Estimating Function Values & Changes with Differentials & L'Hopital's Rule, Study notes of Calculus

Download Estimating Function Values & Changes with Differentials & L'Hopital's Rule and more Study notes Calculus in PDF only on Docsity! Linear Approximations Facts: Let f(x) be a differentiable function, and x0 be a point in the domain of f(x). (1) Recall the tangent line of y = f(x) at x = x0 is y = f(x0) + f ′(x0)(x− x0). (2) The amount ∆x = x− x0 is the x-increment, (also denoted dx), and the amount f(x)− f(x0) is the y-increment,∆y = f(x)− f(x0). (3) The differential of f(x) at x = x0 is df(x0) = f ′(x0)dx. In general, df(x) = f ′(x)dx. When y = f(x), we also use dy for df . (4) The linear approximation of the function f (near the point x = x0) is L(x) = f(x0) + f ′(x0)(x−x0). Linear Approximation L(x) can be used to estimate f(x) when x is near x0. f(x) ' L(x) = f(x0) + f ′(x0)(x− x0). Example 1 Use a linear approximation to estimate the number √ 80. Solution: Let f(x) = √ x. Since we want to estimate √ 80, x = 80. To choose x0, we ask ourselves the question: which number x0 is close to 80 and is a perfect square? The answer is x0 = 81, and so the linear approximation estimation is √ 80 ' √ 81 + 1 18 (−1) = 9− 1 18 = 161 18 . Example 2 Use a linear approximation to estimate the number sin 32o. Solution: Let f(x) = sin(x). Since we want to estimate sin 32o, x = 32o. To choose x0, we ask ourselves the question: which angle x0 is close to 32o and sinx0 is known? The answer is x0 = 30o. When using calculus to deal with trig function values, it is recommended to convert the measure of an angle from degrees to radians. Thus linear approximation estimation is sin 32o ' sin 30o + √ 3 2 · π 90 = 1 2 + π √ 3 180 = 90 + π √ 3 180 . Example 3 Use a linear approximation to estimate the change of the area of a square, when its edge length is decreased from 10 in to 9.8 in. Solution: Let x denote the length of an edge of the square. Then the area is f(x) = x2. Note that x changes from 10 in to 9.8 in, and so we set x0 = 10 and x = 9.8. It follows that ∆x = 9.8 − 10 = −0.2. Note that f ′(x) = 2x. Thus f ′(x0) = f ′(10) = 20. The linear approximation estimated change is ∆f = f ′(10)∆x = (20)(−0.2) = −4. 1