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Magazine Chapter 12: Problem 25
Find the point on the graph of \(y=e^{x}\) at which the curvature is maximum.
Short Answer
Expert verified
The point of maximum curvature is \((0, 1)\) on the graph of \(y=e^{x}\).
Step by step solution
01
Understand the Graph and Curvature
The equation \( y = e^x \) is a simple exponential function where the function increases rapidly. The curvature of a graph at a point is a measure of how quickly the direction of the curve is changing at that point. The curvature can be calculated using the formula for curvature \( \kappa = \frac{|y''|}{(1+(y')^2)^{3/2}} \), where \( y' \) is the first derivative and \( y'' \) is the second derivative of \( y \).
02
Find the First and Second Derivatives
Calculate the first and second derivatives of the given function. - The first derivative \( y' = \frac{d}{dx}e^x = e^x \).- The second derivative \( y'' = \frac{d^2}{dx^2} e^x = e^x \).
03
Plug the Derivatives into the Curvature Formula
Substitute \( y' = e^x \) and \( y'' = e^x \) into the curvature formula:\[ \kappa = \frac{|e^x|}{(1 + (e^x)^2)^{3/2}} \]
04
Simplify the Curvature Formula
Using the equation from Step 3, we expand and simplify the expression inside the fraction:\[ \kappa = \frac{e^x}{(1 + e^{2x})^{3/2}} \]
05
Find Critical Points and Determine Maximum Curvature
To find the maximum curvature, we need to find critical points by differentiating \( \kappa \) with respect to \( x \) and setting it to zero. This is complex, so we use numerical methods for simplification, finding that the maximum curvature occurs at \( x = 0 \).
06
Find the Corresponding y-value
Substitute \( x = 0 \) back into the original function to find the corresponding \( y \):\[ y = e^0 = 1 \].
07
State the Point of Maximum Curvature
From the above calculations, the point at which the curvature is maximized on the graph of \( y = e^x \) is \((0, 1)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
Derivatives are fundamental concepts in calculus that describe how a function changes at any given point. For the function \( y = e^x \), the derivative tells us how the curve's slope changes with each increment of \( x \).
This notion is closely related to the graph's steepness.
This notion is closely related to the graph's steepness.
- The first derivative, denoted as \( y' \), indicates the slope of the tangent line to the curve at any point. For the exponential function \( y = e^x \), the first derivative is also \( e^x \), meaning the slope at any point on the curve is the same as the function's value itself.
- Because the function \( y = e^x \) is continuously increasing, its slope is always positive.
- The second derivative, denoted as \( y'' \), describes the rate at which the slope itself is changing. In this case, \( y'' = e^x \) as well, showing that the slope of the tangent line is increasing at an exponential rate.
Critical Points
Critical points are where a function's first derivative is zero or undefined, indicating potential maxima, minima, or inflection points.
Finding these points helps identify where a function reaches its extreme values. In dealing with curvature, we aim to determine where the curvature itself reaches a maximum. This requires us to:
This fact is important as it locates where \( y = e^x \) bends most steeply.
Finding these points helps identify where a function reaches its extreme values. In dealing with curvature, we aim to determine where the curvature itself reaches a maximum. This requires us to:
- Differentiate the curvature formula \( \kappa = \frac{|y''|}{(1+(y')^2)^{3/2}} \) with respect to \( x \).
- Set this derivative to zero to find the critical points.
This fact is important as it locates where \( y = e^x \) bends most steeply.
Exponential Function
The exponential function \( y = e^x \) is a unique mathematical function where the constant \( e \) serves as the base, making it naturally occur in various growth processes.
It exhibits continuous and rapid growth, which makes its graph an upward-sloping curve.
The point of maximum curvature at \( (0, 1) \) reveals the hallmark feature of exponential functions: a noticeable curve despite a general upward trend. Understanding these functions is key to grasping their widespread applications.
It exhibits continuous and rapid growth, which makes its graph an upward-sloping curve.
- In the context of calculus, exponential functions are notable for their self-replicating properties in derivatives; each derivative of \( e^x \) remains \( e^x \), showcasing its consistent growth.
- This unchanging derivative aspect can simplify calculus computations and is particularly useful in modelling real-world phenomena like population growth and radioactive decay.
The point of maximum curvature at \( (0, 1) \) reveals the hallmark feature of exponential functions: a noticeable curve despite a general upward trend. Understanding these functions is key to grasping their widespread applications.
