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enovo McAfee YouTube Gmail Maps Cart - Cracker Barrel Question 9 \( 0 / 1 \) pt \( 3 \stackrel{\leftrightarrows}{\rightleftarrows} 99 \) Details Convert the point \( (x, y, z)=(-5,-5,2) \) to spherical coordinates. Give answers as positive values, either as expressions, or decimals to one decimal place. \[ (\rho, \theta, \phi)= \] \( \square \) Question Help: Video Submit Question

Ask by Davies Gardner. in the United States
Mar 08,2025

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\[ (\rho, \theta, \phi) = (7.3, 3.9, 1.3) \]

Solución

We start with the Cartesian point \[ (x,y,z)=(-5,-5,2) \] In spherical coordinates, the point is given by \[ (\rho,\theta,\phi) \] where 1. \(\rho=\sqrt{x^2+y^2+z^2}\) 2. \(\theta=\) the angle in the \(xy\)-plane (measured from the positive \(x\)-axis) 3. \(\phi=\) the angle from the positive \(z\)-axis --- **Step 1. Find \(\rho\):** \[ \rho=\sqrt{(-5)^2+(-5)^2+2^2}=\sqrt{25+25+4}=\sqrt{54}=3\sqrt{6} \approx 7.3 \] --- **Step 2. Find \(\theta\):** Using \[ \theta=\arctan\left(\frac{y}{x}\right) \] we get \[ \arctan\left(\frac{-5}{-5}\right)=\arctan(1)=\frac{\pi}{4} \] However, note that the point \((-5,-5)\) lies in the third quadrant where both \(x\) and \(y\) are negative. To account for the correct quadrant, we add \(\pi\): \[ \theta=\frac{\pi}{4}+\pi=\frac{5\pi}{4} \approx 3.9 \] --- **Step 3. Find \(\phi\):** \[ \phi=\arccos\left(\frac{z}{\rho}\right)=\arccos\left(\frac{2}{3\sqrt{6}}\right) \] To approximate, first compute: \[ 3\sqrt{6}\approx 7.3,\quad \frac{2}{7.3}\approx 0.274 \] Then, \[ \phi\approx\arccos(0.274)\approx 1.3 \] --- **Answer:** \[ (\rho,\theta,\phi)=\left(3\sqrt{6}, \frac{5\pi}{4}, \arccos\left(\frac{2}{3\sqrt{6}}\right)\right) \] or, approximated in decimals, \[ (7.3,\,3.9,\,1.3) \]

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To convert the Cartesian coordinates \((-5, -5, 2)\) to spherical coordinates \((\rho, \theta, \phi)\), we first calculate \(\rho\), which is the distance from the origin. It's given by the formula \(\rho = \sqrt{x^2 + y^2 + z^2}\). Plugging in the values: \(\rho = \sqrt{(-5)^2 + (-5)^2 + 2^2} = \sqrt{25 + 25 + 4} = \sqrt{54} \approx 7.35\). Next, we determine \(\theta\), which is the angle in the \(xy\)-plane from the positive \(x\)-axis. It is calculated using \(\tan(\theta) = \frac{y}{x}\), so \(\theta = \tan^{-1}\left(\frac{-5}{-5}\right) = \tan^{-1}(1) = \frac{3\pi}{4}\) (adjusted to the correct quadrant, where both \(x\) and \(y\) are negative). For \(\phi\), the polar angle from the positive \(z\)-axis, we use the formula \(\phi = \cos^{-1}\left(\frac{z}{\rho}\right)\): \(\phi = \cos^{-1}\left(\frac{2}{\sqrt{54}}\right) \approx 1.36\) radians. Thus, the spherical coordinates are: \[ \left(\sqrt{54}, \frac{5\pi}{4}, 1.36\right) \]

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