Linear Approximations and Differentials

Approximate \( f(\theta) = \sin(\theta) \) at \( \theta = 0 \).

Solution:

1. Use the given value for \( a \) in place of \( x \) to find \( f(a) \). This gives you the ordered pair \( (a, f(a)) \).
   • In the context of this problem, you need to solve for \( f(0) \).\[ f(0) = \sin(0) = 0 \]
     • The ordered pair is \( (0, 0) \).
2. Take the derivative of the given function, \( f(x) \), to find the slope of the tangent line.
   • In the context of this problem, you need to find the derivative of \( f(\theta) \).\[ f'(\theta) = \cos(\theta) \]
3. Solve for the derivative at the point \( x = a \).
   • In the context of this problem, you need to plug in the value for \( \theta \) to solve for the derivative at that point.\[ \begin{align}f'(\theta) &= \cos(\theta) \\f'(0) &= \cos(0) \\f'(0) &= 1\end{align} \]
4. Plug the values for \( f(a) \), \( f'(a) \), and \( a \) into the equation: \( L(x) = f(a) + f'(a) (x - a) \).
   • In the context of this problem, you need to plug the value for \( \theta \) and the values you got in steps \( 1–3 \) into the linear approximation equation.\[ \begin{align}L(x) &= f(a) + f'(a) (x - a) \\L(\theta)&= f(0) + f'(0) (\theta - 0) \\&= 0 + 1 (\theta - 0) \\\mathbf{L(\theta)} &= \mathbf{\theta}\end{align} \]

For instance, consider the differentiable function:

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

where \( x = a = 3 \).

1. What is the tangent line to the curve at \( a = 3 \), and
2. Can it be used to approximate the value of \( f(x) \) when \( a = 3.1 \)?

Solution:

1. What is the tangent line to the curve at \( a = 3 \)?
   1. Find \( f(a) \).\[ f(a) = f(3) = \frac{1}{3} \]
   2. Find the derivative of \( f(x) \).\[ f'(x) = -\frac{1}{x^{2}} \]
   3. Plug in the value for \( a \) to solve for the derivative at that point.\[ \begin{align}f'(a) &= -\frac{1}{a^{2}} \\f'(3) &= -\frac{1}{(3)^{2}} \\&= -\frac{1}{9}\end{align} \]
   4. Plug the value for \( a \) and the values you got in steps \( 1–3 \) into the tangent line equation.\[ \begin{align}y &= f(a) + f'(a) (x - a) \\\mathbf{y} &= \mathbf{\frac{1}{3} - \frac{1}{9} (x - 3)}\end{align} \]
2. Can the tangent line be used to approximate the value of \( f(x) \) when \( a = 3.1 \)?
   1. Use the tangent line you found in part A to estimate the value of the function at \( a = 3.1 \).\[ \begin{align}y &= \frac{1}{3} - \frac{1}{9} (3.1 - 3) \\&\approx 0.3222\end{align} \]
   2. Calculate the value of \( f(3.1) \) using the given function.\[ f(3.1) = \frac{1}{3.1} \approx 0.3226\]
   3. Compare the two values from steps \(1-2\). Use the graphs below to help visualize the comparison.Since the difference between the approximation and the actual values are minimal, \[ y \approx 0.3222 \text{ vs. } f(x) \approx 0.3226 \] you can say that the tangent line can be used to approximate the value of \( f(x) \) when \( a = 3.1 \).

Figure 2. The tangent line to the curve of \( f(x) = \frac{1}{x} \) at \( x = 3 \) gives you a good approximation of \( f(x) \) when \( x \) is near \( 3 \).

Figure 3. When \( x = 3.1 \), the value of \( y \) on the tangent line to the curve of \( f(x) = \frac{1}{x} \) is approximately \( 0.3222 \); the actual value of \( f(3.1) \) is \( \frac{1}{3.1} \approx 0.3226 \).

Suppose you have a function, \( f(x) \), that is differentiable at point \( a \). Say the input, \( x \), changes by some small amount called \( \mathrm{d}x \) (this could also be denoted as \( \Delta x \)).

You are interested in how much the output, \( y \), changes based on this tiny change to \( x \).

• If \( x \) changes from \( a \) to \( a + \mathrm{d}x \), then the total change in \( x \) is \( \mathrm{d}x \), and the change in \( y \) is given by the equation:\[ \Delta y = f(a + \mathrm{d}x) - f(a). \]
• So, if \( \mathrm{d}x \) is small,\[ \begin{align}f(a + \mathrm{d}x) &\approx L(a + \mathrm{d}x) \\&= f(a) + f'(a) (a + \mathrm{d}x - a).\end{align} \]
• Subtracting \( f(a) \) from both sides,\[ \begin{align}f(a + \mathrm{d}x) - f(a) &\approx L(a + \mathrm{d}x) - f(a) \\&= f'(a) \mathrm{d}x.\end{align} \]

In other words,

• the actual change in the function, \( f(x) \), (if \( x \) increases from \( a \) to \( a + \mathrm{d}x \)) is approximately the difference between \( L(a + \mathrm{d}x) \) and \( f(a) \),
  • where \( L(x) \) is the linear approximation of \( f(x) \) at \( x = a \).
• By the definition of \( L(x) \), this difference is equal to \( f'(a) \mathrm{d}x \).

To summarize:

\[ \begin{align}\Delta y &= f(a + \mathrm{d}x) - f(a) \\&\approx L(a + \mathrm{d}x) - f(a) \\&= f'(a) \mathrm{d}x \\&= \mathrm{d}y\end{align} \]

This means you can use the differential, \( \mathrm{d}y = f'(a) \mathrm{d}x \), to approximate the change in \( y \) if \( x \) increases from \( x = a \) to \( x = a + \mathrm{d}x \). The graph below shows this in detail.

Figure 4. You use the differential \( \mathrm{d}y = f'(a) \mathrm{d}x \) to approximate the actual change in \( y \) if \( x \) increases from \( a \) to \( a + \mathrm{d}x \).

Approximating Change with Differentials

Given the function:

\[ y = x^{2} + 2x, \]

calculate

1. \( \Delta y \) and
2. \( \mathrm{d}y \)

at \( x = 3 \) if \( \mathrm{d}x = 0.1 \).

Solution:

1. \( \Delta y \) is the actual change in \( y \) if \( x \) changes from \( 3 \) to \( 3.1 \). To find actual change, use the formula: \( \Delta y = f(a + \mathrm{d}x) - f(a) \).\[ \begin{align}\Delta y &= f(a + \mathrm{d}x) - f(a) \\&= f(3.1) - f(3) \\&= \left( (3.1)^{2} + 2(3.1) \right) - \left( 3^{2} + 2(3) \right) \\&= 15.81 - 15 \\&= 0.81\end{align} \]
2. \( \mathrm{d}y \) is the approximate change in \( y \). To find approximate change, use the formula: \( \mathrm{d}y = f'(x) ~\mathrm{d}x \).
   1. Find the derivative of \( f(x) \).\[ f'(x) = 2x + 2 \]
   2. Use the differential formula to solve for the approximate change.\[ \begin{align}\mathrm{d}y &= f'(x) ~\mathrm{d}x \\&= f'(3) ~\mathrm{d}x \\&= (2(3) + 2)(0.1) \\&= 0.8\end{align} \]

Linear Approximation of a Function

Approximate the function:

\[ f(x) = (1 + x)^{n} \]

at \( x = 0 \). Then use the approximation to estimate \( 1.01^{2} \).

Solution:

1. Use the given value for \( a \) in place of \( x \) to find \( f(a) \). This gives you the ordered pair \( (a, f(a)) \).
   • In the context of this problem, you need to solve for \( f(0) \).\[ f(0) = (1 + 0)^{n} = 1 \]
     • The ordered pair is \( (0, 1) \).
2. Take the derivative of the given function, \( f(x) \), to find the slope of the tangent line.
   • In the context of this problem, you need to find the derivative of \( f(x) \).\[ f'(x) = n(1 + x)^{n-1} \]
3. Solve for the derivative at the point \( x = a \).
   • In the context of this problem, you need to plug in the value for \( x \) to solve for the derivative at that point.\[ \begin{align}f'(x) &= n(1 + x)^{n-1} \\f'(0) &= n(1 + 0)^{n-1} \\f'(0) &= n\end{align} \]
4. Plug the values for \( f(a) \), \( f'(a) \), and \( a \) into the equation: \( L(x) = f(a) + f'(a) (x - a) \).
   • In the context of this problem, you need to plug the value for \( x \) and the values you got in steps \( 1–3 \) into the linear approximation equation.\[ \begin{align}L(x) &= f(a) + f'(a) (x - a) \\&= f(0) + f'(0) (x - 0) \\&= 1 + n (x - 0) \\\mathbf{L(x)} &= \mathbf{1 + nx}\end{align} \]
5. Approximate \( 1.01^{2} \) by evaluating \( L(0.01) \) at \( n = 2 \).\[ \begin{align}(1.01)^{2} &= f(1.01) \\&\approx L(1.01) \\&= 1 + n(x - 0) \\&= 1 + 2(0.01) \\\mathbf{(1.01)^{2}} &= \mathbf{1.02}\end{align} \]

Calculating Differentials

Find \( \mathrm{d}y \) and evaluate the following function at \( x = 3 \) and \( \mathrm{d}x = 0.1 \).

\[ y = \cos(x) \]

Solution:

1. Find the derivative of \( f(x) \).\[ f'(x) = -\sin(x) \]
2. Plug the derivative into the differential formula.\[ \begin{align}\mathrm{d}y &= f'(x) ~\mathrm{d}x \\\mathbf{\mathrm{d}y} &= \mathbf{-\sin(x) ~\mathrm{d}x}\end{align} \]
3. Plug in the given values of \( x \) and \( \mathrm{d}x \) and simplify.\[ \begin{align}\mathrm{d}y &= -\sin(3)(0.1) \\\mathbf{\mathrm{d}y} &= \mathbf{-0.1\sin(3)}\end{align} \]