Question

Finding radii of curvature Find the radius of curvature (see Exercise 70 ) of the following curves at the given point. Then write the equation of the circle of curvature at the point. $$y=\sin x \text { at } x=\pi / 2$$

Short answer

Question: Find the radius of curvature and the equation of the circle of curvature for the curve $$y = \sin x$$ at the point $$x = \pi/2$$. Answer: The radius of curvature is $$R = 1$$, and the equation of the circle of curvature is $$(x - \frac{\pi}{2})^2 + (y - 1)^2 = 1$$.

Step-by-step solution

Find the first and second derivatives of $$y=\sin x$$

We start by finding the first derivative of the function. We know that $$\frac{d (\sin x)}{dx} = \cos x$$. Next, we find the second derivative by taking the derivative of the first derivative. We know that $$\frac{d (\cos x)}{dx} = -\sin x$$. So, the first and second derivatives are: $$y' = \cos x$$ $$y'' = -\sin x$$

Find the function value, first derivative, and second derivative at $$x = \pi/2$$

Now we evaluate the function, first derivative, and second derivative at the given point $$x = \pi/2$$. $$y(\pi/2) = \sin(\pi/2) =1$$ $$y'(\pi/2) = \cos(\pi/2) = 0$$ $$y''(\pi/2) = -\sin(\pi/2) = -1$$

Calculate the radius of curvature $$R$$

Using the formula for radius of curvature, $$R = \frac{(1 + (y')^2)^{\frac{3}{2}}}{|y''|}$$, we can find the radius at the point $$x = \pi/2$$. $$R = \frac{(1 + (0)^2)^{\frac{3}{2}}}{|-1|} = 1$$

Write the equation of the circle of curvature

In order to write the equation of the circle of curvature, we need its center and radius. The center of the circle of curvature is given by the coordinates $$(\pi/2, 1)$$. Having calculated the radius of curvature as $$R = 1$$, the equation of the circle of curvature is: $$(x - \frac{\pi}{2})^2 + (y - 1)^2 = 1^2$$ So, the equation of the circle of curvature is: $$(x - \frac{\pi}{2})^2 + (y - 1)^2 =1$$

Key concepts

First Derivative

The first derivative of a function measures how the function's output changes as the input changes. It is essentially the slope or rate of change of the function at any point on its curve.
In calculus notation, if we have a function \( y = f(x) \), the first derivative is denoted by \( y' \) or \( \frac{dy}{dx} \).
To find the first derivative of \( y = \sin x \), we use the fact that \( \frac{d(\sin x)}{dx} = \cos x \). Hence, the first derivative \( y' \) is \( \cos x \).

• The first derivative tells us the slope of the tangent line to the sine curve at any point \( x \).
• At \( x = \frac{\pi}{2} \), \( y' = \cos(\frac{\pi}{2}) = 0 \), which indicates a horizontal tangent.

Second Derivative

The second derivative provides information about the curvature or concavity of a function's graph. It is the derivative of the first derivative. In mathematical terms, if the first derivative is \( y' = f'(x) \), the second derivative is \( y'' = \frac{d^2y}{dx^2} \).
For the function \( y = \sin x \), the first derivative \( y' = \cos x \). Taking the derivative again, \( \frac{d(\cos x)}{dx} = -\sin x \), gives us the second derivative \( y'' = -\sin x \).

• The second derivative helps in understanding how the slope itself is changing at a given point.
• At \( x = \frac{\pi}{2} \), \( y'' = -\sin(\frac{\pi}{2}) = -1 \), suggesting the curve is concave down.

Equation of Circle

In the context of curvature, the equation of a circle is used to determine the circle of curvature at a specific point on a curve. This circle represents how the curve behaves locally around the point.
The general form of a circle's equation is \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center and \(r\) is the radius.

• The center of the circle of curvature on \( y = \sin x \) at \( x = \frac{\pi}{2} \) is \( (\frac{\pi}{2}, 1) \).
• The radius of curvature \( R = 1 \) is calculated using the formula \( R = \frac{(1 + (y')^2)^{\frac{3}{2}}}{|y''|} \). Substituting the values, we find \( R = 1 \).

The equation of this particular circle then becomes \((x - \frac{\pi}{2})^2 + (y - 1)^2 = 1^2\). This equation describes a circle centered at \((\frac{\pi}{2}, 1)\) with a radius of 1.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are fundamental in mathematics and describe relationships in triangles and oscillatory movements.
The function \( \sin x \) models a smooth wave-like curve, repeating its pattern every \(2\pi\). This periodic behavior translates directly into calculus by how the function and its derivatives behave.

• The derivative \( \cos x \) describes the rate of change or slope of \( \sin x \) at every point. It itself is a trigonometric function.
• The second derivative, \(-\sin x\), indicates how the slope, \( \cos x \), is changing, reflecting the curvature of the sine wave.

Understanding these functions and their derivatives is key in problems involving wave motion, circular motion, and harmonic oscillations, where these derivatives explain acceleration and velocities.