The following graph illustrates a function \( y = f(x) \) and its tangent line at a specific point \( (a, f(a)) \). A zoomed-in view shows how the tangent line and the curve become nearly indistinguishable as we get closer to the point of tangency.
Very near \( (a, f(a)) \) the tangent line and the real curve \( f(x) \) are very close to each other.
If \( x \) is very close to \( a \), then we can replace the function with its tangent line without losing much accuracy.
The tangent line approximation gets better the closer we are to \( x = a \).
Let's write out the equation of the tangent line at \( x = a \), \( y = f(a) \):
Then:
\[ y = f(a) + f'(a)(x - a) \approx f(x) \text{ if } x \text{ is near } a \]
This is the linear approximation of \( f(x) \).
linear approximation: \[ L(x) = f(a) + f'(a)(x-a) \approx f(x) \]
this is called the differential
it gives an estimate of the change in \( y = f(x) \) as \( x \) changes from \( a \).
\( \sin(x) \) is hard to evaluate without a calculator here, we will replace it with a line which is much easier to work with.
make a tangent line at \( x = 0 \rightarrow a = 0 \)
\[ L(x) = f(a) + f'(a)(x-a) \]
\[ f(a) = f(0) = \sin(0) = 0 \]
\[ f'(x) = \cos(x) \]
\[ f'(a) = f'(0) = \cos(0) = 1 \]
\[ L(x) = 0 + (1)(x-0) \]
this means that when \( x \) is close to \( a = 0 \), \( x \) and \( \sin(x) \) are very close to each other
for example, \( \sin(0.001) = 0.000999983 \)
we see \( \sin(x) \) and \( x \) are very close
this is another way to explain why \[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
change to radians!
\[ 30^{\circ} = \frac{\pi}{6} \]
\[ \sin(x) \approx x \quad \text{so, we can estimate} \quad \sin(30^{\circ}) = \sin\left(\frac{\pi}{6}\right) \approx \frac{\pi}{6} = 0.52 \]
true value: \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} = 0.5 \)
pretty close, let's calculate the error and percentage error
So, there is a \( 4 \% \) error using linear approximation.
the estimate is good, since we didn't move too far from \( x = 0 \)
try \( \sin(90^{\circ}) = \sin(\frac{\pi}{2}) = 1 \) (true value)
linear approx: \( \sin(x) \approx x \)
so estimate of \( \sin(\frac{\pi}{2}) \) is \( \frac{\pi}{2} = 1.57 \)
Use a linear approximation to estimate \( \sqrt{108} \)
we can look at this as \( f(x) = \sqrt{x} \) with \( x = 108 \)
we can use a linear approximation with an appropriate "a" to estimate this
choose "a" such that we know \( f(a) \) easily and that "a" is not too far from \( x = 108 \)
find the closest number to 108 whose square root we know
so, here, \( a = 100 \) is a good choice
\( L(x) = f(a) + f'(a)(x - a) \)
\( f(a) = f(100) = \sqrt{100} = 10 \)
\( f'(x) = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \)
\( f'(a) = f'(100) = \frac{1}{2\sqrt{100}} = \frac{1}{20} = 0.05 \)
\( L(x) = 10 + 0.05(x - 100) \approx \sqrt{x} \) when \( x \) is near 100
So,
True value of \( \sqrt{108} \) from calculator is \( 10.3923 \).
Use a linear approximation to estimate \( \ln(3) \).
\( f(x) = \ln(x) \) with \( x = 3 \)
Now find an "\( a \)" such that \( f(a) \) is known or easy to calculate.
\( e \) is a good choice: \( e \approx 2.7 \) close to \( 3 \) and \( \ln(e) = 1 \)
From a calculator, true value of \( \ln(3) \) is \( 1.0986 \).
Once again, pretty close because \( 3 \) is close to \( e \).