This lesson covers linear approximation of functions using tangent lines to approximate function values, as well as error and percentage error calculations.
Find the equation of the tangent line to \(f(x) = \sqrt{x}\) at the point \((4, 2)\).
Graph Description: The coordinate plane shows the function \(f(x) = \sqrt{x}\) graphed as a blue curve starting at the origin and increasing with a decreasing slope. At the point \((4, 2)\), marked with a dot, a red tangent line touches the curve. The tangent line has a positive slope and extends beyond the point of tangency in both directions. Dashed lines extend from the point \((4, 2)\) to both axes, marking the coordinates clearly.
Solution:
To find the equation of the tangent line, we need the point \((4, 2)\) and the slope at that point.
First, find the derivative: \(f'(x) = \frac{1}{2\sqrt{x}}\)
Evaluate at \(x = 4\): \(f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}\)
Using point-slope form: \(y - 2 = \frac{1}{4}(x - 4)\)
Simplifying: \(y = \frac{1}{4}x + 1\)
Definition: Linear Approximation
The linear approximation (or tangent line approximation) of a function \(f(x)\) at a point \(x = a\) is given by:
\[L(x) = f(a) + f'(a)(x - a)\]This equation represents the tangent line to \(f(x)\) at the point \((a, f(a))\), and can be used to approximate values of \(f(x)\) for \(x\) near \(a\).
Key Concept: Linear approximation works best when \(x\) is close to \(a\). The farther \(x\) is from \(a\), the less accurate the approximation becomes.
Find the linear approximation of \(f(x) = \sin(x)\) at \(x = 0\).
Solution:
We need \(f(0)\) and \(f'(0)\).
Since \(f(x) = \sin(x)\), we have \(f(0) = \sin(0) = 0\)
The derivative is \(f'(x) = \cos(x)\), so \(f'(0) = \cos(0) = 1\)
The linear approximation is:
\[L(x) = f(0) + f'(0)(x - 0) = 0 + 1 \cdot x = x\]Therefore, \(\sin(x) \approx x\) for \(x\) near 0.
Use the linear approximation of \(f(x) = \sin(x)\) at \(x = 0\) to approximate the values of \(\sin(\pi/6)\), \(\sin(\pi/3)\), and \(\sin(\pi/12)\).
Solution:
From Example 1, we have \(L(x) = x\), so \(\sin(x) \approx x\) for \(x\) near 0.
For \(\sin(\pi/6)\): \(L(\pi/6) = \pi/6 \approx 0.524\)
(Actual value: \(\sin(\pi/6) = 0.5\))
For \(\sin(\pi/3)\): \(L(\pi/3) = \pi/3 \approx 1.047\)
(Actual value: \(\sin(\pi/3) \approx 0.866\))
For \(\sin(\pi/12)\): \(L(\pi/12) = \pi/12 \approx 0.262\)
(Actual value: \(\sin(\pi/12) \approx 0.259\))
Observation: The approximation is most accurate for \(\sin(\pi/12)\) because \(\pi/12\) is closest to 0. The approximation becomes less accurate as we move farther from the point of tangency.
Use linear approximation to estimate the value of \(\sqrt{40}\).
Solution:
Let \(f(x) = \sqrt{x}\). We need to choose a point \(a\) near 40 where we can easily calculate \(f(a)\).
Choose \(a = 36\) (since \(\sqrt{36} = 6\) is easy to compute).
Find the derivative: \(f'(x) = \frac{1}{2\sqrt{x}}\)
Evaluate at \(a = 36\): \(f'(36) = \frac{1}{2\sqrt{36}} = \frac{1}{12}\)
The linear approximation is:
\[L(x) = f(36) + f'(36)(x - 36) = 6 + \frac{1}{12}(x - 36)\]To estimate \(\sqrt{40}\), substitute \(x = 40\):
\[L(40) = 6 + \frac{1}{12}(40 - 36) = 6 + \frac{4}{12} = 6 + \frac{1}{3} \approx 6.333\]Therefore, \(\sqrt{40} \approx 6.333\)
(Actual value: \(\sqrt{40} \approx 6.325\))
Use linear approximation to estimate the value of \(\ln(1.2)\).
Solution:
Let \(f(x) = \ln(x)\). Choose \(a = 1\) since \(\ln(1) = 0\) is easy to compute.
Find the derivative: \(f'(x) = \frac{1}{x}\)
Evaluate at \(a = 1\): \(f'(1) = \frac{1}{1} = 1\)
The linear approximation is:
\[L(x) = f(1) + f'(1)(x - 1) = 0 + 1(x - 1) = x - 1\]To estimate \(\ln(1.2)\), substitute \(x = 1.2\):
\[L(1.2) = 1.2 - 1 = 0.2\]Therefore, \(\ln(1.2) \approx 0.2\)
(Actual value: \(\ln(1.2) \approx 0.182\))
Given \(f(x) = 8 - 3x^2\), use linear approximation at \(x = 2\) to estimate \(f(1.9)\).
Solution:
First, evaluate \(f(2)\): \(f(2) = 8 - 3(2)^2 = 8 - 12 = -4\)
Find the derivative: \(f'(x) = -6x\)
Evaluate at \(x = 2\): \(f'(2) = -6(2) = -12\)
The linear approximation is:
\[L(x) = f(2) + f'(2)(x - 2) = -4 + (-12)(x - 2) = -4 - 12(x - 2)\]To estimate \(f(1.9)\), substitute \(x = 1.9\):
\[L(1.9) = -4 - 12(1.9 - 2) = -4 - 12(-0.1) = -4 + 1.2 = -2.8\]Therefore, \(f(1.9) \approx -2.8\)
(Actual value: \(f(1.9) = 8 - 3(1.9)^2 = 8 - 10.83 = -2.83\))
Estimate the value of \(\cos(11\pi/8)\) using linear approximation of \(f(x) = \cos(x)\) at both \(x = \pi/4\) and \(x = \pi/2\). Which approximation is better?
Solution:
We have \(f(\pi/4) = \cos(\pi/4) = \frac{\sqrt{2}}{2}\)
The derivative is \(f'(x) = -\sin(x)\), so \(f'(\pi/4) = -\sin(\pi/4) = -\frac{\sqrt{2}}{2}\)
Linear approximation:
\[L_1(x) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}(x - \pi/4)\]Substituting \(x = 11\pi/8\):
\[L_1(11\pi/8) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}(11\pi/8 - \pi/4)\] \[= \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}(11\pi/8 - 2\pi/8)\] \[= \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{9\pi}{8}\]We have \(f(\pi/2) = \cos(\pi/2) = 0\)
\(f'(\pi/2) = -\sin(\pi/2) = -1\)
Linear approximation:
\[L_2(x) = 0 + (-1)(x - \pi/2) = -(x - \pi/2)\]Substituting \(x = 11\pi/8\):
\[L_2(11\pi/8) = -(11\pi/8 - \pi/2) = -(11\pi/8 - 4\pi/8) = -\frac{7\pi}{8}\]Comparison: Note that \(11\pi/8 \approx 4.32\), \(\pi/4 \approx 0.785\), and \(\pi/2 \approx 1.571\). Since \(11\pi/8\) is much closer to \(\pi/2\) than to \(\pi/4\), the approximation at \(x = \pi/2\) should be more accurate. However, in practical terms, \(11\pi/8\) is still relatively far from both points, so neither approximation will be highly accurate.
Accuracy of Approximations: The accuracy of a linear approximation depends on:
Definition: Absolute Error
The absolute error in a linear approximation is:
\[\text{Error} = |f(x) - L(x)|\]where \(f(x)\) is the actual value and \(L(x)\) is the approximated value.
Definition: Percentage Error
The percentage error is:
\[\text{Percentage Error} = \frac{|f(x) - L(x)|}{|f(x)|} \times 100\%\]This expresses the error as a percentage of the actual value.
Common Mistake: Don't confuse the error in the approximation with the value \(x - a\). The error \(|f(x) - L(x)|\) measures how far the approximation is from the true value, while \(x - a\) measures the distance from the point of tangency.
Key Takeaways: