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Chapter 3: Problem 74

a. Find an equation of the line tangent to the following curves at the given value of \(x\) b. Use a graphing utility to plot the curve and the tangent line. $$y=\csc x ; x=\frac{\pi}{4}$$

Short Answer

Expert verified
Based on the given information, the equation of the tangent line to the curve \(y = \csc x\) at \(x = \frac{\pi}{4}\) is \(y = -2x + 2 + \frac{\pi}{2}\).

Step by step solution

01

Find the point on the curve

We need to find the point on the curve at \(x = \frac{\pi}{4}\). To do this, we can evaluate \(y = \csc x\) at this value, $$y = \csc\left(\frac{\pi}{4}\right) = 2.$$ So the point on the curve is \(\left(\frac{\pi}{4}, 2\right)\).
02

Find the derivative of the curve

Now we need to find the derivative of the curve with respect to \(x\). The derivative of the cosecant function is, $$\frac{d}{dx}\csc x = -\csc x\cot x.$$
03

Find the slope of the tangent line

We need to evaluate the derivative at \(x = \frac{\pi}{4}\) to find the slope of the tangent line, $$m = -\csc\left(\frac{\pi}{4}\right)\cot\left(\frac{\pi}{4}\right) = -(2)(1) = -2.$$
04

Find the equation of the tangent line

Using the point-slope form of a line, we can write the equation of the tangent line at \(x = \frac{\pi}{4}\), $$y - 2=-2\left(x - \frac{\pi}{4}\right).$$ Alternatively, in slope-intercept form, the equation is $$y = -2x + 2 + \frac{\pi}{2}.$$ This is the equation of the tangent line to the curve \(y = \csc x\) at \(x = \frac{\pi}{4}\).

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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 fundamental to understanding angles and periodic phenomena. These functions relate the angles of a triangle to the lengths of its sides. The primary trigonometric functions include sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)). From these, we can derive other functions such as cosecant (\( \csc \)), secant (\( \sec \)), and cotangent (\( \cot \)). Each of these functions possesses unique properties and can be visualized on the unit circle.

For the function \( y = \csc x \), which is the cosecant function, it is defined as the reciprocal of the sine function:
  • \( \csc x = \frac{1}{\sin x} \)

Since the sine function is periodic with a period of \( 2\pi \), the cosecant function also shares this periodicity, but it's undefined wherever sine is zero. Understanding these properties helps in analyzing the behavior of trigonometric functions.
Derivative of Cosecant
The derivative of the cosecant function is crucial when determining the slope of a tangent line. For any function, the derivative represents the rate at which the function changes with respect to a variable.

In the case of cosecant, the derivative is somewhat more complex than that of the sine or cosine functions. The derivative of \( y = \csc x \) can be found using the chain rule and is given by the formula:
  • \( \frac{d}{dx}\csc x = -\csc x \cdot \cot x \)

This derivative formula tells us that as \( x \) changes, the change in \( \csc x \) is affected not only by \( \csc x \) itself but also by \( \cot x \), another trigonometric function derived from cosine and sine. Comprehending this derivative aids in finding the slope at specific points on the curve of the cosecant function.
Point-Slope Form
The point-slope form is a quick way to write the equation of a line, particularly when we have a point on the line and its slope. It is expressed as:
  • \( y - y_1 = m(x - x_1) \)

where \( (x_1, y_1) \) is a specific point on the line, and \( m \) is the slope. The versatility of the point-slope form allows you to derive the line's equation efficiently from these two components.

In the given exercise, we determined the slope \( m = -2 \) from the derivative evaluation. Using the point \( (\frac{\pi}{4}, 2) \), you can find the equation of the tangent line to the curve at \( x = \frac{\pi}{4} \) using:
  • \( y - 2 = -2(x - \frac{\pi}{4}) \)

This process highlights how the point-slope form simplifies finding tangent lines to curves.

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Most popular questions from this chapter

The total energy in megawatt-hr (MWh) used by a town is given by $$E(t)=400 t+\frac{2400}{\pi} \sin \frac{\pi t}{12}$$ where \(t \geq 0\) is measured in hours, with \(t=0\) corresponding to noon. a. Find the power, or rate of energy consumption, \(P(t)=E^{\prime}(t)\) in units of megawatts (MW). b. At what time of day is the rate of energy consumption a maximum? What is the power at that time of day? c. At what time of day is the rate of energy consumption a minimum? What is the power at that time of day? d. Sketch a graph of the power function reflecting the times when energy use is a minimum or a maximum.

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$x^{3}=\frac{x+y}{x

Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. A damped oscillator The displacement of a mass on a spring suspended from the ceiling is given by \(y=10 e^{-t / 2} \cos \frac{\pi t}{8}\) a. Graph the displacement function. b. Compute and graph the velocity of the mass, \(v(t)=y^{\prime}(t)\) c. Verify that the velocity is zero when the mass reaches the high and low points of its oscillation.

Let \(f(x)=x e^{2 x}\) a. Find the values of \(x\) for which the slope of the curve \(y=f(x)\) is 0 b. Explain the meaning of your answer to part (a) in terms of the graph of \(f\)

A mixing tank A 500 -liter (L) tank is filled with pure water. At time \(t=0,\) a salt solution begins flowing into the tank at a rate of \(5 \mathrm{L} / \mathrm{min.}\) At the same time, the (fully mixed) solution flows out of the tank at a rate of \(5.5 \mathrm{L} / \mathrm{min}\). The mass of salt in grams in the tank at any time \(t \geq 0\) is given by $$M(t)=250(1000-t)\left(1-10^{-30}(1000-t)^{10}\right)$$ and the volume of solution in the tank is given by $$V(t)=500-0.5 t$$ a. Graph the mass function and verify that \(M(0)=0\) b. Graph the volume function and verify that the tank is empty when \(t=1000 \mathrm{min}\) c. The concentration of the salt solution in the tank (in \(\mathrm{g} / \mathrm{L}\) ) is given by \(C(t)=M(t) / V(t) .\) Graph the concentration function and comment on its properties. Specifically, what are \(C(0)\) \(\underset{t \rightarrow 1000^{-}}{\operatorname{and}} C(t) ?\) d. Find the rate of change of the mass \(M^{\prime}(t),\) for \(0 \leq t \leq 1000\) e. Find the rate of change of the concentration \(C^{\prime}(t),\) for \(0 \leq t \leq 1000\) f. For what times is the concentration of the solution increasing? Decreasing?
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