The linear approximation<\/strong> of a function at a given point, also known as the linearization<\/strong>, is equivalent to the equation of the tangent line to the graph of the function at that point.<\/p>\r\n Since the slope of a function [latex]f(x)[\/latex] at [latex](x_0, y_0)[\/latex] is given by [latex]f'(x_0)[\/latex], the linear approximation of a function [latex]f(x)[\/latex] at point [latex](x_0, y_0)[\/latex] is:<\/p>\r\n [latex]L(x) = y_0 + f'(x_0)(x - x_0)[\/latex]<\/p>\r\n For values near the point [latex](x_0, y_0)[\/latex], we have [latex]L(x) \\approx f(x)[\/latex]. In other words, [latex]L(x)[\/latex] can predict function values near [latex](x_0, y_0)[\/latex].<\/p>\r\n The actual change in function output, [latex]\\Delta y[\/latex], can be approximated using the slope at [latex](x_0, y_0)[\/latex]:<\/p>\r\n [latex]\\Delta y \\approx f'(x_0) \\Delta x[\/latex]<\/p>\r\n\r\n<\/section>To use a direction field, start by choosing any point in the field. The line segment at that point serves as a signpost telling you what direction to go from there.\r\n\r\nLet's continue with our equation [latex]y^{\\prime} = 3x + 2y - 4[\/latex]. Suppose a solution to this differential equation passes through point [latex](0, 1)[\/latex]. The slope of the solution at that point is:\r\n [latex]y^{\\prime} = 3(0) + 2(1) - 4 = -2[\/latex]<\/p>\r\n Now let [latex]x[\/latex] increase slightly to [latex]x = 0.1[\/latex]. Using linear approximation:<\/p>\r\n [latex]L(x) = y_0 + f'(x_0)(x - x_0) = 1 + (-2)(x - 0) = 1 - 2x[\/latex]<\/p>\r\n Substituting [latex]x = 0.1[\/latex] gives an approximate [latex]y[\/latex] value of [latex]0.8[\/latex].<\/p>\r\n At this new point, the slope changes according to the differential equation. You can keep progressing, recalculating the slope as you take small steps to the right, and watch how the solution behaves. Figure 3 shows the solution passing through point [latex](0, 1)[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"459\"] This curve is the graph of the solution to the initial-value problem<\/strong>:<\/p>\r\n [latex]y^{\\prime} = 3x + 2y - 4, \\quad y(0) = 1[\/latex]<\/p>\r\n This curve is called a solution curve<\/strong> passing through the point [latex](0, 1)[\/latex]. The exact solution to this initial-value problem is:<\/p>\r\n [latex]y = -\\frac{3}{2}x + \\frac{5}{4} - \\frac{1}{4}e^{2x}[\/latex]<\/p>\r\n The graph of this exact solution is identical to the curve in Figure 3.<\/p>\r\n\r\n Create a direction field for the differential equation [latex]y^{\\prime} ={x}^{2}-{y}^{2}[\/latex] and sketch a solution curve passing through the point [latex]\\left(-1,2\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558898\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n Use [latex]x[\/latex] and [latex]y[\/latex] values ranging from [latex]-5[\/latex] to [latex]5[\/latex]. For each coordinate pair, calculate [latex]y^{\\prime} [\/latex] using the right-hand side of the differential equation.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n![\"A](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233932\/CNX_Calc_Figure_08_02_002.jpg\")
Figure 3. Direction field for the differential equation [latex]y^{\\prime} =3x+2y - 4[\/latex] with the solution passing through the point [latex]\\left(0,1\\right)[\/latex].[\/caption]\r\n![\"A](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233936\/CNX_Calc_Figure_08_02_003.jpg\")
Figure 4.[\/caption]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>