin q}{\cos q} \) simplifies to \( \tan q \) since it's the definition of the tangent function.
  • \( \frac{\cos q}{\cos q} \) simplifies to 1.
By recognizing and applying these trigonometric identities, we are left with a much simpler expression: \( \tan q + 1 \). This simplification reduces the complexity of finding the derivative, allowing us to focus on applying our derivative rules with ease.
Derivatives
Derivatives represent the rate of change of a function with respect to a variable. They allow us to analyze how a function behaves as inputs change, and they are fundamental in calculus.
For the derivative of our simplified expression \( p = \tan q + 1 \), we follow these steps:
  • The derivative of \( \tan q \) is \( \sec^2 q \). This is a result of the chain rule and basic trigonometric derivative rules.
  • The derivative of a constant, such as 1 in our expression, is 0. Constants do not change, so their rate of change is zero.
Therefore, the resulting derivative \( \frac{dp}{dq} = \sec^2 q \) shows us how \( p \) changes as \( q \) changes. This powerful tool gives us insights into the behavior of our function under different conditions.

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Most popular questions from this chapter

In Exercises \(81-86,\) find a parametrization for the curve. the lower half of the parabola \(x-1=y^{2}\)

The linearization is the best linear approximation This is why we use the linearization.) Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the erroo \(E(x)=f(x)-g(x)\) were small enough near \(x=a,\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=\) \(f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions 1\. \(E(a)=0\) 2\. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\)

In Exercises \(19-30,\) find \(d y\) $$ y=2 \cot \left(\frac{1}{\sqrt{x}}\right) $$

In Exercises \(81-86,\) find a parametrization for the curve. the ray (half line) with initial point \((-1,2)\) that passes through the point \((0,0)\)

Use a CAS to perform the following steps on the parametrized curves in Exercises \(113-116 .\) a. Plot the curve for the given interval of \(t\) values. b. Find \(d y / d x\) and \(d^{2} y / d x^{2}\) at the point \(t_{0}\) . c. Find an equation for the tangent line to the curve at the point defined by the given value \(t_{0} .\) Plot the curve together with the tangent line on a single graph. $$ \begin{array}{l}{x=2 t^{3}-16 t^{2}+25 t+5, \quad y=t^{2}+t-3, \quad 0 \leq t \leq 6} \\ {t_{0}=3 / 2}\end{array} $$
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