Calc 3.9a

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  • x – c is called the change in x and is denoted delta x. When delta x is small, the change in y, denoted delta y, can be approximated by
  • Calc 3.9a

    1. 1. Understand the concept of a tangent line approximation Compare the value of the differential, dy, with the actual change in y, ∆y
    2. 2. Consider the equation for the tangent line for a differentiable function at the point (c, f(c)) or This tangent line is called the tangent line approximation of f(x) at c. By choosing values of x sufficiently close to c, the values of y can be used as approximations for values of f. In other words, as x->c, the limit of y is f(c). slope
    3. 3. Ex. 1 p. 235 Using a tangent line approximation Find the tangent line approximation for f(x) = 1 + sin x at the point (0, 1). Then use the table to compare the y-values of the linear function with those of f(x) on an open interval containing x = 0. f ‘ (x) = cos x. Slope at (0, 1) = cos 0 = 1 Tangent line is y – f(0) = f ‘(0)(x – 0) y – 1 = 1(x – 0) y = x + 1 x -0.5 -0.1 -0.01 0 0.01 0.1 0.5 f(x) = 1 + sin x 0.521 0.9002 0.9900002 1 1.0099998 1.0998 1.479 y = x + 1 0.5 0.9 0.99 1 1.01 1.1 1.5
    4. 4. Differentials When ∆x is small, ∆y = f(c + ∆x) – f(c) is ≈ f ‘(c) ∆x which we’ll call dy
    5. 5. For such an approximation, the quantity ∆x = dx, and is called the differential of x. Then ∆y ≈ dy as defined below. In many types of applications, the differential of y can be used as an approximation of the change in y. That is, ∆y ≈ dy, or ∆y ≈ f ‘(x) dx
    6. 6. Ex 2. p 236 Comparing ∆y and dy Let y = x 3 Find dy when x = 1 and dx = 0.01. Compare this value with ∆y for x = 1 and ∆x = 0.01 Solution: f ‘(x) = 3x 2 . dy = f ‘(x)dx = f ‘(1)(0.01) = 3(0.01) = 0.03 Now, using ∆x = 0.01, the change in y is ∆ y = f(x + ∆x) – f(x) = f(1 + 0.01) – f(1) = (1.01) 3 – 1 3 = 1.030301 – 1 = 0. 030301 In this example, the tangent line at (1,1) is y = 3x – 2 or g(x) = 3x – 2 . Choosing x-values near 1, f (1.01) = 1.030301 and g(1.01) = 1.03
    7. 7. Error Propagation Physicists and engineers make liberal use of approximation dy to replace ∆y. In practice this is in estimation of errors propagated by measuring devices. If x represents the measured value of variable, and x + ∆ x is the exact value, then ∆x is the error in measurement . If the measured value x is used to compute another value f(x), the difference between f(x + ∆x) and f(x) is the propagated error . f(x + ∆x) – f(x) = ∆y
    8. 8. Example 3 p.237 Estimation of Error The radius of a ball bearing is measured to be 0.7 inches. If the measurement is correct to within 0.01 inch, estimate the propagated error in the volume of the ball bearing. The formula for volume is where r is the radius of the sphere. So r = 0.7 (measured radius) and Differentiate V to obtain dV/dr = 4 π r 2 ∆ V ≈ dV = 4 π r 2 dr = 4 π (.7) 2 (  0.01) ≈  0.06158 cubic inches Is that a lot of change in volume?
    9. 9. The answer is best given in relative terms by comparing dV with V. This is called the relative error . The corresponding percent error is approximately 4.29%
    10. 10. 3.9a p. 240/ 1-5 odd, 7-11 all, 13-33 every other odd

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