Problem 32 - Calculus with Concepts in Calculus

Question

Find all inflection points (if any) of the graph of the function. Then sketch the graph of the function. $$ f(t)=t-\cos t $$

Step-by-Step Solution

Answer

Inflection points occur at \( t = \frac{\pi}{2} + k\pi \), \( k \in \mathbb{Z} \).

Step 1: Find the First Derivative

To find the inflection points, we first need to determine the first derivative of the function. The given function is \( f(t) = t - \cos t \). Differentiating with respect to \( t \), we get: \[ f'(t) = \frac{d}{dt}(t) - \frac{d}{dt}(\cos t) = 1 + \sin t \]

Step 2: Find the Second Derivative

The next step is to find the second derivative of the function \( f(t) \). From the first derivative \( f'(t) = 1 + \sin t \), differentiate again with respect to \( t \):\[ f''(t) = \frac{d}{dt}(1 + \sin t) = \cos t \]

Step 3: Set the Second Derivative to Zero to Find Critical Points

Inflection points occur where the second derivative changes sign. Set the second derivative equal to zero to find potential inflection points:\[ \cos t = 0 \] This occurs when:\[ t = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z} \]

Step 4: Determine Sign Change to Confirm Inflection Points

To confirm that these points are indeed inflection points, check the sign change of \( f''(t) \) around \( t = \frac{\pi}{2} + k\pi \). Since \( \cos t \) changes sign at these points (from positive to negative or vice versa), they are indeed inflection points.

Step 5: Sketch the Graph

Using the function \( f(t) = t - \cos t \) and the inflection points \( t = \frac{\pi}{2} + k\pi \):- Recognize that the graph is a combination of a linear function and the cosine wave.- Inflection points are where the curvature changes, identifiable by the periodic nature where \( \cos t = 0 \).- Sketch a line with periodic 'wiggles' following the shape of the cosine wave, noting inflection at these critical points.

Key Concepts

First Derivative

The first derivative of a function helps us understand how the function is changing at every point. It tells us about the slope of the tangent line to the graph of the function at any given point. When we're looking at the function \( f(t) = t - \cos t \), we derive it with respect to \( t \) to find the first derivative.With \( f(t) = t - \cos t \), the first derivative is given by:

• \( t \) becomes 1, since the derivative of \( t \) with respect to \( t \) is 1.
• \( \cos t \) changes to \( +\sin t \), because the derivative of \( -\cos t \) is \( \sin t \).

Therefore, the first derivative is \( f'(t) = 1 + \sin t \).This function, \( 1 + \sin t \), shows us where the slope is increasing (positive) or decreasing (negative), which will be essential when we look for the inflection points later.

Second Derivative

The second derivative of a function gives insight into the curvature or concavity of the function. Essentially, it tells us how the slope itself is changing. For the given function, we need to take a second derivative to find potential inflection points.To get it, we differentiate the first derivative \( f'(t) = 1 + \sin t \):

• The derivative of 1 is 0, as constants disappear when derived.
• Differentiate \( \sin t \) to get \( \cos t \). This involves recognizing the patterns of trigonometric derivatives, where sine turns to cosine.

Thus, the second derivative is \( f''(t) = \cos t \).Inflection points occur where this second derivative, \( \cos t \), changes sign. When \( \cos t = 0 \), the function might have an inflection point. We solved \( \cos t = 0 \) to find \( t = \frac{\pi}{2} + k\pi \), with \( k \) being any integer. These are our candidate points for inflection. Once we find where \( \cos t \) switches sign, we confirm them as true inflection points.

Trigonometric Functions

Trigonometric functions like sine and cosine are foundational in the study of calculus, especially when dealing with derivatives and integrals. They have repeating, predictable shapes which makes analyzing points of interest, such as inflection points, substantially more manageable.For \( f(t) = t - \cos t \), we're dealing with the cosine function shifted vertically and diagonally by the linear portion \( t \).

• Cosine starts at its maximum (1) when \( t = 0 \) and reaches zero at \( \pi/2 + k\pi \), making those points interesting for inflection discussion.
• Given \( \cos t \) in \( f''(t) = \cos t \), the periodic nature allows us to predict these zero-crossings periodically at every \( \frac{\pi}{2} + k\pi \).

Understanding these periodic properties can help greatly when sketching graphs, as we see with \( f(t) = t - \cos t \). Knowing these zero points are crucial: it tells where the behavior of the function bending upwards or downwards changes. This is expressed in the function as the inflection points identified earlier, reinforcing the utility of understanding trigonometric derivatives in calculus.