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Eliminating the parameter Eliminate the parameter to express the following parametric equations as a single equation in \(x\) and \(y\). $$x=2 \sin 8 t, y=2 \cos 8 t$$

Short Answer

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Question: Eliminate the parameter \(t\) from the given parametric equations \(x = 2 \sin 8t\) and \(y = 2 \cos 8t\). Answer: The equation in terms of \(x\) and \(y\) is \(x^2 + y^2 = 4\).

Step by step solution

01

Find a relation between sin and cos

We will use the trigonometric identity: \(\sin^2 \theta + \cos^2 \theta = 1\). In our case, taking \(\theta=8t\), we have \(\sin^2 (8t) + \cos^2 (8t) = 1\).
02

Solve for sin and cos in terms of \(x\) and \(y\)

From the parametric equations, $$x = 2 \sin (8t) \Rightarrow \sin (8t)=\frac{x}{2}$$ and $$y = 2 \cos (8t) \Rightarrow \cos (8t)=\frac{y}{2}$$
03

Substitute sin and cos in the relation from Step 1

Now, we will substitute the expressions we found in Step 2 into the trigonometric identity we mentioned in Step 1. We get: $$\left(\frac{x}{2}\right)^2 + \left(\frac{y}{2}\right)^2 = 1$$
04

Simplify the equation and express it in terms of \(x\) and \(y\)

To express the relation as a single equation in \(x\) and \(y\), we just need to get rid of the fractions and simplify the expression. This gives us: $$\frac{x^2}{4} + \frac{y^2}{4} = 1 \Rightarrow x^2 + y^2 = 4$$ So the equation in terms of \(x\) and \(y\) is: $$x^2 + y^2 = 4$$

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Key Concepts

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

Parametric Equations
Parametric equations allow us to express a set of related quantities as functions of an independent parameter, often denoted as 't' in the context of motion or curves. Imagine plotting the path of a fly buzzing around a lamp; the path can curve and twist in all directions. To describe its position, we could use parametric equations to express the coordinates of the fly at any time 't'.

In the exercise, the parametric equations are given as
\[x = 2 \sin(8t)\] and
\[y = 2 \cos(8t)\].
These equations define the x and y coordinates of a point at any given time 't', establishing a relationship between the two coordinates through their dependence on the parameter 't'. In more practical terms, parametric equations often provide a convenient way to represent curves and shapes in a more controllable format than standard y=f(x) functions.
Trigonometric Identity
Trigonometric identities are equations involving trigonometric functions that are true for all possible values of the variable within their domains. A fundamental trigonometric identity used in countless scenarios, including the one echoed in our exercise, is the Pythagorean identity:
\[\sin^2(\theta) + \cos^2(\theta) = 1\].
This identity derives from the Pythagorean theorem and relates the sine and cosine of an angle to the length of the sides of a right-angled triangle.

Applying this identity, as seen in the step-by-step solution, allows us to interlink the original parametric equations, binding 'x' and 'y' together without the parameter 't'. The power of this identity lies in its ability to simplify expressions and resolve complex problems. It's a keystone in trigonometry that enables us to solve many types of equations where angles and their respective sine or cosine values are present.
Express Parametric Equations in Terms of x and y
Converting from parametric to Cartesian form involves finding a single equation in 'x' and 'y' that represents the same set of points as the original parametric equations but eliminates the parameter 't'. This conversion is particularly useful since many scenarios require a relationship between 'x' and 'y' without the intervention of a parameter.

In the given exercise, by isolating \(\sin(8t)\) and \(\cos(8t)\) in terms of 'x' and 'y', and then plugging these into the trigonometric identity, we effectively eliminate 't'. This step is the bridge that links the parametric world with the Cartesian, creating a standalone equation (\(x^2 + y^2 = 4\)) that elegantly describes the same curve without a direct reference to the parameter. It's a form of 'translation' from a time-based description to a spatial one, stripping away the temporal element and leaving us with a clear geometric picture.
Trigonometry in Calculus

Integrating Trigonometric Functions

Trigonometric functions and identities are not only central to geometry and algebra but also to calculus. In differentiation and integration, trigonometry's role is pivotal. For example, when deriving or integrating functions involving \(\sin(x)\) or \(\cos(x)\), we rely on the periodic properties and derivatives of trigonometric functions.

Moreover, when dealing with polar coordinates or parametric equations, calculus utilizes trigonometry to find slopes (derivatives) and areas (integrals) of curves that are not well described by mere x-y points. Thus, understanding trigonometric identities, like the one used to eliminate the parameter in our given exercise, is crucial for students not only to solve geometric problems but also to handle complex calculus tasks involving changes over time and space.

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Most popular questions from this chapter

Let \(C\) be the curve \(x=f(t)\), \(y=g(t),\) for \(a \leq t \leq b,\) where \(f^{\prime}\) and \(g^{\prime}\) are continuous on \([a, b]\) and C does not intersect itself, except possibly at its endpoints. If \(g\) is nonnegative on \([a, b],\) then the area of the surface obtained by revolving C about the \(x\)-axis is $$S=\int_{a}^{b} 2 \pi g(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$$. Likewise, if \(f\) is nonnegative on \([a, b],\) then the area of the surface obtained by revolving C about the \(y\)-axis is $$S=\int_{a}^{b} 2 \pi f(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$$ (These results can be derived in a manner similar to the derivations given in Section 6.6 for surfaces of revolution generated by the curve \(y=f(x)\).) Consider the curve \(x=3 \cos t, y=3 \sin t+4,\) for \(0 \leq t \leq 2 \pi\) a. Describe the curve. b. If the curve is revolved about the \(x\) -axis, describe the shape of the surface of revolution and find the area of the surface.

The region bounded by the parabola \(y=a x^{2}\) and the horizontal line \(y=h\) is revolved about the \(y\) -axis to generate a solid bounded by a surface called a paraboloid (where \(a \geq 0\) and \(h>0\) ). Show that the volume of the solid is \(3 / 2\) the volume of the cone with the same base and vertex.

Find an equation of the line tangent to the following curves at the given point. $$y^{2}-\frac{x^{2}}{64}=1 ;\left(6,-\frac{5}{4}\right)$$

Suppose the function \(y=h(x)\) is nonnegative and continuous on \([\alpha, \beta],\) which implies that the area bounded by the graph of h and the x-axis on \([\alpha, \beta]\) equals \(\int_{\alpha}^{\beta} h(x) d x\) or \(\int_{\alpha}^{\beta} y d x .\) If the graph of \(y=h(x)\) on \([\alpha, \beta]\) is traced exactly once by the parametric equations \(x=f(t), y=g(t),\) for \(a \leq t \leq b,\) then it follows by substitution that the area bounded by h is $$\begin{array}{l}\int_{\alpha}^{\beta} h(x) d x=\int_{\alpha}^{\beta} y d x=\int_{a}^{b} g(t) f^{\prime}(t) d t \text { if } \alpha=f(a) \text { and } \beta=f(b) \\\\\left(\text { or } \int_{\alpha}^{\beta} h(x) d x=\int_{b}^{a} g(t) f^{\prime}(t) d t \text { if } \alpha=f(b) \text { and } \beta=f(a)\right)\end{array}$$. Show that the area of the region bounded by the ellipse \(x=3 \cos t, y=4 \sin t,\) for \(0 \leq t \leq 2 \pi,\) equals \(4 \int_{\pi / 2}^{0} 4 \sin t(-3 \sin t) d t .\) Then evaluate the integral.

A focal chord of a conic section is a line through a focus joining two points of the curve. The latus rectum is the focal chord perpendicular to the major axis of the conic. Prove the following properties. Let \(L\) be the latus rectum of the parabola \(y^{2}=4 p x\) for \(p>0 .\) Let \(F\) be the focus of the parabola, \(P\) be any point on the parabola to the left of \(L\), and \(D\) be the (shortest) distance between \(P\) and \(L\) Show that for all \(P, D+|F P|\) is a constant. Find the constant.

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