Showing posts with label Integral Calculus. Show all posts
Showing posts with label Integral Calculus. Show all posts

## Thursday, April 10, 2014

### Finding Equation - Curve, 14

Category: Differential Equations, Integral Calculus, Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the equation of the curve for which y" = 2, and which has a slope of -2 at its point of inflection (1, 3).

Solution:

The concavity of a curve is equal to the second derivative of a curve with respect to x. In this case, y" = d²y/ dx². Let's consider the given concavity of a curve

We can rewrite the above equation as follows

Multiply both sides of the equation by dx, we have

Integrate on both sides of the equation, we have

The point of inflection is a point where the direction of the concavity of a curve will start to change. In this case (1, 3) is the point of inflection of a curve. Since it is also included in the curve, then we can use it to substitute the value of x and y later in the problem.

Substitute the value of the given slope and the point of inflection to the above equation in order to solve for the value of a constant, we have

Hence, the above equation becomes

Multiply both sides of the equation by dx, we have

Integrate on both sides of the equation, we have

In order to solve for the value of a constant, substitute the value of x and y from the coordinates of a given point of inflection to the above equation, we have

Therefore, the equation of a curve is