Lesson 3-11. Linear Approximations and Differentials. Objectives. Use linear approximation and differentials to estimate functional values and changes. Vocabulary. Linear approximation – linear approximation to a function at a point, also known as a tangent line approximation
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∆y f(x) – f(a)
f’(a) = lim ----- = lim --------------
∆x x - a
Linear Approximation L(x) = f(a) + f’(a)(x – a)
= f(a) + f’(a)(∆x)
Makes a line (L) using the slope of the function at point a to estimate small changes around a. It uses the alternate form of the derivative:
∆y = f(b) – f(a)
dy = L(b) – f(a)
----- = f’(a)
dy = f’(a) dx
dx = ∆x
For small changes in x (∆x close to 0), differential dy approximates ∆y
dy = (3x2 – 3)dx
dy = (2x + 3)dx / (x² + 3x)½
dy = (4x3 – 6x) sin (x4 – 3x² + 11) dx
It is known that y = 3sin (2t) + 4cos² t. If t is measured as 1.13 ± 0.05, calculate y and give an estimate for the error.
y = 3sin (2t) + 4cos² t
y = 3sin (2.26) + 4cos² (1.13) = 3.0434
dy = (6cos (2t) - 8cos t (sin t))dt
∆y ≈ dy = (6cos (2(1.13)) - 8cos (1.13) (sin (1.13))(0.05)
Comparing ∆y and dy. ∆y is the actual change in y. dy is the approximate change in y.
Let y = x³. Find dy when x = 1 and dx = 0.05. Compare this value to ∆y when x = 1 and ∆x = 0.05.
y = x³
dy = 3x² dx
dy = 3(1) (0.05) = 0.15
y(1) = 1 y(1.05) = 1.157625 so ∆y = 0.157625
We also know we can use the tangent line to approximate a curve. The tangent line y = f(a) + f’(a)(x – a) is called the linear approximation of f at a.
Find the linear approximation to f(x) = 5x³ + 6x at x = 2. Then approximate f(1.98) for the function.
f(x) = 5x³ + 6x
f’(x) = 15x² + 6
f(2) = 52
f’(2) = 66
f(1.98) ≈ f(2) + f’(2) (1.98 – 2)
≈ 52 + 66 (-0.02)
≈ 50.68 f(1.98) = 50.692
Tachykara, maker of volleyballs for USVBA, is required to make volleyballs that are 10 cm in radius with an error no more than 0.05 cm. For a special event, the event director fills the volleyballs with a mixture of regular air and helium. She only has a small amount of the mixture and is concerned about the error in volume. Find the maximum error in volume.
V = 4/3 πr³
dV = 4 πr² dr
dV = 4 π(10)² (0.05)
= 62.832 cm³ (maximum error in volume)
dV 4 πr² dr dr
----- = -------------- = 3 ----- Relative Error
V 4/3 πr³ r