Problem 49
Question
(a) Find the equation of the tangent line to \(y=\ln x\) at \(x=1\) (b) Use it to calculate approximate values for \(\ln (1.1)\) and \(\ln (2)\). (c) Using a graph, explain whether the approximate values are smaller or larger than the true values. Would the same result have held if you had used the tangent line to estimate \(\ln (0.9)\) and \(\ln (0.5) ?\) Why?
Step-by-Step Solution
Verified Answer
The tangent line at \(x=1\) is \(y = x - 1\). Approximate values: \(\ln(1.1) \approx 0.1\), \(\ln(2) \approx 1\). Tangent line estimates are under true values beyond \(x=1\) and over at \(x<1\).
1Step 1: Differentiation
Find the derivative of the function \( y = \ln x \). The derivative is \( \frac{dy}{dx} = \frac{1}{x} \). This tells us the slope of the tangent line at any point \( x \).
2Step 2: Find the Slope at x=1
Substitute \( x = 1 \) into the derivative to find the slope of the tangent line at \( x = 1 \). \[ \text{Slope} = \frac{1}{1} = 1 \]
3Step 3: Determine the Point of Tangency
Substitute \( x = 1 \) into the original function \( y = \ln x \) to find the y-coordinate. \[ y = \ln(1) = 0 \] The point of tangency is \( (1, 0) \).
4Step 4: Equation of the Tangent Line
Use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), with \( m = 1 \), \( x_1 = 1 \), and \( y_1 = 0 \) to write the equation of the tangent line. \[ y - 0 = 1(x - 1) \] Simplify to get \( y = x - 1 \).
5Step 5: Approximate \( \ln(1.1) \)
Use the tangent line equation \( y = x - 1 \) to approximate \( \ln(1.1) \). Substitute \( x = 1.1 \) into the equation. \[ y = 1.1 - 1 = 0.1 \] Hence, \( \ln(1.1) \approx 0.1 \).
6Step 6: Approximate \( \ln(2) \)
Use the tangent line equation \( y = x - 1 \) to approximate \( \ln(2) \). Substitute \( x = 2 \) into the equation. \[ y = 2 - 1 = 1 \] Hence, \( \ln(2) \approx 1 \).
7Step 7: Graph Explanation
The graph of \( y = \ln x \) is concave down. This means the tangent line lies below the curve for \( x > 1 \) and above the curve for \( x < 1 \). Thus, estimates using \( x = 1 \) are lower than the true values for \( \ln(1.1) \) and \( \ln(2) \). Conversely, estimates for \( \ln(0.9) \) and especially \( \ln(0.5) \), using the tangent line at \( x = 1 \), would be higher than the actual values because the line is above these segments of the curve.
Key Concepts
Tangent LineNatural LogarithmDifferentiationApproximationCurve Sketching
Tangent Line
A tangent line is a straight line that touches a curve at a single point, without crossing it. This line represents the instantaneous rate of change of the function at that point. It provides a linear approximation to the curve around the point of tangency. In our problem, we are interested in the tangent line to the function \( y = \ln x \) at the point \( x = 1 \).
To find this tangent line, we first need the slope of the line at the tangency point. This requires differentiation of the function.
To find this tangent line, we first need the slope of the line at the tangency point. This requires differentiation of the function.
Natural Logarithm
The natural logarithm, denoted by \( \ln x \), is a logarithmic function with base \( e \), an irrational constant approximately equal to 2.71828. The function is defined for \( x > 0 \) and is one of the most important functions in calculus due to its natural appearance in calculus and exponential growth contexts.
It is particularly easy to analyze because its derivative and integral have simple expressions. For example, the derivative \( \frac{d}{dx} \ln x = \frac{1}{x} \) tells us how rapidly \( \ln x \) is increasing as \( x \) increases.
It is particularly easy to analyze because its derivative and integral have simple expressions. For example, the derivative \( \frac{d}{dx} \ln x = \frac{1}{x} \) tells us how rapidly \( \ln x \) is increasing as \( x \) increases.
Differentiation
Differentiation is a fundamental tool in calculus used to compute the derivative of a function. The derivative provides information about the function's rate of change. In practical terms, it tells us the slope of the tangent line to the graph of the function at any point.
For the function \( y = \ln x \), its derivative is \( \frac{1}{x} \). To find the slope of the tangent line at the point \( x = 1 \), we substitute \( x = 1 \) into the derivative. This yields a slope of 1. Consequently, the tangent line at \( x = 1 \) has a slope of 1 and touches the curve at the point \((1, 0)\).
For the function \( y = \ln x \), its derivative is \( \frac{1}{x} \). To find the slope of the tangent line at the point \( x = 1 \), we substitute \( x = 1 \) into the derivative. This yields a slope of 1. Consequently, the tangent line at \( x = 1 \) has a slope of 1 and touches the curve at the point \((1, 0)\).
Approximation
Approximation using tangents involves using the equation of the tangent line to estimate the value of the function near the point of tangency. This method is particularly useful for functions that are difficult to evaluate exactly.
In our problem, the tangent line equation \( y = x - 1 \) is used to approximate \( \ln(1.1) \) and \( \ln(2) \). These approximations are close to the actual values as the points we chose are near the point \( x = 1 \), where we calculated the tangent. For \( \ln(1.1) \), substituting into the tangent line gives 0.1, while for \( \ln(2) \), it gives 1.
In our problem, the tangent line equation \( y = x - 1 \) is used to approximate \( \ln(1.1) \) and \( \ln(2) \). These approximations are close to the actual values as the points we chose are near the point \( x = 1 \), where we calculated the tangent. For \( \ln(1.1) \), substituting into the tangent line gives 0.1, while for \( \ln(2) \), it gives 1.
Curve Sketching
Curve sketching involves analyzing the graph of a function to understand its shape and behavior. When we sketch \( y = \ln x \), we notice it's a smooth, increasing curve that flattens out as \( x \) increases. The curve has a vertical asymptote at \( x = 0 \) and is concave down everywhere.
When using the tangent line at \( x = 1 \) to approximate values on \( y = \ln x \), we must consider that for \( x > 1 \), the tangent line underestimates the values on the curve since it lies below the actual curve. However, for \( x < 1 \), the line is above the curve, resulting in overestimates. This behavior illustrates the limitations of using tangent lines for approximation, particularly when moving further from the point of tangency.
When using the tangent line at \( x = 1 \) to approximate values on \( y = \ln x \), we must consider that for \( x > 1 \), the tangent line underestimates the values on the curve since it lies below the actual curve. However, for \( x < 1 \), the line is above the curve, resulting in overestimates. This behavior illustrates the limitations of using tangent lines for approximation, particularly when moving further from the point of tangency.
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