Explanation Given: The following figure is given: The following equation is given: − 1 y ' = − g ( ω 2 x ) Where ω is the angular velocity. Calculation: Along the surface of the water, the slope of the tangent line, at any point, is given by the value of y ' at that point. Thus, the slope of the line perpendicular to the surface of the water will be − 1 y ' . The slope of the resultant force generated by the gravitational force and the centrifugal force is: − m g m ω 2 x = − g ω 2 x Therefore, the curve obtained by taking the vertical cross-section of the water surface is determined by the equation: − 1 y ' = − g ω 2 x Consider the equation: − 1 y ' = − g ( ω 2 x ) y ' = ...

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ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 10.1, Problem 67E

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If $y = f (x)$ is the equation of the curve obtained by taking a vertical cross section through the axis, then $- \frac{1}{y'} = - \frac{g}{(ω^{2} x)}$ . Also, show that $y = f (x)$ is a parabola.

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Chapter 10.1, Problem 66E

Chapter 10.1, Problem 68E

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Chapter 10 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

Ch. 10.1 - Prob. 1PQ Ch. 10.1 - Prob. 2PQ Ch. 10.1 - Prob. 3PQ Ch. 10.1 - Prob. 4PQ Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E

Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - Prob. 16E Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - Prob. 19E Ch. 10.1 - Prob. 20E Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - Prob. 24E Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29E Ch. 10.1 - Prob. 30E Ch. 10.1 - Prob. 31E Ch. 10.1 - Prob. 32E Ch. 10.1 - Prob. 33E Ch. 10.1 - Prob. 34E Ch. 10.1 - Prob. 35E Ch. 10.1 - Prob. 36E Ch. 10.1 - Prob. 37E Ch. 10.1 - Prob. 38E Ch. 10.1 - Prob. 39E Ch. 10.1 - Prob. 40E Ch. 10.1 - Prob. 41E Ch. 10.1 - Prob. 42E Ch. 10.1 - Prob. 43E Ch. 10.1 - Prob. 44E Ch. 10.1 - Prob. 45E Ch. 10.1 - Prob. 46E Ch. 10.1 - Prob. 47E Ch. 10.1 - Prob. 48E Ch. 10.1 - Prob. 49E Ch. 10.1 - Prob. 50E Ch. 10.1 - Prob. 51E Ch. 10.1 - Prob. 52E Ch. 10.1 - Prob. 53E Ch. 10.1 - Prob. 54E Ch. 10.1 - 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Prob. 3PQ Ch. 10.3 - Prob. 4PQ Ch. 10.3 - Prob. 1E Ch. 10.3 - Prob. 2E Ch. 10.3 - Prob. 3E Ch. 10.3 - Prob. 4E Ch. 10.3 - Prob. 5E Ch. 10.3 - Prob. 6E Ch. 10.3 - Prob. 7E Ch. 10.3 - Prob. 8E Ch. 10.3 - Prob. 9E Ch. 10.3 - Prob. 10E Ch. 10.3 - Prob. 11E Ch. 10.3 - Prob. 12E Ch. 10.3 - Prob. 13E Ch. 10.3 - Prob. 14E Ch. 10.3 - Prob. 15E Ch. 10.3 - Prob. 16E Ch. 10.3 - Prob. 17E Ch. 10.3 - Prob. 18E Ch. 10.3 - Prob. 19E Ch. 10.3 - Prob. 20E Ch. 10.3 - Prob. 21E Ch. 10.3 - Prob. 22E Ch. 10.3 - Prob. 23E Ch. 10.3 - Prob. 24E Ch. 10.3 - Prob. 25E Ch. 10.3 - Prob. 26E Ch. 10.3 - Prob. 27E Ch. 10.3 - Prob. 28E Ch. 10.3 - Prob. 29E Ch. 10.4 - Prob. 1PQ Ch. 10.4 - Prob. 2PQ Ch. 10.4 - Prob. 3PQ Ch. 10.4 - Prob. 1E Ch. 10.4 - Prob. 2E Ch. 10.4 - Prob. 3E Ch. 10.4 - Prob. 4E Ch. 10.4 - Prob. 5E Ch. 10.4 - Prob. 6E Ch. 10.4 - Prob. 7E Ch. 10.4 - Prob. 8E Ch. 10.4 - Prob. 9E Ch. 10.4 - Prob. 10E Ch. 10.4 - Prob. 11E Ch. 10.4 - Prob. 12E Ch. 10.4 - Prob. 13E Ch. 10.4 - Prob. 14E Ch. 10.4 - Prob. 15E Ch. 10.4 - Prob. 16E Ch. 10.4 - Prob. 17E Ch. 10.4 - Prob. 18E Ch. 10.5 - Prob. 1PQ Ch. 10.5 - Prob. 2PQ Ch. 10.5 - Prob. 3PQ Ch. 10.5 - Prob. 4PQ Ch. 10.5 - Prob. 1E Ch. 10.5 - Prob. 2E Ch. 10.5 - Prob. 3E Ch. 10.5 - Prob. 4E Ch. 10.5 - Prob. 5E Ch. 10.5 - Prob. 6E Ch. 10.5 - Prob. 7E Ch. 10.5 - Prob. 8E Ch. 10.5 - Prob. 9E Ch. 10.5 - Prob. 10E Ch. 10.5 - Prob. 11E Ch. 10.5 - Prob. 12E Ch. 10.5 - Prob. 13E Ch. 10.5 - Prob. 14E Ch. 10.5 - Prob. 15E Ch. 10.5 - Prob. 16E Ch. 10.5 - Prob. 17E Ch. 10.5 - Prob. 18E Ch. 10.5 - Prob. 19E Ch. 10.5 - Prob. 20E Ch. 10.5 - Prob. 21E Ch. 10.5 - Prob. 22E Ch. 10.5 - Prob. 23E Ch. 10.5 - Prob. 24E Ch. 10.5 - Prob. 25E Ch. 10.5 - Prob. 26E Ch. 10.5 - Prob. 27E Ch. 10.5 - Prob. 28E Ch. 10.5 - Prob. 29E Ch. 10.5 - Prob. 30E Ch. 10.5 - Prob. 31E Ch. 10.5 - Prob. 32E Ch. 10.5 - Prob. 33E Ch. 10.5 - Prob. 34E Ch. 10.5 - Prob. 35E Ch. 10.5 - Prob. 36E Ch. 10.5 - Prob. 37E Ch. 10.5 - Prob. 38E Ch. 10.5 - Prob. 39E Ch. 10.5 - Prob. 40E Ch. 10.5 - Prob. 41E Ch. 10.5 - Prob. 42E Ch. 10.5 - 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Prob. 43CRE Ch. 10 - Prob. 44CRE Ch. 10 - Prob. 45CRE Ch. 10 - Prob. 46CRE Ch. 10 - Prob. 47CRE Ch. 10 - Prob. 48CRE Ch. 10 - Prob. 49CRE Ch. 10 - Prob. 50CRE Ch. 10 - Prob. 51CRE Ch. 10 - Prob. 52CRE

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If y = f ( x ) is the equation of the curve obtained by taking a vertical cross section through the axis, then − 1 y ' = − g ( ω 2 x ) . Also, show that y = f ( x ) is a parabola.

Solution Summary: The author proves that if y = f ( x ) is the equation of the curve obtained by taking a vertical cross section through the axis, then 1'= g' and

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If $y = f (x)$ is the equation of the curve obtained by taking a vertical cross section through the axis, then $- \frac{1}{y'} = - \frac{g}{(ω^{2} x)}$ . Also, show that $y = f (x)$ is a parabola.

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Chapter 10 Solutions

If y = f ( x ) is the equation of the curve obtained by taking a vertical cross section through the axis, then − 1 y ' = − g ( ω 2 x ) . Also, show that y = f ( x ) is a parabola.

Solution Summary: The author proves that if y = f ( x ) is the equation of the curve obtained by taking a vertical cross section through the axis, then 1'= g' and

Videos

If y=f(x) is the equation of the curve obtained by taking a vertical cross section through the axis, then −1y'=−g(ω2x). Also, show that y=f(x) is a parabola.

Want to see the full answer?

Chapter 10 Solutions

If $y = f (x)$ is the equation of the curve obtained by taking a vertical cross section through the axis, then $- \frac{1}{y'} = - \frac{g}{(ω^{2} x)}$ . Also, show that $y = f (x)$ is a parabola.