/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Find the radius of curvature at ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the radius of curvature at \((x, y)\) of the curve $$ y=\tanh ^{-1} x \quad(|x|<1) $$

Short Answer

Expert verified
The radius of curvature is \( R = \left| \frac{(1 + x^2)^{3/2}}{2x} \right| \).

Step by step solution

01

Differentiate the Curve

Given the curve \( y = \tanh^{-1} x \), we need to find the first derivative \( \frac{dy}{dx} \). It is known that \( \frac{d}{dx}[\tanh^{-1} x] = \frac{1}{1-x^2} \). Thus, the first derivative is \( \frac{dy}{dx} = \frac{1}{1-x^2} \).
02

Differentiate Again for the Second Derivative

To find the radius of curvature, we also need the second derivative \( \frac{d^2y}{dx^2} \). Differentiate \( \frac{1}{1-x^2} \) with respect to \( x \):\[\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{1}{1-x^2} \right) = \frac{2x}{(1-x^2)^2}\]
03

Apply the Radius of Curvature Formula

The radius of curvature \( R \) is given by:\[R = \left| \frac{(1 + (\frac{dy}{dx})^2)^{3/2}}{\frac{d^2y}{dx^2}} \right|\]Substitute the derivatives found:\[R = \left| \frac{(1 + (\frac{1}{1-x^2})^2)^{3/2}}{\frac{2x}{(1-x^2)^2}} \right|\]
04

Simplify the Expression

Simplify the expression for \( R \):\[R = \left| \frac{(1 + \frac{1}{(1-x^2)^2})^{3/2}}{\frac{2x}{(1-x^2)^2}} \right| = \left| \frac{((1-x^2)^2 + 1)^{3/2}}{2x} \right|\]
05

Final Expression for Radius of Curvature

This simplifies further to:\[R = \left| \frac{(1 + x^2)^{3/2}}{2x} \right|\]This is the radius of curvature at any point \((x, y)\) on the curve \(y = \tanh^{-1} x\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Radius of Curvature
The radius of curvature is a measure that represents the curve's "bending" at a particular point. Imagine driving on a winding road. The radius of curvature would be like the size of the circle that best fits the bend of the road at any point.
For a curve described by the equation \(y = f(x)\), we use a specific formula to find the radius of curvature:
  • \(R = \left| \frac{(1 + (\frac{dy}{dx})^2)^{3/2}}{\frac{d^2y}{dx^2}} \right| \)

This formula requires both the first \(\frac{dy}{dx}\) and the second derivatives \(\frac{d^2y}{dx^2}\). Remember, the more sharply the curve bends, the smaller the radius of curvature. Conversely, a gentle curve would have a larger radius as it approximates more of a straight line. To find it for a given curve, like \(y = \tanh^{-1} x\) in this case, we substitute the derivatives into the formula and simplify.
Derivatives
Derivatives are a fundamental concept in calculus that represent the rate of change. For a function \(y = f(x)\), the first derivative \(\frac{dy}{dx}\) measures how \(y\) changes with a small change in \(x\).
This is essentially the slope of the tangent line to the curve at any given point. If \(\frac{dy}{dx}\) is positive, the function is increasing; if negative, the function is decreasing. When calculating the radius of curvature, the first and second derivatives are both necessary
In our exercise, the first derivative of \(y = \tanh^{-1} x\) is \(\frac{dy}{dx} = \frac{1}{1-x^2}\). The second derivative tells us how the slope is changing, giving deeper insight into the curve's "bending." For \(y = \tanh^{-1} x\), the second derivative is \(\frac{d^2y}{dx^2} = \frac{2x}{(1-x^2)^2}\). These derivatives were used to calculate the radius of curvature.
Inverse Hyperbolic Functions
Inverse hyperbolic functions, like \(\tanh^{-1} x\), are interesting counterparts to the inverse trigonometric functions. They are representations of the inverse functions of hyperbolic functions such as hyperbolic sine and hyperbolic cosine.
The notation \(\tanh^{-1} x\) specifically represents the inverse of the hyperbolic tangent function. Its domain is \(|x| < 1\), which ensures that the function behaves nicely and is well-defined. Hyperbolic functions themselves originate from a hyperbola, similar to how trigonometric functions originate from a circle.
This connection to hyperbolas makes them useful in various applications, including calculus, where modelling of real-world situations often occurs using these functions due to their natural appearance in the solutions of certain differential equations. Understanding inverse hyperbolic functions helps streamline working with functions like \(y = \tanh^{-1} x\) in calculus, especially when calculating derivatives and curvatures.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the indefinite integrals (a) \(\int\left(3+2 x-x^{2}\right) \mathrm{d} x\) (b) \(\int \frac{\mathrm{d} x}{\sqrt{\left(x^{2}-6 x+5\right)}} \mathrm{d} x\) (c) \(\int \frac{\mathrm{d} x}{\sqrt{\left(x^{2}-4 x+8\right)}} \mathrm{d} x\) (d) \(\int \frac{x+3}{\sqrt{\left(x^{2}+4 x+13\right)}} \mathrm{d} x\) (e) \(\int x\left(3+2 x-x^{2}\right) \mathrm{d} x\)

Find the values of the constants \(a\) and \(b\) such that $$ \frac{3 x+2}{x^{2}+2 x+5}=\frac{a(2 x+2)}{x^{2}+2 x+5}+\frac{b}{x^{2}+2 x+5} $$ and hence find its integral. (Note that \(\left.(d / d x)\left(x^{2}+2 x+5\right)=2 x+2 .\right)\)

If \(y=3 \mathrm{e}^{2 x} \cos (2 x-3)\), verify that \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}-4 \frac{\mathrm{d} y}{\mathrm{~d} x}+8 y=0\)

The rate at which a solute diffuses through a membrane is proportional to the area and to the concentration difference across the membrane. A solution of concentration \(C\) flows down a tube with constant velocity \(v\). The solute diffuses through the wall of the tube into an ambient solution of the same solute of a lower fixed concentration \(C_{0} .\) If the tube has constant circular cross-section of radius \(r\), show that at distance \(x\) along the tube the concentration \(C(x)\) satisfies the differential equation $$ \frac{\mathrm{d} C}{\mathrm{~d} x}=-\frac{2 k}{r v}\left(C-C_{0}\right) $$ where \(k\) is a constant.

Find the stationary values of the following functions, distinguishing carefully between them. In each case sketch a graph of the function. (a) \(f(x)=\frac{3 x}{(x-1)(x-4)}\) (b) \(f(x)=2 \mathrm{e}^{-x}(x-1)^{3}\) (c) \(f(x)=x^{2} \mathrm{e}^{-x}\) (d) \(f(x)=\frac{1}{x^{2}}+\frac{8}{(1-x)^{2}}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.