Problem 45 - Calculus Early Transcendentals

Question

Find an equation for the tangent line to the graph at the specified value of \(x .\) $$ y=\sec ^{3}\left(\frac{\pi}{2}-x\right), x=-\frac{\pi}{2} $$

Step-by-Step Solution

Answer

The tangent line is \(y = -1\).

Step 1: Understand the Function

The function given is \(y = \sec^3\left(\frac{\pi}{2}-x\right)\). The secant function, \(\sec(\theta)\), is the reciprocal of the cosine function: \(\sec(\theta) = \frac{1}{\cos(\theta)}\). The function is being evaluated at \(x = -\frac{\pi}{2}\).

Step 2: Simplify Function at Given x

To find the value of \(y\) at \(x = -\frac{\pi}{2}\), substitute \(-\frac{\pi}{2}\) into the function: \[y = \sec^3\left(\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)\right) = \sec^3(\pi)\]. \(\sec(\pi) = -1\), thus \(y = (-1)^3 = -1\).

Step 3: Differentiate the Function

Find the derivative \(\frac{dy}{dx}\) by using the chain rule. Let \(u = \frac{\pi}{2} - x\), so \(y = (\sec(u))^3\). The derivative of \(\sec(u)\) is \(\sec(u)\tan(u)\), hence using the chain and product rules:\[\frac{dy}{dx} = 3\sec^2(u) \cdot \sec(u) \cdot \tan(u) \cdot \frac{du}{dx}\]\(\frac{du}{dx} = -1\), so:\[\frac{dy}{dx} = -3\sec^3(u)\tan(u)\]

Step 4: Evaluate the Derivative at x

Substitute \(u = \pi\) into \(\frac{dy}{dx}\):\[\frac{dy}{dx} = -3\sec^3(\pi) \tan(\pi) = -3(-1)^3 \cdot 0 = 0\]Thus, the slope of the tangent line at \(x = -\frac{\pi}{2}\) is 0.

Step 5: Write the Tangent Line Equation

Use the point-slope form of a line: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = \left(-\frac{\pi}{2}, -1\right)\) and \(m = 0\). Substitute the values: \[y - (-1) = 0(x + \frac{\pi}{2})\]Simplify to get \(y = -1\).

Key Concepts

Tangent Line

The tangent line to a curve at a specific point is a straight line that just "touches" the curve at that point. It represents the instantaneous rate of change of the function, similar to how speed tells us how fast something is moving at a particular moment.
For this exercise, our task was to find the tangent line to the function \(y = \sec^3\left(\frac{\pi}{2} - x\right)\) at the point \(x = -\frac{\pi}{2}\).

• Start by identifying the specific point on the curve. Here, it was \((-\frac{\pi}{2}, -1)\).
• The slope of this tangent line is given by the derivative of the function at \(x = -\frac{\pi}{2}\).
• Knowing that the slope is 0 at that point, we concluded the tangent line is horizontal: \(y = -1\).

Chain Rule

The chain rule is a core principle in calculus used to differentiate composite functions. It states that to differentiate a function \( y(x) \) made up of another function \( u(x) \), you need to multiply the derivative of the outer function by the derivative of the inner function. In the problem, the function \( y = \sec^3(u) \) with \( u = \frac{\pi}{2} - x \) is a perfect example of a composite function.

• First, determine the derivative of the outer function: \( 3\sec^2(u) \cdot \sec(u) \cdot \tan(u) \).
• Next, compute the derivative of the inner function: \( \frac{du}{dx} = -1 \).
• Finally, multiply the outcomes: \( \frac{dy}{dx} = -3\sec^3(u)\tan(u) \).

Using the chain rule makes it efficient and straightforward to handle derivatives of complex functions encountered frequently in calculus.

Derivative

A derivative measures how a function's value changes as its input changes—essentially, it's a rate of change. In a geometric sense, the derivative at a point is the slope of the tangent line to the function's graph at that point.
Using derivatives, we were able to determine the slope of the tangent line for the provided function at the specified point \(x = -\frac{\pi}{2}\).

• First, simplify and identify the point of interest.
• Compute the derivative, which in this scenario included differentiating \(y = \sec^3(u)\).
• Evaluate the derivative at the specific x-value to find the slope of the tangent.

Here, the derivative was found to be 0, indicating a horizontal tangent line. This concept is crucial, as it allows the analysis of how changes in one quantity affect another in intricate mathematical relationships.