The 0.0001 Time Rate Approach, an Extended Application

In the previous post, I introduced to you the 0.0001 Time Rate approach This post is an extended application of the said technique on curvatures and radius of curvatures.

Curvature is defined (by The Facts on File Dictionary of Mathematics) as the rate of change of the slope of the tangent to a curve. For each point on a smooth curve there is a circle that has the same tangent and the same curvature at that point. The radius of this circle called the “radius of curvature”, is the reciprocal of the curvature, and its center is known as the center of curvature. If the graph of a function y=f(x) is a continuous curve, the slope of the tangent at any point is given by the derivative dy/dx and the curvature is given by

C = \displaystyle\frac{\displaystyle\frac{d^2y}{dx^2}}{(1+(\displaystyle\frac{dy}{dx})^2)^{3/2}}

and the radius of curvature is given by

RC = \frac{1}{C} = \displaystyle \frac{(1+(\displaystyle \frac{dy}{dx})^2)^{3/2}}{\displaystyle \frac{d^2y}{dx^2}}

where: \frac{dy}{dx} is the derivative of y with respect to x and \frac{d^2x}{dx^2}  is the second derivative of y with respect to x or in simpler terms, it is the derivative of \frac{dy}{dx}.

For example, we solve this problem.

What is the radius of curvature y + \ln cos x = 0 of at point (1,1)?

We will first try to solve this using the usual approach. I will assume here that you already know how to get the derivative of some basic functions.

Solving for \frac{dy}{dx} of the curve, we get

Differentiating: y + \ln \cos x = 0

We have

\displaystyle \frac{dy}{dx} + \displaystyle \frac{- \sin x}{\cos x} = 0.

\displaystyle \frac{dy}{dx} = \tan x.

Solving for \frac{d^2y}{dx^2}, we get

Differentiating \displaystyle \frac{dy}{dx} = \tan x

We have  \frac{d^2y}{dx^2} = (\sec x)^2.

Substituting \frac{dy}{dx} and \frac{d^2y}{dx^2} to the formula for Radius of Curvature, we have

RC = \frac{1}{C} = \displaystyle \frac{(1+(\displaystyle \frac{dy}{dx})^2)^{3/2}}{\displaystyle \frac{d^2y}{dx^2}} = \frac{(1 + \tan x^2)^{3/2}}{(\sec x)^2}

But we know that 1 + (\tan x)^2 = (\sec x)^2. Thus,

RC = \displaystyle \frac{((\sec x)^2)^{3/2}}{(\sec x)^2}

RC = \displaystyle \frac{(\sec x)^3}{(\sec x)^2} = \sec x.

So, the Radius of Curvature of y + \ln \cos x = 0  at (1,1) is

\sec x \cong 1.850815718.   (Answer).

The solution would quite be lengthy for starters. As a remedy, I present to you an easier approach using the 0.0001 Time Rate Approach.

Solving the problem using 0.0001 Time Rate Approach

Our calculators can solve exact derivatives directly and since we are looking for the Radius of Curvature at a given point, we can use x=1, y=1.

\frac{dy}{dx} = (-\ln \cos x |_{x=1}

NOTE: When you are differentiating/integrating using your calculator, always use the radian mode, you may arrive at a different answer when you use the degree mode.

 \frac{dy}{dx}= 1.557407725

For \frac{d^2y}{dx^2}, we use the “0.0001 Time Rate Approach”.

Since \frac{d^2y}{dx^2} is the change of \frac{dy}{dx} with respect to x we can represent that change as a very minute change or differential change with dt = 0.0001.

So,

\displaystyle \frac{d^2y}{dx^2} = \displaystyle \frac{\frac{dy}{dx_2} - \displaystyle \frac{dy}{dx_1}}{dt}

\displaystyle\frac{dy}{dx_1} is the initial derivative of the function,

\displaystyle \frac{dy}{dx_1} = \frac{dy}{dx} = 1.557407725.

\frac{dy}{dx_2} is the derivative of the function after the differential time of 0.0001.

So, \frac{dy}{dx_2} is

\frac{dy}{dx_2} = ( - \ln \cos|_{x=1 + 0.0001 = 1.0001}

\frac{dy}{dx_2} = 1.55775033.

Again, I advise you not to round off decimals to maintain accuracy.

So,

\frac{d^2y}{dx^2} = \frac{\frac{dy}{dx_2} - \frac{dy}{dx_1}}{dt}= \frac{1.55775033 - 1.557407725}{0.0001}

\frac{d^2y}{dx^2} = 3.42605.

Radius of Curvature is

RC = \displaystyle \frac{(1+(\displaystyle \frac{dy}{dx})^2)^{3/2}}{\displaystyle \frac{d^2y}{dx^2}} = \frac{(1 + 1.557407725)^2)^{3/2}}{3.42605}.

Now, we arrived at the same answer using the “0.0001 Time Rate Approach”. However, this technique can only be applied if we are looking for exact derivatives. The “0.0001 Time Rate Approach” is a “board examination primer” for engineering students and aspiring Math Quizzers. I hope this one and the previous post will help you ace your Math subjects and future exams on Differential Calculus.

Stay tuned for more! Enjoy your learning and share your knowledge to others.

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