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Q2*) In question P3 we showed that a minimal surface of revolution is given by revolution (about the x-axis) of the catenary, with equation y = C cosh ((x – B)/C). - (a) Suppose, without loss of generality, that the catenary passes through the initial point P = (x1,y1) = (0, 1). First deduce an expression for the one-parameter family of catenaries passing through point P. Next calculate the value of x at which y takes its minimum value. By using the inequality cosh > √2 (you might like to think about how to prove this), show that there are points Q for which it is impossible to find a catenary passing through both P and Q. In particular, show that it is impossible to find a catenary joining the points (0, 1) and (2, 1). (b) A minimal surface of revolution can be realised experimentally by soap films attached to circular wire frames (see this link and this link for examples). The physical reason for this is that the surface tension, which is proportional to the area, is being minimised. Given the result in part (a), what do you think will happen as the rings are moved slowly further apart? What will be the minimal surface when the rings are well separated?

Q2*) In question P3 we showed that a minimal surface of revolution is given by revolution (about the x-axis) of the catenary, with equation y = C cosh ((x – B)/C). - (a) Suppose, without loss of generality, that the catenary passes through the initial point P = (x1,y1) = (0, 1). First deduce an expression for the one-parameter family of catenaries passing through point P. Next calculate the value of x at which y takes its minimum value. By using the inequality cosh > √2 (you might like to think about how to prove this), show that there are points Q for which it is impossible to find a catenary passing through both P and Q. In particular, show that it is impossible to find a catenary joining the points (0, 1) and (2, 1). (b) A minimal surface of revolution can be realised experimentally by soap films attached to circular wire frames (see this link and this link for examples). The physical reason for this is that the surface tension, which is proportional to the area, is being minimised. Given the result in part (a), what do you think will happen as the rings are moved slowly further apart? What will be the minimal surface when the rings are well separated?

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 39RE
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Q2*) In question P3 we showed that a minimal surface of revolution is given by revolution (about the x-axis) of the catenary, with equation y = C cosh ((x – B)/C). - (a) Suppose, without loss of generality, that the catenary passes through the initial point P = (x1,y1) = (0, 1). First deduce an expression for the one-parameter family of catenaries passing through point P. Next calculate the value of x at which y takes its minimum value. By using the inequality cosh > √2 (you might like to think about how to prove this), show that there are points Q for which it is impossible to find a catenary passing through both P and Q. In particular, show that it is impossible to find a catenary joining the points (0, 1) and (2, 1). (b) A minimal surface of revolution can be realised experimentally by soap films attached to circular wire frames (see this link and this link for examples). The physical reason for this is that the surface tension, which is proportional to the area, is being minimised. Given the result in part (a), what do you think will happen as the rings are moved slowly further apart? What will be the minimal surface when the rings are well separated?
Transcribed Image Text:Q2*) In question P3 we showed that a minimal surface of revolution is given by revolution (about the x-axis) of the catenary, with equation y = C cosh ((x – B)/C). - (a) Suppose, without loss of generality, that the catenary passes through the initial point P = (x1,y1) = (0, 1). First deduce an expression for the one-parameter family of catenaries passing through point P. Next calculate the value of x at which y takes its minimum value. By using the inequality cosh > √2 (you might like to think about how to prove this), show that there are points Q for which it is impossible to find a catenary passing through both P and Q. In particular, show that it is impossible to find a catenary joining the points (0, 1) and (2, 1). (b) A minimal surface of revolution can be realised experimentally by soap films attached to circular wire frames (see this link and this link for examples). The physical reason for this is that the surface tension, which is proportional to the area, is being minimised. Given the result in part (a), what do you think will happen as the rings are moved slowly further apart? What will be the minimal surface when the rings are well separated?
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