Believe it or not, there was a time in the past when people had to solve math problems without Calculus because it had not yet been discovered. Today, everyone uses the derivative of a function to find a tangent line at a certain point. We will be looking at how we can determine the tangent line at a certain point when given the curve and that point. The method that we will be using is to find a circle which is tangent to the curve at that point and then use some algebraic methods to find the equation of the tangent line to the curve at that point. Click here for the GSP file with which we will be working.
Consider finding the tangent line to the curve
at the point (1,2). We know that we can construct some circle with center (c,0). We can know that by moving the center we will come across a value of c for which the circle is tangent to the curve at the point (1,2).We can construct a right triangle at that point where the hypotenuse is the radius of the circle (See figure below).
What we need to find is the slope of the hypotenuse of the triangle (radius of the circle) since the tangent line to the curve at that point would be perpendicular to the radius of the circle. The legs of the right triangle have lengths of 2 and c-1. So by using the Pythagorean Theorem we get an equation for r.
If we knew the center of the circle then we would be able to find the slope of the radius. We can now use our equation for r to find the equation of the circle.
Yet we know from the equation of our curve that
By arranging the terms in this way, we can see the coefficients of each term of our quadratic equation which are often referred to as a, b, and c for the quadratic equation
. We will be using c' for the c in the quadriatic equation since the center of our circle is (c,0). For our equation we find:
.
(Be sure to note that we used c' for our coefficient since we were already using c for the x value of the center of our circle.)
By using the GSP file, we see that by moving the center of our circle we could have more than one intersection with the curve (see figure below).
If we want to find the center for which there is only one intersection with the curve at the point (1,2) then we need know that the discriminant
for the equation of the circle must equal 0. So we can take our coefficients and find when the value of c for which the discriminant is zero.
We found that the circle intersects the curve at only the point (1,2) when the center of the circle is
.
We can now use the point (1,2) and
Now since the tangent line to the curve at that point will be perpendicular to r then the slope of the tangent line will be the negative reciprocal of the slope of r or
.
We can now use point-slope form in order to find the equation of our tangent line.
We have now found the tangent line to the curve at the point (1,2) without using any Calculus!
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