Problem 38 - Calculus of a Single Variable

Question

In Exercises 31–38, find the slope of the graph of the function at the given point. Use the derivative feature of a graphing utility to confirm your results. $$ g(t)=-2 \cos t+5 \quad(\pi, 7) $$

Step-by-Step Solution

Answer

The slope of the graph of the function at the given point (\(\pi\), 7) is 0.

Step 1: Find the derivative

The derivative of the function \(g(t)\) will be calculated first. It's evident that \(g(t)=-2 \cos t+5.\) The derivative of a cosine function is -sin, so after differentiating, the derivative will be \(g'(t) = 2\sin t\).

Step 2: Evaluate the derivative at the given point

Now the derivative of the function is known, it needs to be evaluated at the specified x-coordinate to obtain the slope at the point. For \(t=\pi\), the value of \(\sin \pi\) is 0. Therefore, \(g'(\pi) = 2\sin\pi = 0.\)

Step 3: Verify results using a graphing utility

To confirm the results, plot the original function and the derivative function in a graphing utility. The slope of the function at \(t=\pi\) should equal to the y-value of the derivative function at the same t-coordinate

Key Concepts

Derivative of Cosine

Understanding the derivative of the cosine function is pivotal in calculus, especially when dealing with the rate of change in periodic functions. The cosine function, denoted as \(\cos(x)\), has a well-established derivative, which is \( -\sin(x)\). In practice, if we have a function \(f(x) = \cos(x)\), then after differentiating, we get \(f'(x) = -\sin(x)\).

When a constant is multiplied with the cosine function, for instance, \(g(t) = -2\cos(t)\), the derivative reflects this constant as well. Each part of the function is differentiated separately, leading to \(g'(t) = -2\sin(t)\). This result shows the rate at which the function's value changes concerning \(t\), which directly informs us about the slope of the tangent line at any given point.

Evaluating Derivatives

Evaluating derivatives at a specific point is essentially the process of finding the slope of the tangent line at that point. After finding the derivative function, you plug in the x-coordinate of the point in question into this derivative function. This act of substitution gives us the slope of the tangent line to the original function at that particular point.

For the function \(g(t) = -2\cos(t) + 5\) at the point \((\pi, 7)\), evaluating the derivative means calculating \(g'(\pi)\). Since we have the derivative \(g'(t) = 2\sin(t)\), substituting \((t = \pi)\) gives us \(g'(\pi) = 2\sin(\pi)\), which equals zero. Thus, at \(t = \pi\), the slope of the graph of the function \(g(t)\) is zero, implying that the tangent line is horizontal at that point.

Graphing Utility Confirmation

With the advent of technology, confirming calculus results is easier than ever. A graphing utility, whether software or an online tool, can visually represent functions and their derivatives. This graphical representation augments understanding and provides confirmation of analytical results.

After evaluating the derivative and finding the slope at a specific point algebraically, students can use a graphing utility to plot the original function and its derivative. The visual confirmation should align with the analytical findings. If you plot the function \(g(t) = -2\cos(t) + 5\) and its derivative, and then inspect the graph at \(t = \pi\), the slope of the tangent line on the graph will match the evaluated derivative, which in this case is zero.

Slope of Tangent Line

The slope of the tangent line to a curve at any given point is a fundamental concept in calculus. It represents the instantaneous rate of change of the function at that specific point. The slope can be found by taking the derivative of the function and then evaluating it at the desired point.

For instance, the slope of the tangent line to the function \(g(t)\) at the point \(\pi\) tells us how steeply the curve rises or falls at that exact location. When the slope is zero, as it is in our example of \(g'(\pi) = 0\), it indicates that there is no immediate change in the function's value — in other words, the function is flat, or horizontally tangent, at that point. Conversely, a nonzero slope would imply that the function is either increasing or decreasing at that location.