Question

Find equations for the lines that are tangent and normal to the curve y=\sqrt{2} \cos x\( at the point \)(\pi / 4,1)

Short answer

The equation of the tangent line is \(y = -x + 1 + \pi/4\), and the equation of the normal line is \(y = x + 1 - \pi/4\).

Step-by-step solution

Differentiation

Differentiate the given function \(y = \sqrt{2} cos(x)\) with respect to x using the chain rule. The derivative of \(cos(x)\) is \(-sin(x)\), giving us \(dy/dx = -\sqrt{2} sin(x)\).

Find the slope of the tangent line

Substitute \(x = \pi/4\) into \(dy/dx\) to get the slope (\(m_t\)) of the tangent line: \(m_t = -\sqrt{2} sin(\pi /4) = -\sqrt{2}/\sqrt{2} = -1\).

Find the slope of the normal line

The slope of the normal line (\(m_n\)) is the negative reciprocal of the slope of the tangent line. As \(m_t = -1\), the slope of the normal line is \(m_n = -1/(-1) = 1\).

Use point-slope form to find the tangent line equation

Using the point-slope form of a line \(y - y_1 = m (x - x_1)\), where \((x_1, y_1) = (\pi / 4, 1)\) is the point of tangency, and \(m_t = -1\), the equation of the tangent line is \(y - 1 = -1 (x - \pi/4)\). Simplifying gives \(y = -x + 1 + \pi/4\).

Use point-slope form to find the normal line equation

Again using the point-slope form, but now with \(m_n = 1\), the equation of the normal line is \(y - 1 = 1 (x - \pi/4)\). Simplifying gives \(y = x + 1 - \pi/4\).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. It's essentially calculating the slope of the tangent line to a function's curve at a specific point. To differentiate a function, various rules and techniques such as the power rule, product rule, quotient rule, and the chain rule can be applied. In the context of the exercise, differentiation was applied to the trigonometric function \(y = \sqrt{2} \cos(x)\) using the chain rule. This resulted in \(\frac{dy}{dx} = -\sqrt{2} \sin(x)\), which tells us how the function's rate of change varies with respect to \(x\). This step is vital as it leads to understanding the slope of the function at any given point, which is the next integral concept.

Slope of a Line

The slope of a line is a measure of its steepness and direction. In calculus, the slope at a point on a curve is found by taking the derivative of the function at that point. For a linear function \(y=mx+b\), \(m\) represents the slope. A positive slope means the line is ascending, while a negative slope indicates a descending line. For non-linear functions, we can speak of the slope at a specific point by examining the value of the first derivative at that point. In our example, the slope of the tangent line \(m_t\) was calculated as \(m_t = -\sqrt{2} \sin(\pi /4) = -1\), giving a negative slope, meaning the line slopes downwards.

Point-Slope Form

The point-slope form is an equation used to express the line's slope passing through a specific point. This form is particularly useful when the slope, \(m\), and a single point on the line \( (x_1, y_1) \) are known. The generalized equation is \(y - y_1 = m(x - x_1)\). This format allows for a swift determination of the equation of a line, as seen in the textbook example. The tangent and normal lines' equations were found using the point-slope form with their respective slopes and the given point of tangency, leading to the final equations.

Chain Rule

The chain rule is a technique in calculus for differentiating compositions of functions. It is used when dealing with functions like \(f(g(x))\), stating that the derivative of \(f(g(x))\) with respect to \(x\) is the product of the derivative of \(f\) with respect to \(g\) and the derivative of \(g\) with respect to \(x\). Symbolically, it's written as \(\frac{df}{dx} = \frac{df}{dg} \frac{dg}{dx}\). In the provided exercise, the chain rule was used to differentiate \(y = \sqrt{2} \cos(x)\). Since \(\cos(x)\) is nested within a square root, the derivative of the \(\cos\) function was calculated first, and then it was multiplied by the derivative of the square root.

Differentiation of Trigonometric Functions

Trigonometric functions are periodic functions that have their own specific derivatives. Differentiation of trigonometric functions is a key part of calculus, especially when analyzing waveforms or oscillatory movement. The main trigonometric functions—sine, cosine, and tangent—have derivatives that are cyclically related. For example, the derivatives are as follows: \(\frac{d}{dx}\sin(x) = \cos(x)\), \(\frac{d}{dx}\cos(x) = -\sin(x)\), and \(\frac{d}{dx}\tan(x) = \sec^2(x)\). These derivatives are used to find the slope of the tangent line to the curve of a trigonometric function at a certain point, like the \(\cos(x)\) function used in our original exercise.