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Chapter 7: Problem 49

Find the polar coordinates of the points in \(\mathbb{R}^{2}\) whose Cartesian coordinates are as follows: (i) \((1,1)\), (ii) \((0,3)\), (iii) \((2,2 \sqrt{3})\), (iv) \((2 \sqrt{3}, 2)\).

Short Answer

Expert verified
The polar coordinates for the given points are: (i) \((1,1)\): \((\sqrt{2}, \frac{\pi}{4})\) (ii) \((0,3)\): \((3, \frac{\pi}{2})\) (iii) \((2, 2\sqrt{3})\): \((4, \frac{\pi}{3})\) (iv) \((2\sqrt{3}, 2)\): \((4, \frac{\pi}{6})\)

Step by step solution

01

Find the Radius

Use the distance from the origin formula to find the radius: \[r = \sqrt{x^2 + y^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\]
02

Find the Angle

Use the angle formula to find the angle 胃: \[\theta = \tan^{-1}(\frac{y}{x}) = \tan^{-1}(\frac{1}{1}) = \tan^{-1}(1) = \frac{\pi}{4}\] The polar coordinates for the point \((1,1)\) are \(\boxed{(\sqrt{2}, \frac{\pi}{4})}\). (ii) Point \((0, 3)\):
03

Find the Radius

Use the distance from the origin formula to find the radius: \[r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = 3\]
04

Find the Angle

Use the angle formula to find the angle 胃: \[\theta = \tan^{-1}(\frac{y}{x}) = \tan^{-1}(\frac{3}{0}) = \frac{\pi}{2}\] The polar coordinates for the point \((0,3)\) are \(\boxed{(3, \frac{\pi}{2})}\). (iii) Point \((2, 2\sqrt{3})\):
05

Find the Radius

Use the distance from the origin formula to find the radius: \[r = \sqrt{x^2 + y^2} = \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{16} = 4\]
06

Find the Angle

Use the angle formula to find the angle 胃: \[\theta = \tan^{-1}(\frac{y}{x}) = \tan^{-1}(\frac{2\sqrt{3}}{2}) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\] The polar coordinates for the point \((2, 2\sqrt{3})\) are \(\boxed{(4, \frac{\pi}{3})}\). (iv) Point \((2\sqrt{3}, 2)\):
07

Find the Radius

Use the distance from the origin formula to find the radius: \[r = \sqrt{x^2 + y^2} = \sqrt{(2\sqrt{3})^2 + 2^2} = \sqrt{16} = 4\]
08

Find the Angle

Use the angle formula to find the angle 胃: \[\theta = \tan^{-1}(\frac{y}{x}) = \tan^{-1}(\frac{2}{2\sqrt{3}}) = \tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}\] The polar coordinates for the point \((2\sqrt{3}, 2)\) are \(\boxed{(4, \frac{\pi}{6})}\).

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Most popular questions from this chapter

Let \(f:(0, \infty) \rightarrow \mathbb{R}\) be continuous and satisfy $$ \int_{1}^{x y} f(t) d t=y \int_{1}^{x} f(t) d t+x \int_{1}^{y} f(t) d t \quad \text { for all } x, y \in(0, \infty) $$ Show that \(f(x)=f(1)(1+\ln x)\) for all \(x \in(0, \infty)\). (Hint: Consider \(F(x):=\) \(\left(\int_{1}^{x} f(t) d t\right) / x\) for \(x \in(0, \infty)\) and use Exercise 5.)

Let \(\alpha \in \mathbb{R}\) and \(f:(0, \infty) \rightarrow \mathbb{R}\) be a differentiable function such that \(f^{\prime}(x)=\alpha / x\) for all \(x \in(0, \infty)\) and \(f(1)=0\). Show that \(f(x)=\alpha \ln x\) for all \(x \in(0, \infty)\). (Compare Exercise 4 of Chapter 4.)

Show from first principles that the function cos is differentiable at \(\pi / 2\) and its derivative at \(\pi / 2\) is \(-1\).

Prove the following for all \(x_{1}, x_{2} \in \mathbb{R}:\) (i) \(\sin x_{1}+\sin x_{2}=2 \sin \left(\left(x_{1}+x_{2}\right) / 2\right) \cos \left(\left(x_{1}-x_{2}\right) / 2\right)\) (ii) \(\sin x_{1}-\sin x_{2}=2 \cos \left(\left(x_{1}+x_{2}\right) / 2\right) \sin \left(\left(x_{1}-x_{2}\right) / 2\right)\), (iii) \(\cos x_{1}+\cos x_{2}=2 \cos \left(\left(x_{1}+x_{2}\right) / 2\right) \cos \left(\left(x_{1}=x_{2}\right) / 2\right)\), (iv) \(\cos x_{1}-\cos x_{2}=2 \sin \left(\left(x_{1}+x_{2}\right) / 2\right) \sin \left(\left(x_{2}-x_{1}\right) / 2\right)\).

Let \(\alpha, \beta \in \mathbb{R}\). Suppose \(f, g: \mathbb{R} \rightarrow \mathbb{R}\) are differentiable functions such that $$ f^{\prime}=\alpha f+\beta g, \quad g^{\prime}=\alpha g-\beta f, \quad f(0)=0, \quad \text { and } \quad g(0)=1 $$ Show that \(f(x)=e^{\alpha x} \sin \beta x\) and \(g(x)=e^{\alpha x} \cos \beta x\) for all \(x \in \mathbb{R}\). (Hint: Consider \(h: \mathbb{R} \rightarrow \mathbb{R}\) given by \(h(x):=\left(f(x)-e^{\alpha x} \sin \beta x\right)^{2}+(g(x)-\) \(\left.e^{\alpha x} \cos \beta x\right)^{2}\). Find \(h^{\prime}\).) (Compare Exercise 6 of Chapter 4.)
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