Lesson 28 (11/04/26)

Page 1

Lesson 28 (11/04/26)

Linear Approximation (4.6)

Warmup: Find equation of tangent line to \( f(x) = \sqrt{x} \) at \( (4, 2) \).

A coordinate plane graph showing two curves. A blue curve representing the square root function starts at the origin (0,0) and curves upwards with a decreasing slope. A purple straight line represents the tangent line, which touches the blue curve at the point (4, 2). The tangent line intersects the blue curve at this point and extends above and below it, clearly showing the linear approximation at that specific coordinate.

Visual Description: A coordinate plane graph showing two curves. A blue curve representing the square root function starts at the origin (0,0) and curves upwards with a decreasing slope. A purple straight line represents the tangent line, which touches the blue curve at the point (4, 2). The tangent line intersects the blue curve at this point and extends above and below it, clearly showing the linear approximation at that specific coordinate.

Slope of tangent line at \( (4, 2) \) = Derivative at \( x = 4 \)

\[ f'(x) = \frac{d}{dx} x^{1/2} = \frac{1}{2} x^{-1/2} \]

\[ f'(4) = \frac{1}{2} (4)^{-1/2} = \frac{1}{4} \]

Tangent line = Line through \( (4, 2) \) with slope \( 1/4 \)

\[ (y - 2) = \frac{1}{4}(x - 4) \implies y = 2 + \frac{1}{4}(x - 4) \]

Page 2

Linear Approximation: Approximating function value using tangent line

A coordinate plane graph showing a curved function y=f(x) in blue and a straight tangent line L in purple. The line L touches the curve at the point (a, f(a)). Vertical dashed lines drop from the point (a, f(a)) down to the x-axis, marking the position of 'a'.

Visual Description: A coordinate plane graph showing a curved function y=f(x) in blue and a straight tangent line L in purple. The line L touches the curve at the point (a, f(a)). Vertical dashed lines drop from the point (a, f(a)) down to the x-axis, marking the position of 'a'.

L = tangent line at (a, f(a))

= line through (a, f(a)) with slope f'(a)

y = f(a) + f'(a)(x - a)

Linear Approximation of f(x) at x = a is L(x) = f(a) + f'(a)(x - a)

Note: The term f'(a)(x - a) is labeled as the differential.

ERROR = |Actual - Approx|

% ERROR = \(\frac{|Actual - Approx|}{|Actual|} \times 100\)

Page 3

\( f(x) = \sqrt{x} \)

\( L(x) = \) tangent line

\( = f(4) + f'(4)(x - 4) \)

\[ L(x) = 2 + \frac{1}{4}(x - 4) \]

we \( L(x) \) to approximate \( f(x) \)

A coordinate plane showing a graph of f(x) = sqrt(x) in blue, which is a curve passing through the origin. A tangent line L(x) in purple touches the curve at the point (4, 2). Vertical dashed lines extend down to the x-axis marking points 4, 4.1, and 5. The tangent line L(x) is shown to be above the curve f(x) at x=5, highlighting the difference between the approximation and the actual value.

Visual Description: A coordinate plane showing a graph of f(x) = sqrt(x) in blue, which is a curve passing through the origin. A tangent line L(x) in purple touches the curve at the point (4, 2). Vertical dashed lines extend down to the x-axis marking points 4, 4.1, and 5. The tangent line L(x) is shown to be above the curve f(x) at x=5, highlighting the difference between the approximation and the actual value.

1. Approximation examples

\( f(4.1) = \sqrt{4.1} \approx L(4.1) = 2 + \frac{1}{4}(4.1 - 4) = 2.025 \)

Actual \( = \sqrt{4.1} = 2.0248 \)

Note: The difference between these two lines is bracketed with the text "ERROR = |Actual - Approx| = 0.0002".

2. Further approximation

\( f(5) = \sqrt{5} \approx L(5) = 2 + \frac{1}{4}(5 - 4) = 2.25 \) or \( 2 + \frac{1}{4} = 9/4 \)

Actual \( = \sqrt{5} = 2.236 \)

Note: The lines for the second example are bracketed with the text "ERROR = 0.014".

Page 4

Example: Use linear approximation to estimate the value of \(\sqrt{40}\)

A graph on a Cartesian coordinate system showing a concave down curve (blue) starting from the origin. There are two straight lines intersecting the curve. A green tangent line touches the curve at x = 36. A purple secant line passes through x = 4 and another point on the curve. Vertical dashed lines are drawn from the x-axis at values 4, 36, and 40 to the corresponding points on the curve and lines. The graph illustrates that the linear approximation at x = 4 is far from the actual value of the function at x = 40, while the linear approximation at x = 36 is much closer.

Visual Description: A graph on a Cartesian coordinate system showing a concave down curve (blue) starting from the origin. There are two straight lines intersecting the curve. A green tangent line touches the curve at x = 36. A purple secant line passes through x = 4 and another point on the curve. Vertical dashed lines are drawn from the x-axis at values 4, 36, and 40 to the corresponding points on the curve and lines. The graph illustrates that the linear approximation at x = 4 is far from the actual value of the function at x = 40, while the linear approximation at x = 36 is much closer.

L.A. at \(x=4\):

\[L(x) = 2 + \frac{1}{4}(x-4)\]

If we approximate \(\sqrt{40}\) using this:

\[L(40) = 2 + \frac{1}{4}(40-4) = 11\]

Note: NOT a good approximation.

Instead we Linear Approx. at \(x=36\) to approximate \(\sqrt{40}\).

Page 5

\( f(x) = \sqrt{x} \), use Linear Approx. at \( x=36 \) to approximate \( \sqrt{40} \)

\[ L(x) = f(36) + f'(36)(x-36) \]

\[ f(36) = \sqrt{36} = 6 \]

\[ f'(x) = \frac{1}{2\sqrt{x}} \implies f'(36) = \frac{1}{2\sqrt{36}} = \frac{1}{12} \]

\[ L(x) = 6 + \frac{1}{12}(x-36) \]

\[ \sqrt{40} \approx L(40) = 6 + \frac{1}{12}(40-36) = 6.333 \]

Page 6

Example: Use linear approximation to estimate the value of \(\ln(1.2)\)

Note: We know \(\ln 1 = 0\).

Use linear approximation at \(x=1\) to approximate \(\ln(1.2)\).

\[ L(x) = f(1) + f'(1)(x-1) \]

\[ f(1) = \ln 1 = 0 \]

\[ f'(x) = \frac{1}{x} \implies f'(1) = \frac{1}{1} = 1 \]

\[ L(x) = x - 1 \]

\[ \ln(1.2) \approx L(1.2) = 1.2 - 1 = 0.2 \]

Note: Using a calculator, \(\ln(1.2) = 0.1823\).

\[ \text{ERROR} = 0.0177 \]

\[ \% \text{ ERROR} = \frac{\text{ERROR}}{\text{Actual}} \times 100 = \left( \frac{0.0177}{0.1823} \right) \times 100 \approx 10\% \]

Page 7

Example: Linear Approximation

Eg: \( f(x) = 8 - 3x^2 \)

Use linear approximation at \( x = 2 \) to estimate \( f(1.9) \)

\[ L(x) = f(2) + f'(2)(x-2) \]

\[ f(2) = 8 - 3(4) = -4 \]

\[ f'(x) = -6x \implies f'(2) = -12 \]

\[ L(x) = -4 - 12(x-2) \]

\[ f(1.9) \approx L(1.9) = -4 - 12(1.9 - 2) = -4 + 1.2 = -2.8 \]

\[ \text{Actual} = f(1.9) = 8 - 3(1.9)^2 = -2.83 \]

\[ \text{ERROR} = 0.03 \]

\[ \% \text{ ERROR} = \frac{\text{ERROR}}{|\text{Actual}|} \times 100 = \frac{0.03}{2.83} \times 100 \approx 1\% \]

Page 8

Example: Find linear approximation of \( f(x) = \sin x \) at \( x = 0 \).

\[ L(x) = f(0) + f'(0)(x - 0) \]

\[ f(0) = \sin 0 = 0 \]

\[ f'(x) = \cos x \implies f'(0) = \cos 0 = 1 \]

\[ L(x) = 0 + 1(x - 0) = x \]

When \( x \) is close to \( 0 \), \( f(x) = \sin x \approx L(x) = x \).

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

A Cartesian coordinate system showing a plot of f(x) = sin(x) as a blue curve oscillating through the origin. A straight line representing the linear approximation L(x) = x is drawn in purple, tangent to the sine curve at the origin (0,0), visually demonstrating that for small values of x, the curve and the line are nearly identical.

Visual Description: A Cartesian coordinate system showing a plot of f(x) = sin(x) as a blue curve oscillating through the origin. A straight line representing the linear approximation L(x) = x is drawn in purple, tangent to the sine curve at the origin (0,0), visually demonstrating that for small values of x, the curve and the line are nearly identical.

Page 9

Linear Approximation of \(f(x) = \sin x\) at \(x = 0\) is \(L(x) = x\)

Let use this to approximate \(\sin(\pi/3)\), \(\sin(\pi/6)\), \(\sin(\pi/2)\)

at \(x = \pi/6\):

\[L(\pi/6) = \pi/6 \approx 0.52\]

\[\sin(\pi/6) = 0.5\]

Note: The grouping indicates an error calculation: \(\text{ERROR} = 0.02\)

\[\% \text{ ERROR} = \frac{0.02}{0.5} \times 100 = 4\%\]

at \(x = \pi/3\):

\[L(\pi/3) = \pi/3 \approx 1.047\]

\[\sin(\pi/3) = \frac{\sqrt{3}}{2} \approx 0.866\]

Note: The grouping indicates a percentage error calculation:

\[\% \text{ ERROR} = 100 \times \frac{|1.047 - 0.866|}{0.866} \approx 20\%\]

at \(x = \pi/2\):

\[L(\pi/2) = \pi/2 = 1.57\]

\[\sin(\pi/2) = 1\]

Note: The grouping indicates a percentage error calculation:

\[\% \text{ ERROR} = 100 \times \frac{|1.57 - 1|}{1} \approx 57\%\]