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Chapter 7: Problem 50

If \(x, y \in \mathbb{R}\) are not both zero and \((r, \theta)\) are the polar coordinates of \((x, y)\), then determine the polar coordinates of (i) \((y, x)\), and (ii) \((t x, t y)\), where \(t\) is any positive real number.

Short Answer

Expert verified
(i) The polar coordinates of \((y,x)\) are \((r, \theta + n\pi)\). (ii) If \(t\) is a positive real number, the polar coordinates of \((tx, ty)\) are \((|t|r, \theta)\).

Step by step solution

01

Expressing \(y\) and \(x\) in terms of \(r\) and \(\theta\)

Recall that \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), substitute \(x\) and \(y\) in \(( y, x)\) which gives us: \((y, x) = (r \sin(\theta), r \cos(\theta))\)
02

Determine the Radius of the Point \((r, \theta)\)

The distance \(R\) of the point \((y, x)\) from the origin in the \(x-y\) plane, which is the radius in polar coordinates, is given by \(R = \sqrt{x^2+y^2} = \sqrt{(r \sin(\theta))^2 + (r \cos(\theta))^2} = r\)
03

Determine the Angle of the Point \((r, \theta)\)

The angle \(\Theta\) of the point is calculated as follows: \(\Theta = \arctan{\frac{y}{x}} = \arctan{\frac{r \sin(\theta)}{r \cos(\theta)}} = \arctan{\tan(\theta)} = \theta + n\pi\) This implies that the polar coordinates of \((y, x)\) are \((r, \theta + n\pi)\). ##PART (ii)##:
04

Expressing \(tx\) and \(ty\) in terms of \(r\) and \(\theta\)

Given that \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), substitute \(x\) and \(y\) in \((tx, ty)\) to give: \((tx, ty) = (tr \cos(\theta), tr \sin(\theta))\)
05

Determine the Radius of the Point \((tr, \theta)\)

The distance \(R\) of the point \((tx, ty)\) from the origin in the \(x-y\) plane, which is the radius in polar coordinates, is given by \(R = \sqrt{{(tx)}^2+{(ty)}^2} = \sqrt{(tr \cos(\theta))^2 + (tr \sin(\theta))^2} = |t|r\)
06

Determine the Angle of the Point \((tr, \theta)\)

The angle \(\Theta\) of the point is calculated as follows: \(\Theta = \arctan{\frac{ty}{tx}} = \arctan{\frac{tr \sin(\theta)}{tr \cos(\theta)}} = \arctan{\tan(\theta)} = \theta\) This implies that the polar coordinates of \((tx, ty)\) are \((|t|r, \theta)\) if \(t\) is a positive real number.

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Most popular questions from this chapter

Consider the function \(f: \mathbb{R} \backslash\\{0\\} \rightarrow \mathbb{R}\) defined by \(f(x)=(\sin (1 / x)) / x\). Show that the amplitude of the oscillation of the function \(f\) increases without any bound as \(x\) tends to 0 .

Let \(r\) be a positive real number and \(\theta \in(-\pi, \pi]\) and \(\alpha \in \mathbb{R}\) be such that \(\theta+\alpha \in(-\pi, \pi] .\) If \(P\) and \(P_{\alpha}\) denote the points with polar coordinates \((r, \theta)\) and \((r, \theta+\alpha)\), respectively, then find the Cartesian coordinates of \(P_{\alpha}\) in terms of the Cartesian coordinates of \(P\). [Note: The transformation \(P \mapsto P_{\alpha}\) corresponds to a rotation of the plane by the angle \(\alpha\).]

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Let \(n \in \mathbb{N}\). By applying L'H么pital's rule \(n\) times, prove the following: (i) \(\lim _{x \rightarrow 0} \frac{\exp x-\sum_{k=0}^{n} x^{k} / k !}{x^{n+1}}=\frac{1}{(n+1) !}\), (ii) \(\lim _{x \rightarrow 1} \frac{\ln x-\sum_{k=1}^{n}(-1)^{k}(x-1)^{k} / k}{(x-1)^{n+1}}=\frac{(-1)^{n}}{(n+1)}\), (iii) \(\lim _{x \rightarrow 0} \frac{\sin x-\sum_{k=0}^{\lceil(n-2) / 2\rceil}(-1)^{k} x^{2 k+1} /(2 k+1) !}{x^{n+1}}=\left\\{\begin{array}{c}\frac{(-1)^{n / 2}}{(n+1) !} \text { if } n \text { is even, } \\ 0 \quad \text { if } n \text { is odd }\end{array}\right.\) (iv) \(\lim _{x \rightarrow 0} \frac{\cos x-\sum_{k=0}^{\lfloor n / 2\rfloor}(-1)^{k} x^{2 k} /(2 k) !}{x^{n+1}}=\left\\{\begin{array}{c}\frac{(-1)^{(n+1) / 2}}{(n+1) !} \text { if } n \text { is odd, } \\ 0 & \text { if } n \text { is even. }\end{array}\right.\)

Prove the following for all \(x_{1}, x_{2} \in \mathbb{R}\) : (i) \(\sin x_{1}=\sin x_{2} \Longleftrightarrow x_{2}=m \pi+(-1)^{m} x_{1}\), where \(m \in \mathbb{Z}\). (ii) \(\cos x_{1}=\cos x_{2} \Longleftrightarrow x_{2}=2 m \pi \pm x_{1}\), where \(m \in \mathbb{Z}\). (iii) \(\sin x_{1}=\sin x_{2}\) and \(\cos x_{1}=\cos x_{2} \Longleftrightarrow x_{2}=2 m \pi+x_{1}\), where \(m \in \mathbb{Z}\). (Hint: Exercise 28 and solutions of the equations \(\sin x=0, \cos x=0 .\) )
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