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

## Tuesday, April 8, 2014

### Finding Equation - Curve, 12

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

"Published in Newark, California, USA"

Find the equation of the curve for which y" = 4/x³, and which is tangent to the line at the point (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

In order to get the value of arbitrary constant, we need the values of x and dy/dx. Since dy/dx is not given in the problem but the equation of the tangent line is given, then we can use the slope of the tangent line as the slope of a curve which is dy/dx. The point of tangency which is (1, 3) is also a point in the curve which is the intersection of the curve and a tangent line. Write the equation of a tangent line in slope intercept form as follows

The slope of a tangent line which is also equal to the slope of a curve is equal to

Next, substitute the values of x and dy/dx in order to get the value of arbitrary 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 get the value of arbitrary constant, substitute the value of the given point which is (1, 3) to the above equation, we have

Therefore, the equation of a curve is