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# Proximity questions a. What point on the line y = + 4 is

ISBN: 9780321570567 2

## Solution for problem 55E Chapter 4.4

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition

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Problem 55E

Proximity questions a. What point on the line y ? ? = ? ? + 4 is closest to the origin? b. What point on the parabola ?y? = 1 ? x i 2 ? s closest to the point (1, 1)? c. Find the point on the graph of y = x t? ? hat is nearest the point (p,0) if (i)p > 2 and (ii)0 < p < 2 .

Step-by-Step Solution:

Solution 55E Step 1 a) Consider the liney = 3x+4. Let the required point be (x,y) on this line and consider the distance from this point to the (0,0)is D. Therefore, D = (x0) +(y0) 2 = x +(3x+4) 2 = 10x +24x+16 For the convenience of solution, consider the square of the distance. Find the critical point by equating the derivative to zero. dD2 dx = 20x+24 0 = 20x+24 20x = 24 x = 6 5 Substitute to get the value of y = 5 2 It is found that second derivative of D is a positive value. Therefore the distance is minimized at these points of and y. 6 2 Thus the closest point on the liney = 3x+4 to the origin is5( 5 ) b) Consider the parabolay = 1x . Let the required point be (x,y) on this line. And consider the distance from this point to the point (1,1) .D Therefore, D = (x1) +(y1) 2 2 2 2 = (x1) +(1x 1) = (x1) +(x ) 22 = x +x 2x+1 For the convenience of solution, consider the square of the distance. Find the critical point by equating the derivative to zero. 2 ddx = 4x +2x2 3 0 = 4x +2x2 x = 0.59 Substitute to get the value of y 2 2 2 It is found that second derivative of D ,D = 12x +2 is a positive value. Therefore the distance is minimized at these points of and y. Thus the closest point on the parabola y = 1x to the point (1,1)is (0.59,0.65) c) Consider the parabolay = x Let the required point be (x,y) on this line. And consider the distance from this point to the point (p,0) i . Therefore, 2 2 D = (xp) +(y0) = (y p) +(y) 2 = y +p 2py +y 2 2 For the convenience of solution, consider the square of the distance. Find the critical point by equating the derivative to zero. 2 dD = 4y +4py+2y dx 3 0 = 4y +4py+2y Step 2 : i)Substitute to get the value of y It is found that second derivative of D that is D = 12y 4p+2 is is a positive value. Therefore the distance is minimized at these points of x 1 holds only if p > 2 Thus the closest point on the parabola y = x to the point (p,0)whenp > (p ,2 p )2

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