LovelyHead0 From: US Posts: 8 Votes: 32 |
Find, correct to two decimal places, the coordinates of the pointon the curve y =tan(x) that is closest to the point (1,1). ***I don't know where to start; our professor never showed usany example problems like this.*** |

Anonymous From: - Posts: - Votes: - |
(2x-2 +2*tan(x)*sec2(x) - 2*sec2(x)) = 0 This seems to be when x = 0.82, which is when y = 1.08 ------------------------------- Follow up: I didn't show you how to obtain the 0.82, 1.08. You can get that byusing a nonlinear equation solver (EXCEL), but I doubt the teacherwill let you do that. You can try the method of Newton-Raphson: f(x) = (2x-2+ 2*tan(x)*sec2(x) - 2*sec2(x)) f '(x) = 2 + 2*sec4(x) + 2*tan(x)*2*sec(x)*sec(x)*tan(x)- 2*2*sec(x)*sec(x)*tan(x) f ' (x) = 2 + 2*sec4(x) +4*tan2(x)*sec2(x) -4*sec2(x)*tan(x) x_n+1 = x_n - f(x_n)/f '(x_n+1) x_n = 1 (initial guess) f(x_n) = 3.818821 f '(x_n) = =37.36328 x_n+1 = 0.897792 Let x_n+1 = 0.897782 f(x_n+1) = 1.105375 f ' (x_n+1) = 18.53371 x_n+2 = 0.838151 x_n+2 = 0.828151 f(x_n+2) = 0.174756 f '(x_n+2) = 13.10218 x_n+3 = 0.824813 (I will just give the x_n+k output now) x_n+4 = 0.82431009 x_n+5 = 0.824309 (value has converged) So x = 0.824309 y = tan(0.824309) = 1.0810173 (x,y) = (0.82,1.08) |

Anonymous From: - Posts: - Votes: - |
Distance between curve and point= sqrt((x-1)^2 + (y-1)^2) distance = sqrt((x-1)^2 + (tan(x)-1)^2)) minimize distance by differentiating with respect to x and set toderivative to 0 to find critical points. d(distance)/dx = d((x^2 - 2x + 1 + tan2(x)-2*tan(x) +1)0.5/dx derivative = (0.5)*(2x-2 + 2*tan(x)*sec2(x) -2*sec2(x))/(x^2 - 2x + 1 + tan2(x)-2*tan(x)+1)0.5=0 (2x-2 + 2*tan(x)*sec2(x) - 2*sec2(x)) = 0 Find when derivative = 0 This seems to be when x = 0.82, which is when y = 1.08 |

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