Non-Calculus approach to Invariable Calculus Project

I shared my posts (here and here) on the Invariable Calculus Project in the AP Calculus Community.  Gary Litvin posted a response within the Community offering there a great non-calculus alternative solution to the original problem of the area of the triangle formed by the x- and y-axes and any Quadrant I tangent line to \displaystyle \frac{1}{x}.  Here’s a paraphrase of Gary’s approach.

Let \displaystyle \frac{x}{a} + \frac{y}{b}=1 be any Quadrant I tangent line to \displaystyle y=\frac{1}{x}.  (In case you don’t recognize it, this tangent equation uses the intercept form of a line–a is the x-intercept and b is the y-intercept.)  Because the line intersects the parabola in a single point, we can find that point by solving the system of equations defined by the two equations.  Substituting for y gives

\displaystyle \frac{x}{a} + \frac{\frac{1}{x}}{b}=1.

This is equivalent to x^2 - a \cdot x+\frac{a}{b}=0, a quadratic.  We could determine the value of x using the quadratic formula.  Because there is only one solution to this equation (there is only one point of intersection, the point of tangency), the discriminant must be zero.  That means

\displaystyle (-a)^2 - 4\cdot \left( \frac{a}{b} \right)=0

which can be rearranged to give ab=4 (a=0 is extraneous).  Therefore, the area of the triangle formed by the tangent line to \displaystyle y=\frac{1}{x} and the coordinate axes is \displaystyle Area=\frac{1}{2} ab=2 no matter what the point of tangency.

Shiny.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s