1)What is linear approximation used for? Why would you use the linear approximat
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1)What is linear approximation used for? Why would you use the linear approximation of a function rather than the actual function? 1B)State the general formula for La(x), the linear approximation of a function around the point x = a. Use this formula to find the linear approximation Lo(x) of the function f(x) =root 1 ? x at a = 0. 1 c.) Use the linear approximation that you found in part (b) to approximate root 0.99. 1d)sketch and label the graphs of both f(x) and Lo(x) and label the point where they touch. Use these graphs to determine if your estimate in part (c) is greater or less than the actual value of root 0.99. Explain your choice by referring to the graphs of f(x) and Lo(x)Explanation / Answer
Linear Approximations This approximation is crucial to many known numerical techniques such as Euler's Method to approximate solutions to ordinary differential equations. The idea to use linear approximations rests in the closeness of the tangent line to the graph of the function around a point. Let x0 be in the domain of the function f(x). The equation of the tangent line to the graph of f(x) at the point (x0,y0), where y0 = f(x0), is If x1 is close to x0, we will write , and we will approximate by the point (x1,y1) on the tangent line given by If we write , we have In fact, one way to remember this formula is to write f'(x) as and then replace d by . Recall that, when x is close to x0, we have Example. Estimate . Let . We have . Using the above approximation, we get We have So . Hence or . Check with your calculator and you'll see that this is a pretty good approximation for . Remark. For a function f(x), we define the differential df of f(x) by Example. Consider the function y = f(x) = 5x2. Let be an increment of x. Then, if is the resulting increment of y, we have On the other hand, we obtain for the differential dy: In this example we are lucky in that we are able to compute exactly, but in general this might be impossible. The error in the approximation, the difference between dy(replacing dx by ) and , is , which is small compared to . r this reason, the linear function whose graph is the tangent line to y = f(x) at a specified point (a, f(a)) is called the linear approximation of f(x) near x = a. Q What is the formula for the linear approximation? A All we need is the equation of the tangent line at a specified point (a, f(a)). Since the tangent line at (a, f(a)) has slope f'(a), we can write down its equation using the point-slope formula: y = y0 + m(x - x0) = f(a) + f'(a)(x - a) Thus, the the linear approximation to f(x) near x = a is given by L(x) = f(a) + f'(a)(x - a). Q The above argument is based on geometry: the fact that the tangent line is close to the original graph near the point of tangency. Is there an algebriac way of seeing why this is true? A Yes. This links to an algebraic derivation of the linear approximation. Linear Approximation of f(x) Near x = a If x is close to a, then f(x) f(a) + (x-a)f'(a). The right-hand side, L(x) = f(a) + (x-a)f'(a), which is a linear function of x, is called the linear approximation of f(x) near x = a. Example 1 Linear Approximation of the Square Root Let f(x) = x1/2. Find the linear approximation of f near x = 4 (at the point (4, f(4)) = (4, 2) on the graph), and use it to approximate 4.1. Solution Since f'(x) = 1/(2x1/2), f'(4) = 1/(2 . 41/2) = 1/4. so the linear approximation is L(x) = f(4) + (x-4)f'(4) = 2 + (x-4)/4 = 0.25x + 1. We can use L(x) to approximate the square root of any number close to 4 very easily without using a calculator. For example, 4.1 L(4.1) = 0.25(4.1) + 1 = 2.025 Q 3.82 Q The Linear approximation of the same function, f(x) = x1/2, near x = 9 is given by L(x) = Example 2 Linear Approximation of the Logarithm Use linear approximation to approximate ln(1.134). Solution Here, we are not given a value for a. The key is to use a value close to 1.134 whose natural logarithm we know. Since we know that ln(1) = 0, we take a to be 1. Now use the formula for linear approximation: L(x) = f(a) + (x-a)f'(a). Substituting and simplifying gives (numerical answers should be accurate to 4 decimal places): Q L(x) = ln(1.134) The actual value is ln(1.134) is The error in the approximation is The approximation is accurate to Before we go on... You can use L(x) = x-1 to find approximations to the natural logarithm of any number close to 1: for instance, ln(0.843) 0.843 - 1 = -0.157, ln(0.999) 0.999 - 1 = -0.001. Error Estimation When a physical measurement is made, there is always some uncertainty about it accuracy. For instance, if you are measuring the radius of a ball bearing, you might measure it repeatedly and obtain slightly differing results. Rather than concluding, say, that the radius of the ball bearing is exactly 1.2mm, you may instead conclude that the radius is 1.2 mm 0.1 mm. (The actual calculation of the range 0.1 mm is often given by a statistical formula based on the standard deviation of all the separate measurements.) Once you have an error estimate for the radius, you might wonder how this error might effect the calculation of the volume id the ball bearing. In other words, if the radius is off by 0.1 mm, by how much is the volume off? To answer the question, think of the error of the radius as a change, r, in r, and then compute the associated change, V, in the volume V. The general question is therefore: Q If x changes by x, and y is a function of x, what is the associated change y in y? To answer this question, let us go back to our linear approximation formula: We saw above that, near x = a, f(x) f(a) + (x-a)f'(a), or f(x) - f(a) (x-a)f'(a) The quantity f(x) - f(a) represents the change in f corresponding to a change in the independent variable from a to x. In other words, Change in f Change in x f'(a). Using the delta notation, this becomes f x f'(x). If y = f(x), we can write this formula as y x dy dx Notice how the dx and x appear to cancel Estimating the error of y = f(x) If x = a, with a possible error of x, and y = f(x),then y = f(a), with a possible error of y, given by y x dy dx x=a Example Suppose y = x2 + 3x, and x = 2, accurate to 0.2, then the associated value of y is 22 + 3(2) = 10, and is accurate to within y, where y x dy dx x=2 = (0.2)(7) Since dy/dx = 2x+3, which is 7 when x = 2 = 1.4. Therefore, even though the error in x is only 0.2, the error in y is much larger; approximately 1.4. The next examples discuss functions specified algebraically rather than geometrically. Example 3 Measurement Error Precision Corp. manufactures ball bearings with a radius of 1.2 millimeter, varying by 0.1 millimeters. What is the volume of the ball bearings, and by how much can it vary? Solution The volume of a sphere and its derivative are given by V = 4 3 r3. dV dr = 4r3 Evaluating these quantities at r = 1.2 gives V = 4 3 (1.2)3 7.24 mm3 dV dr r=1.2 = 4(1.2)3 18.10 Thus, V r dV dr r=1.2 = (0.1)(18.1) = 1.81 Thus, the volume of the ball bearings is 7,24 1.81 mm3 Example 4 Little Cones The Little Cones Operad Co. manfactures cone-shaped ornaments of various colors. All the ornaments have height 10 mm and radius of base 2 mm. The radius of the cones is known to be accurate to within 0.15 mm. (Note: The volume of a cone of height h and radius of base r is V = 1 3 r2h.)
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