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Determine the minimum distance from the point (5, –1) to the parabola given byy = x², using Lagrange multipliers in the following way. (a) Let f(x, y) denote the square of the distance between the point (5, –1) and a point (x, y). Write down an expression for f(x, y). (b) Using g(x, y) = x² – y, write down the three simultaneous equations that x, y and the Lagrange multiplier 2 must satisfy at an extreme value of f(x, y) subject to the constraint g(x, y) = 0. (c) Use the simultaneous equations from part (b) to expressx and y in terms of å. Then substitute these expressions into the constraint g(x, y) = 0 to obtain a cubic equation for 2. (d) Show that À = -4 is the only solution to the cubic equation from part (c). Hence, find the point on the parabola that is closest to (5, –1) and calculate the corresponding minimum distance.

Determine the minimum distance from the point (5, –1) to the parabola given byy = x², using Lagrange multipliers in the following way. (a) Let f(x, y) denote the square of the distance between the point (5, –1) and a point (x, y). Write down an expression for f(x, y). (b) Using g(x, y) = x² – y, write down the three simultaneous equations that x, y and the Lagrange multiplier 2 must satisfy at an extreme value of f(x, y) subject to the constraint g(x, y) = 0. (c) Use the simultaneous equations from part (b) to expressx and y in terms of å. Then substitute these expressions into the constraint g(x, y) = 0 to obtain a cubic equation for 2. (d) Show that À = -4 is the only solution to the cubic equation from part (c). Hence, find the point on the parabola that is closest to (5, –1) and calculate the corresponding minimum distance.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 94E
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Determine the minimum distance from the point (5, – 1) to the parabola given byy = x², using Lagrange multipliers in the following way. x2, (a) Let f(x, y) denote the square of the distance between the point (5, –1) and a point (x, y). Write down an expression for f(x, y). (b) Using g(x, y) = x² – y, write down the three simultaneous equations that x, y and the Lagrange multiplier å must satisfy at an extreme value of f(x, y) subject to the constraint g(x, y) = 0. (c) Use the simultaneous equations from part (b) to expressx and y in terms of 1. Then substitute these expressions into the constraint g(x, y) = 0 to obtain a cubic equation for 2. (d) Show that = -4 is the only solution to the cubic equation from part (c). Hence, find the point on the parabola that is closest to (5, –1) and calculate the corresponding minimum distance. Paragraph BI Path: p II !!!
Transcribed Image Text:Determine the minimum distance from the point (5, – 1) to the parabola given byy = x², using Lagrange multipliers in the following way. x2, (a) Let f(x, y) denote the square of the distance between the point (5, –1) and a point (x, y). Write down an expression for f(x, y). (b) Using g(x, y) = x² – y, write down the three simultaneous equations that x, y and the Lagrange multiplier å must satisfy at an extreme value of f(x, y) subject to the constraint g(x, y) = 0. (c) Use the simultaneous equations from part (b) to expressx and y in terms of 1. Then substitute these expressions into the constraint g(x, y) = 0 to obtain a cubic equation for 2. (d) Show that = -4 is the only solution to the cubic equation from part (c). Hence, find the point on the parabola that is closest to (5, –1) and calculate the corresponding minimum distance. Paragraph BI Path: p II !!!
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