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Chapter 15: Problem 31

Exer. 29-32: Find the points on the graph of the equation at which the curvature is \(0 .\) $$ y=\sinh x $$

Short Answer

Expert verified
The point of zero curvature is (0, 0).

Step by step solution

01

Find the First Derivative

The given function is \( y = \sinh x \). The first derivative of \( y \) with respect to \( x \) is found using the derivative of hyperbolic sine function. We have: \( \frac{dy}{dx} = \cosh x \).
02

Find the Second Derivative

Next, we compute the second derivative which is the derivative of \( \frac{dy}{dx} = \cosh x \). So, \( \frac{d^2y}{dx^2} = \sinh x \).
03

Calculate the Curvature

Curvature \( \kappa \) for a function \( y = f(x) \) is given by the formula: \[ \kappa = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}} \] Substituting \( f'(x) = \cosh x \) and \( f''(x) = \sinh x \), the curvature is \( \kappa = \frac{|\sinh x|}{(1 + (\cosh x)^2)^{3/2}} \).
04

Set Curvature to Zero

For the curvature to be zero, the numerator of the curvature formula must be zero. Therefore, \( |\sinh x| = 0 \). Since \( \sinh x = 0 \) when \( x = 0 \), the point on the graph where the curvature is zero is when \( x = 0 \).
05

Find the Corresponding y-coordinate

When \( x = 0 \), substitute back into the original function to find \( y \): \( y = \sinh(0) = 0 \). Thus, the point of zero curvature on the graph is \( (0, 0) \).

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Key Concepts

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

Hyperbolic Functions
Hyperbolic functions may seem a bit unfamiliar at first, but they are quite analogous to trigonometric functions. The hyperbolic sine function, denoted as \( \sinh x \), is one such function and can be described as:
  • Related to the exponential function: \( \sinh x = \frac{e^x - e^{-x}}{2} \).
  • Unlike trigonometric functions, hyperbolic functions are defined for all real numbers.
  • The graph of \( \sinh x \) wiggles around the x-axis but doesn't oscillate between fixed values like sine does.
Understanding \( \sinh x \) helps us in problems dealing with the curvature of graphs and in situations modeling hyperbolic movements. The analogous function to cosine in hyperbolic terms is \( \cosh x \). These functions work together similarly to sine and cosine.
Derivative
The derivative is a fundamental tool in calculus. It provides us with the rate at which a function is changing at any point. For hyperbolic functions, such as \( y = \sinh x \), knowing how to find the derivative is crucial. Here鈥檚 the key takeaway:
  • The derivative of \( \sinh x \) is \( \cosh x \).
  • This results from the definition of hyperbolic functions, where \( \cosh x = \frac{e^x + e^{-x}}{2} \).
The derivative helps us determine the slope of the tangent line to the function at any given point. This concept is essential when analyzing functions for curvature, as it allows us to understand how the graph is bending or changing direction as \( x \) changes.
Second Derivative
If the first derivative tells you the slope, the second derivative gives information about how that slope itself is changing. This is particularly important in understanding curvature.
  • For \( y = \sinh x \), the second derivative is the derivative of \( \cosh x \).
  • Deriving \( \cosh x \) gives us \( \sinh x \) again, coming full circle.
The second derivative is tremendously informative as it tells us not just about the rate of change of the function, but about the nature of its graph. For curvature calculations, the second derivative is a key player.
Graph Analysis
Graph analysis is not only about sketching a function but understanding its properties such as curvature. For the function \( y = \sinh x \), here's how we use derivatives:
  • Curvature \( \kappa \) is calculated using both the first and the second derivatives.
  • For \( y = \sinh x \), \( \kappa \) becomes zero when \( \sinh x = 0 \), specifically at \( x = 0 \).
  • At this point, the graph has no bending, lying flat against the tangent.
Curvature is important in fields like physics and engineering, where understanding the path and shape of an object is critical. Through graph analysis, we can see how the theoretical structure brings about practical applications.

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Most popular questions from this chapter

Find general formulas for the tangential and normal components of acceleration and for the curvature of the curve \(C\) determined by \(\mathbf{r}(t)\). $$ \mathbf{r}(t)=\cosh t \mathbf{i}+\sinh t \mathbf{j} $$

Find general formulas for the tangential and normal components of acceleration and for the curvature of the curve \(C\) determined by \(\mathbf{r}(t)\). $$ \mathbf{r}(t)=t^{2} \mathbf{i}+(3 t+2) \mathbf{j} $$

Exer. \(31-34:\) Find \(r(t)\) subject to the given conditions. $$ \begin{aligned} \mathbf{r}^{\prime}(t)=t^{2} \mathbf{i}+(6 t+1) \mathbf{j}+8 t^{3} \mathbf{k} & \\ \mathbf{r}(0)=2 \mathbf{i}-3 \mathbf{j}+\mathbf{k} \end{aligned} $$

If \(C\) is the graph of \(y=\cosh x,\) find an equation of the circle of curvature for the point \(P(0,1)\)

Evaluate the integral. $$ \int_{0}^{1}\left[t e^{\left(t^{2}\right)} \mathbf{i}+\sqrt{t} \mathbf{j}+\left(t^{2}+1\right)^{-1} \mathbf{k}\right] d t $$
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