Calculus

The function given is: We are tasked with finding the coefficient of in the Taylor series expansion of this function about and . The Taylor series expansion of a function around a point is given by: We expand this for around , so we need to compute the derivatives of at . First, we calculate : Next, compute the partial derivative of with respect to : At , we have: Thus, the coefficient of in the Taylor expansion is 5.

To find the centre of the circle of curvature at a given point on a curve, we use the formula for the radius of curvature : where is the first derivative and is the second derivative of the curve. The equation of the curve is: Step 1: Find the first and second derivatives of .
- The first derivative - The second derivative Step 2: Evaluate and at .
At : - - Step 3: Compute the radius of curvature .
Substitute and into the formula for : Thus, the radius of curvature at is . Step 4: Find the coordinates of the centre of the circle of curvature.
The centre of the circle lies on the normal to the curve at the given point, and the radius is directed perpendicular to the tangent at the point. The slope of the tangent line at is , and the slope of the normal is the negative reciprocal of the slope of the tangent: Thus, the equation of the normal line at is: Now, the distance from to the centre of the circle is the radius . The centre of the circle lies on this normal line. By using geometry and solving for the centre, we find the coordinates of the centre of the circle to be .
Thus, the centre of the circle of curvature is .

We are tasked with evaluating the integral: Step 1: Simplify the integrand.
We can use the double-angle identity for cosine: So the integral becomes: This expression can be split into two separate integrals: Simplifying each term: Step 2: Evaluate each integral.
1. The first term: 2. The second term requires a bit more work. Using the identity , we recognize that: This integral evaluates to: Step 3: Combine the results.
Thus, the total integral becomes: Therefore, the integral is:

The curvature of a curve at any point is given by the formula: where is the first derivative of the curve and is the second derivative of the curve.
For a straight line, the second derivative is always 0 because the slope of the line does not change. Therefore, the curvature of any straight line is 0.
For the given straight line , the first derivative (the slope) is 2, and the second derivative is 0.
Thus, the curvature of the straight line at any point is: Therefore, the curvature at point is 0.