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Fill in the blank below. The shortest and least-curved arc between two points on a surface is a The shortest and least-curved arc between two points on a surface is a sphere. plane. geodesic. pseudosphere. line.

          Fill in the blank below.
The shortest and least-curved arc between two points on a surface is a
The shortest and least-curved arc between two points on a surface is a
sphere.
plane.
geodesic.
pseudosphere.
line.
        
Fill in the blank below.
The shortest and least-curved arc between two points on a surface is a
The shortest and least-curved arc between two points on a surface is a
sphere.
plane.
geodesic.
pseudosphere.
line.

Added by Stephen M.

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Elementary and Intermediate Algebra
Elementary and Intermediate Algebra
Alan S. Tussy, R. David Gustafson 5th Edition
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The shortest distance between two points on a surface is called a "geodesic". In spirit, a geodesic is locally a straight line on the surface, but globally it bends with the surface. For example, the lines of constant longitude and lattitude are geodesics of the earth's surface. (a) Use spherical polar coordinates to show that the length of a path on the surface of a sphere is L = R ∫_{θ_1}^{θ_2} √(1 + sin² θ ϕ'(θ)²) dθ viewing the path as the function ϕ(θ) where θ is the polar angle and ϕ the azimuthal angle of a point on the sphere. (b) Prove that the shortest distance between two points on the sphere is a great circle. Hint place your initial point on the north pole.

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The shortest path between two points on a curved surface, such as the surface of a sphere, is called a geodesic. To find a geodesic, one has first to set up an integral that gives the length of a path on the surface in question. This will always be similar to the integral (6.2) but may be more complicated (depending on the nature of the surface) and may involve different coordinates than $x$ and $y$. To illustrate this, use spherical polar coordinates $(r, \theta, \phi)$ to show that the length of a path joining two points on a sphere of radius $R$ is $L=R \int_{\theta_{1}}^{\theta_{2}} \sqrt{1+\sin ^{2} \theta \phi^{\prime}(\theta)^{2}} d \theta$ if $\left(\theta_{1}, \phi_{1}\right)$ and $\left(\theta_{2}, \phi_{2}\right)$ specify the two points and we assume that the path is expressed as $\phi=\phi(\theta)$ (You will find how to minimize this length in Problem $6.16 .)$

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Transcript

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00:01 Hello students, here we have given points that is x1, y1 and x2, y2 and the path y is equal to y of x.
00:17 Now to find the path y of x that has the shortest length that is in a straight line, we consider length of the path is equal to dx square plus dy square power 1 by 2 that is root.
00:38 Now dy is equal to dy by dx that is y dash of x into dx.
00:47 Now substituting this in the above equation, ds is equal to 1 plus y dash of x square power 1 by 2 as it is.
01:02 Here we take common as dx from both, hence we write dx because dx square power 1 by 2 is dx.
01:14 The shortest distance of the path between points 1 and 2 is given by this equation.
01:21 Now in polar coordinates, we consider x is equal to r cos theta, y is equal to r sin theta, then dx is equal to r minus sin theta d theta and dy is equal to r cos theta d theta.
01:49 Hence, we can write y dash of x is equal to dy by dx dividing r cos theta by r sin theta, rr get cancelled and we get minus cot theta because cos theta by sin theta is cot theta.
02:13 Hence we obtain the equation.
02:16 Next to calculate distance, let us consider starting from a and ending to b at t is equal to 1...
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