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In Exercises \(27-32,\) find \(d p / d q\) $$ p=\frac{\tan q}{1+\tan q} $$

Short Answer

Expert verified
\( \frac{dp}{dq} = \frac{\sec^2 q}{(1 + \tan q)^2} \)

Step by step solution

01

Identify the function structure

Given the function \( p = \frac{\tan q}{1 + \tan q} \), we need to identify it as a quotient function, \( \frac{u(q)}{v(q)} \), where \( u(q) = \tan q \) and \( v(q) = 1 + \tan q \). From here, we know that the method to find the derivative will involve the quotient rule.
02

Differentiate using the quotient rule

The quotient rule states that for \( \frac{u(q)}{v(q)} \), the derivative \( \frac{dp}{dq} \) is given by:\[ \frac{dp}{dq} = \frac{v(q) \cdot \frac{d}{dq}(u(q)) - u(q) \cdot \frac{d}{dq}(v(q))}{[v(q)]^2} \]Applying this to our function, we have:- \( u(q) = \tan q \), so \( \frac{d}{dq}(u(q)) = \sec^2 q \).- \( v(q) = 1 + \tan q \), so \( \frac{d}{dq}(v(q)) = \sec^2 q \).
03

Substitute derivatives back into the quotient rule

Substituting these into the formula, we get:\[ \frac{dp}{dq} = \frac{(1 + \tan q) \cdot \sec^2 q - \tan q \cdot \sec^2 q}{(1 + \tan q)^2} \]Upon simplification, the numerator becomes \( \sec^2 q \), since \((1 + \tan q) \sec^2 q - \tan q \sec^2 q = \sec^2 q\).
04

Simplify the expression

Simplify the expression obtained:\[ \frac{dp}{dq} = \frac{\sec^2 q}{(1 + \tan q)^2} \]This is the derivative of \( p \) with respect to \( q \).
05

Express the final result

We have obtained the derivative as:\[ \frac{dp}{dq} = \frac{\sec^2 q}{(1 + \tan q)^2} \]This represents the rate of change of \( p \) with respect to \( q \).

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Key Concepts

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

Derivative
A derivative represents the rate at which a function changes as its input changes. In simpler terms, it's a way of measuring how a function's output value reacts to changes in the input value. For example, if you have a function describing the height of a rocket over time, the derivative of that function would tell you how fast the rocket's height is changing at any given moment.
In calculus, the derivative of a function is denoted by \( \frac{d}{dx} \), where \( x \) is the input variable. This notation is read as "the derivative of the function with respect to \( x \)."
The process of finding a derivative involves different rules such as the product rule, chain rule, and, most importantly for this task, the quotient rule. Each rule helps tackle different types of problems depending on the function's structure.
Trigonometric Functions
Trigonometric functions are mathematical functions that relate the angles of a triangle to the lengths of its sides. The most common trigonometric functions include sine, cosine, and tangent, which are foundational in the study of periodic phenomena like waves. These functions will pop up often when dealing with calculus problems involving oscillating quantities.
The function \( \tan q = \frac{\sin q}{\cos q} \) represents the tangent. When differentiating trigonometric functions, specific derivatives need to be recognized. For instance, the derivative of \( \tan q \) with respect to \( q \) is \( \sec^2 q \), which arises often in calculus.
Understanding how to differentiate these trigonometric functions is crucial as you've seen in the original exercise. Successfully applying these derivative relationships allows us to solve complex calculus problems.
Calculus
Calculus is an extensive field of mathematics focused on rates of change and the accumulation of quantities. There are two main branches of calculus: differential calculus (concerned with derivatives) and integral calculus (concerned with areas under curves). These concepts together allow mathematicians and scientists to describe and solve real-world problems regarding change and motion.
In the exercise at hand, we make use of differential calculus to compute how the function \( p = \frac{\tan q}{1 + \tan q} \) changes as \( q \) changes. This involves applying the quotient rule, a pivotal method within calculus used for finding the derivative of a fraction of two functions.
  • Identifying the structure of the function is crucial to applying the right calculus techniques, such as identifying \( \tan q \) as part of a trigonometric function.
  • Understanding and applying specific rules, like the quotient rule, allows us to express complex changing relationships in a systematic way.
  • Ultimately, calculus is about breaking down these changing relationships and representing them mathematically, as demonstrated in the solution provided.

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Most popular questions from this chapter

Running machinery too fast Suppose that a piston is moving straight up and down and that its position at time \(t\) sec is $$s=A \cos (2 \pi b t)$$' with \(A\) and \(b\) positive. The value of \(A\) is the amplitude of the motion, and \(b\) is the frequency (number of times the piston moves up and down each second). What effect does doubling the frequency have on the piston's velocity, acceleration, and jerk? (Once you find out, you will know why some machinery breaks when you run it too fast.)

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} x+\tan \left(\frac{y}{x}\right)=2, \quad P\left(1, \frac{\pi}{4}\right) \end{equation}

The best quantity to order One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is $$A(q)=\frac{k m}{q}+c m+\frac{h q}{2}$$ where \(q\) is the quantity you order when things run low (shoes, TVs, brooms, or whatever the item might be); \(k\) is the cost of placing an order (the same, no matter how often you order); \(c\) is the cost of one item (a constant); \(m\) is the number of items sold each week (a constant) ;and \(h\) is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security. Find \(d A / d q\) and \(d^{2} A / d q^{2}\) .

If \(x^{2}+y^{2}=25\) and \(d x / d t=-2,\) then what is \(d y / d t\) when \(x=3\) and \(y=-4 ?\)

In Exercises \(57-60,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: $$ \begin{array}{l}{\text { a. Plot the function } f \text { over } I} \\ {\text { b. Find the linearization } L \text { of the function at the point } a \text { . }} \\ {\text { c. Plot } f \text { and } L \text { together on a single graph. }} \\ {\text { d. Plot the absolute error }|f(x)-L(x)| \text { over } I \text { and find its max- }} \\ {\text { imum value. }}\end{array} $$ $$ \begin{array}{l}{\text { e. From your graph in part (d), estimate as large a } \delta>0 \text { as you }} \\ {\text { can, satisfing }}\end{array} $$ $$ \begin{array}{c}{|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon} \\\ {\text { for } \epsilon=0.5,0.1, \text { and } 0.01 . \text { Then check graphically to see if }} \\ {\text { your } \delta \text { -estimate holds true. }}\end{array} $$ $$ f(x)=x^{3}+x^{2}-2 x, \quad[-1,2], \quad a=1 $$

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