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Chapter 2: Problem 62

Find the equation of the tangent line to the graph of \(y=1+x \sin 3 x\) at \(\left(\frac{\pi}{3}, 1\right)\). Where does this line cross the \(x\) -axis?

Short Answer

Expert verified
The tangent line is \(y = -\pi x + \frac{\pi^2}{3} + 1\) and it crosses the \(x\)-axis at \(x = \frac{\pi}{3} + \frac{1}{\pi}\).

Step by step solution

01

Find the Derivative

To find the equation of the tangent line, we start by finding the derivative of the function \(y=1+x \sin 3x\). Using the product rule for differentiation (since the term involves \(x \sin 3x\)), the derivative of \(y\) is: \(\frac{dy}{dx} = 1 \cdot \sin 3x + x \cdot 3 \cos 3x = \sin 3x + 3x \cos 3x\).
02

Evaluate the Derivative at the Point

Now, evaluate the derivative at the given point \(x = \frac{\pi}{3}\). Substitute this value into the derivative to find the slope of the tangent:\(\frac{dy}{dx} \bigg|_{x = \frac{\pi}{3}} = \sin 3\frac{\pi}{3} + 3\frac{\pi}{3} \cos 3\frac{\pi}{3} = \sin \pi + \pi \cos \pi\). Since \(\sin \pi = 0\) and \(\cos \pi = -1\), the slope becomes \(0 - \pi = -\pi\).
03

Write the Equation of the Tangent Line

The equation of the tangent line can be found using the point-slope form \(y - y_1 = m (x - x_1)\). Here, \((x_1, y_1) = (\frac{\pi}{3}, 1)\) and the slope \(m = -\pi\):\[y - 1 = -\pi \left(x - \frac{\pi}{3}\right)\].Simplifying gives:\[y = -\pi x + \frac{\pi^2}{3} + 1\].
04

Find the x-intercept of the Tangent Line

The line crosses the \(x\) axis where \(y = 0\). Substitute \(y = 0\) in the tangent equation and solve for \(x\):\[0 = -\pi x + \frac{\pi^2}{3} + 1\].Rearrange to solve for \(x\):\[-\pi x = -\frac{\pi^2}{3} - 1 \quad \Rightarrow \quad x = \frac{\pi^2}{3\pi} + \frac{1}{\pi} = \frac{\pi}{3} + \frac{1}{\pi}\].
05

Finalize the Solution

Thus, the \(x\)-intercept of the tangent line is \(x = \frac{\pi}{3} + \frac{1}{\pi}\). The equation of the tangent line is \(y = -\pi x + \frac{\pi^2}{3} + 1\), and it crosses the \(x\)-axis at \(x = \frac{\pi}{3} + \frac{1}{\pi}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
Calculating the derivative of a function allows us to determine the slope of the tangent line at any given point. In essence, the derivative tells us how a function is changing at each point in its domain. For a function like \(y = 1 + x \sin 3x\), we apply differentiation techniques to find its derivative function.
The original function has two parts: a constant \(1\) and a product \(x\sin 3x\), which involves both a variable and a trigonometric function.
  • The derivative of a constant \(1\) is \(0\) since it does not change.
  • For the product \(x \sin 3x\), we utilize the product rule in differentiation. This rule is essential when differentiating two functions multiplied together.
Combining these, we use the derivative information to find the slope of our tangent line.
Product Rule
The product rule is a handy formula in calculus for finding the derivative of functions that are multiplied together. It鈥檚 especially useful with functions like our example: \(x \sin 3x\).
The product rule states that if you have two functions \(u(x)\) and \(v(x)\), their derivative, \((u \, v)'\), is given by:\[(uv)' = u'v + uv'\] Here, \(u(x) = x\) and \(v(x) = \sin 3x\). Therefore:
  • \(u'(x) = 1\) because the derivative of \(x\) is \(1\).
  • \(v'(x) = 3 \cos 3x\), which can be found by applying the chain rule to \(\sin 3x\).
  • Using the product rule: \(x' \sin 3x + x \cdot 3 \cos 3x = \sin 3x + 3x \cos 3x\).
This gives us the expression for the derivative: \(\sin 3x + 3x \cos 3x\), which we use to determine the slope of our tangent line.
Point-Slope Form
Once we've determined the slope from the derivative at a specific point, we can use the point-slope formula to write the equation of the tangent line. The point-slope form is very practical when you know the slope of a line and a point through which it passes.
In our example, the point-slope form is:\[y - y_1 = m(x - x_1)\] Here:
  • The slope \(m\) is \(-\pi\), calculated previously.
  • The point \((x_1, y_1) = \left(\frac{\pi}{3}, 1\right)\) is where the tangent touches the curve.
By inserting these into the point-slope formula, we construct the equation of the tangent line. Simplifying the expression yields \(y = -\pi x + \frac{\pi^2}{3} + 1\). This formula describes the tangent line, which visually represents the instantaneous rate of change (slope) at the specific point on the curve.

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