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Chapter 27: Problem 51

Solve the given problems. Find the slope of a line tangent to the curve of \(y=\frac{2 \sin 3 x}{x},\) where \(x=0.15 .\) Verify the result by using the derivative-evaluating feature of a calculator.

Short Answer

Expert verified
The slope of the tangent is approximately -33.173.

Step by step solution

01

Understanding the Question

We need to find the slope of the line tangent to the curve defined by the function \( y = \frac{2 \sin(3x)}{x} \) at the point where \( x = 0.15 \). The tangent slope is equivalent to the derivative of the function at this point.
02

Differentiating the Function

To find the slope, differentiate the function \( y = \frac{2\sin(3x)}{x} \). We will use the quotient rule, which states that if \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^2} \). Let \( u = 2\sin(3x) \) and \( v = x \).
03

Finding Derivatives of Components

First, find the derivatives of \( u \) and \( v \):- \( u = 2\sin(3x) \rightarrow u' = 6\cos(3x) \) by using the chain rule.- \( v = x \rightarrow v' = 1 \).
04

Applying the Quotient Rule

Substitute the components into the quotient rule:\[\frac{d}{dx}\left(\frac{2\sin(3x)}{x}\right) = \frac{(6\cos(3x))(x) - (2\sin(3x))(1)}{x^2} = \frac{6x\cos(3x) - 2\sin(3x)}{x^2}\]
05

Evaluating the Derivative at x=0.15

Now, evaluate the derivative at \( x = 0.15 \):Substitute \( x = 0.15 \) into the derivative:\[\frac{6(0.15)\cos(0.45) - 2\sin(0.45)}{(0.15)^2}\]Compute the values using a calculator:\( \cos(0.45) \approx 0.9004 \) and \( \sin(0.45) \approx 0.434.\)The numerator becomes approximately \( 0.135 \times 0.9004 - 2 \times 0.434 \approx 0.121554 - 0.868 \approx -0.7464 \).The denominator is \( (0.15)^2 = 0.0225 \).Thus, the slope is approximately \( \frac{-0.7464}{0.0225} \approx -33.173 \).
06

Verifying with a Calculator

Use the derivative-evaluating feature of a calculator to find the numerical derivative of \( y = \frac{2 \sin(3x)}{x} \) at \( x = 0.15 \). Enter this function in your calculator and use the numerical derivative function to confirm the result. You should find it matches approximately \(-33.173\).

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Key Concepts

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

Tangent Line
A tangent line is a straight line that touches a curve at exactly one point. This means it has the same slope as the curve at that point. Bisecting the curve, it gives us a very good approximation of the curve's behavior at that exact spot. When you think about how a curve bends, the tangent line at a specific point gives the immediate direction of the curve there.
  • Immediate Touch: The tangent doesn't cut through the curve, just touches it.
  • Slope Representation: The slope of the tangent line equals the derivative of the function at that precise point.
Finding a tangent line involves determining this slope, which in our initial exercise was calculated using the derivative of the given function.
Derivative
The derivative of a function is a measure of how the function's value changes as its input changes. Essentially, it's a way to compute the slope of a function at any given point.
  • Rate of Change: The derivative provides the rate of change of the function, reflecting how steep the curve is at a particular point.
  • Notation: It's often denoted as \( f'(x) \) for a function \( f(x) \), meaning the derivative of \( f \) with respect to \( x \).
In our exercise, the goal was to find the derivative of the function \( y=\frac{2\sin(3x)}{x} \). This derivative gives the slope of the tangent line at any point \( x \), particularly at \( x=0.15 \) in this case.
Chain Rule
The chain rule is a fundamental theorem used in calculus to differentiate composite functions. When you have a function inside another function, the chain rule helps in finding the derivative.
  • Composite Function: If you have a function \( y=f(g(x)) \), the chain rule expresses the derivative as \( y' = f'(g(x)) \cdot g'(x) \).
  • Practical Use: By breaking down the function \( u = 2\sin(3x) \), we used the chain rule to find \( u' = 6\cos(3x) \).
Using this rule, we can efficiently simplify complicated derivatives that involve nested functions, making it indispensable in many calculus problems like ours.
Quotient Rule
The quotient rule is a method for finding the derivative of a function that is the ratio of two differentiable functions. Given a function \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the derivative is given by the formula:
  • Formula: \[ y' = \frac{u'v - uv'}{v^2} \]
  • Step by Step: First, find the derivatives of both the numerator \( u \) and the denominator \( v \).
  • Our Example: We used it to differentiate \( y=\frac{2\sin(3x)}{x} \). Calculating \( u' \) and \( v' \), the quotient rule allowed us to assemble the derivative and find the slope at \( x = 0.15 \).
Understanding the quotient rule is crucial when dealing with function expressions involving divisions, as it systematically manages the changes in the outputs due to changes in both components, \( u \) and \( v \).

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