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Chapter 11: Problem 18

Assuming that the equations define \(x\) and \(y\) implicitly as differentiable functions \(x=f(t), y=g(t)\) , find the slope of the curve \(x=f(t), y=g(t)\) at the given value of \(t\) . $$ x \sin t+2 x=t, \quad t \sin t-2 t=y, \quad t=\pi $$

Short Answer

Expert verified
The slope at \(t = \pi\) is \(-2\).

Step by step solution

01

Differentiate the First Equation

Start with the equation \(x \sin t + 2x = t\). Differentiating both sides with respect to \(t\) gives: \(\frac{dx}{dt} \sin t + x \cos t + 2\frac{dx}{dt} = 1\). This simplifies to \((\sin t + 2) \frac{dx}{dt} + x \cos t = 1\).
02

Differentiate the Second Equation

Next, differentiate the equation \(t \sin t - 2t = y\) with respect to \(t\). This gives: \(\sin t + t \cos t - 2 = \frac{dy}{dt}\).
03

Compute \(\frac{dy}{dx}\)

To find the slope of the curve, use \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\). Plug in the derivatives obtained: \(\frac{dy}{dx} = \frac{\sin t + t \cos t - 2}{(\sin t + 2) \frac{dx}{dt} + x \cos t}\).
04

Evaluate Derivatives at \(t = \pi\)

At \(t = \pi\), substitute the values in the differentiated equations:1. From Step 1: \((0 + 2) \frac{dx}{dt} + x (-1) = 1\) which simplifies to \(2\frac{dx}{dt} - x = 1\).2. From Step 2: \(0 + \pi(-1) - 2 = \frac{dy}{dt}\) which simplifies to \(-\pi - 2 = \frac{dy}{dt}\).
05

Solve for \(x\) and \(\frac{dx}{dt}\)

To find \(x\), substitute \(t = \pi\) in the original equation: \(x \cdot 0 + 2x = \pi\), so \(2x = \pi\) gives \(x = \frac{\pi}{2}\). Substitute this value back into the equation for \(\frac{dx}{dt}\): \(2\frac{dx}{dt} - \frac{\pi}{2} = 1\), thus \(2\frac{dx}{dt} = 1 + \frac{\pi}{2}\), which gives \(\frac{dx}{dt} = \frac{1 + \frac{\pi}{2}}{2}\).
06

Compute the Slope

Substitute the values at \(t = \pi\) into the formula for \(\frac{dy}{dx}\): \[\frac{dy}{dx} = \frac{-\pi - 2}{\frac{1 + \frac{\pi}{2}}{2}}\].Simplifying, the slope of the curve at \(t = \pi\) is \(-\frac{2(2 + \pi)}{2 + \pi}\), resulting in a slope of \(-2\).

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Key Concepts

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

Differentiable Functions
Differentiable functions are those that can be differentiated, meaning we can determine their derivative at any point within their domain. In simpler terms, these functions have well-defined tangent lines at every point, capturing a smooth transition without any breaks or sharp corners.

Commonly represented in mathematical terms as a function contains variables, such as time or another parameter, that relate to each other smoothly. For implicit functions like in this exercise, the relationship between the variables isn't given explicitly (like \(y = f(x)\)), but instead, both dependent variables are functions of a common independent variable, often denoted by \(t\).

Implicitly, two equations are used to express how two variables relate as functions of \(t\). When differentiating these rules, we capture the inherent relationship enforced by their parameters, allowing us to understand how changes in one dimension impact another.
Slope of the Curve
The slope of a curve at a particular point is a measure of how steep the tangent line to the curve is at that point. In mathematical terms, it's expressed as the derivative \( \frac{dy}{dx} \). This value indicates the rate of change of \(y\) with respect to \(x\), helping us understand the behavior of the curve.

To find the slope using implicit differentiation, we differentiate each equation with respect to \(t\), then use the ratio of differential changes \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) to form a single derivative \( \frac{dy}{dx} \). This derivative gives the slope of the curve at any point along it. It's like merging the separate motions along \(t\) into a single movement describing \(y\) in terms of \(x\). This step is crucial as it helps translate the implicit relationship into an understandable form to determine the instantaneous rate of change.
Derivative Evaluation
Evaluating derivatives means calculating the derivative values at specific points, like \(t = \pi\) in this exercise. This step involves plugging specific values into the derivatives obtained from the implicit differentiation.

At \(t = \pi\), both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are evaluated to give the numerical slope \( \frac{dy}{dx} \) at that precise point. By substituting \(t = \pi\) and solving the equations, you find exact values for the variables and their derivatives. This meticulous process leads to an actual slope measure, offering insights on whether the curve at \(t = \pi\) is increasing, decreasing, or constant.

For example, a negative slope of \(-2\) indicates the curve is decreasing at that point, implying a negative rate of change of \(y\) with \(x\) around \(t = \pi\). Such practical evaluations are vital in connecting theoretical calculus with real-world applications.

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Most popular questions from this chapter

Exercises \(29-36\) give the eccentricities of conic sections with one focus at the origin along with the directrix corresponding to that focus. Find a polar equation for each conic section. $$e=1 / 4, \quad x=-2$$

Find polar equations for the circles in Exercises \(57-64 .\) Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations. $$x^{2}+y^{2}+y=0$$

The witch of Maria Agnesi The bell-shaped witch of Maria Agnesi can be constructed in the following way. Start with a circle of radius \(1,\) centered at the point \((0,1),\) as shown in the accompanying figure. Choose a point \(A\) on the line \(y=2\) and connect it to the origin with a line segment. Call the point where the segment crosses the circle \(B\) . Let \(P\) be the point where the vertical line through \(A\) crosses the horizontal line through \(B\) . The witch is the curve traced by \(P\) as \(A\) moves along the line \(y=2\) . Find parametric equations and a parameter interval for the witch by expressing the coordinates of \(P\) in terms of \(t\) the radian measure of the angle that segment \(O A\) makes with the positive \(x\) -axis. The following equalities (which you may assume) will help. $$\begin{array}{ll}{\text { a. } x=A Q} & {\text { b. } y=2-A B \sin t} \\\ {\text { c. } A B \cdot O A=(A Q)^{2}}\end{array}$$

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ x^{2}-y^{2}-2 x+4 y=4 $$

Vertical and horizontal lines a. Show that every vertical line in the \(x y\) -plane has a polar equation of the form \(r=a \sec \theta .\) b. Find the analogous polar equation for horizontal lines in the \(x y-\) -plane.
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