which is free of logical issues. Unfortunately Descartes’ technique is also very difficult to use and as we will see is of limited applicability.
Descartes’s method is often called the Method of Normals because what he actually finds is the line normal to (perpendicular to) a curve. Once the line normal to the curve at a point is obtained, the line perpendicular to the normal line will be tangent to the curve. Descartes’ Method relies solely on the fact (from Geometry) that at each point of a circle the radius and tangent line through the point are perpendicular. We’ll illustrate Descartes’ idea with the following example.
Example3.5.2.
To find the slope of the normal (and, eventually the tangent) line to the graph of the curve \(y=\sqrt{2x}\) at the point \((2,2)\) Descartes’ approach was to look at the family of circles with centers on the \(x\)-axis and passing through the point \((2,2)\text{.}\) The sketch below displays several members of of that family of circles.
Notice that each circle crosses the parabola at the point \((2,2)\text{.}\) We are searching for the center of the solid black circle which touches (is tangent to) the parabola at \((2,2)\text{.}\) If we can find the center of the black circle then the radial line through the point \((2,2)\) will be normal to the parabola at \((2,2)\) and the slope of the tangent line will be the negative reciprocal of the slope of the normal.
We now want to turn the geometric problem of finding the intersection of our circle and parabola into an algebraic problem. Let \((a,0)\) denote the coordinates of the center of a circle in that family, then the equation of the circle with center \((a,0)\) is \((x-a)^2+y^2=r^2\) where \(r\) is the length of the radius of the circle and \(a\) is a parameter. Since we require our circle to pass through the point \((2,2)\) the length of this radius will be the distance from \((2,2)\) to \((a,0)\text{,}\)\(r=\sqrt{(2-a)^2+2^2}\text{.}\) Thus we have
At this point it is tempting to use the Quadratic Formula to solve for \(x\) and get (typically) two distinct solutions for \(x\) in terms of \(a\text{.}\) But the fact is we really don’t care about the variable \(x\text{.}\) We want to find the value of \(a\) which ensures that the circle and the curve intersect exactly once.
Think about this for a moment. When we use the Quadratic Formula to solve this equation we get a single solution precisely when the discriminant (the part under the square root) is zero.
Problem3.5.3.
Use the Quadratic Formula to show that the discriminant of equation (3.5) is \((2-2a)^2-4(4a-8)\text{.}\) Setting this equal to zero and solving, we get
So the center of the circle touching the curve only once is at \((3, 0)\) and the line segment from \((3,0)\) to \((2,2)\) is normal (perpendicular) to the graph of \(y=\sqrt{2x}\text{.}\)
Problem3.5.4.
Descartes would not have approached this problem using the idea of a discriminant as we just did. He knew that the only way the circle and the parabola could be tangent at \(x=2\) is if \(2\) is a double root of equation (3.5). Since a quadratic polynomial only has two roots, this means that equation (3.5) must be
Like any powerful tool, Analytic Geometry must be used carefully. Sometimes, as in this problem, the transition from Geometry to Algebra can generate algebraic solutions that have no corresponding geometric solution. An extraneous root is the most common example of this.
Problem3.5.5.Extraneous Roots.
Show that the roots of \(x^2+(2-2a)x+(4a-8)=0\) are \(x=2\) and \(x=2a-4\text{.}\) Notice that when \(a\ge2\) and \(a\ne3\) the parabola and circle will have two intersection points (as seen in the graph in Example 3.5.2). When \(a\lt{}2\) we get one positive and one negative root, but the parabola and the circle cross only at \((2,2)\text{.}\) How do we make peace with this apparent contradiction?
Hint.
Look closely at the equation for the parabola. Can \(x\) be negative?
Example 3.5.2 illustrates that there is a significant difference between curves touching (curves that are tangent) and curves crossing (curves that intersect). Finding the line tangent to a curve is actually much more subtle than you would expect at first. But this is not the appropriate place to start that conversation so we will hold off until we reach Section 5.1.
Problem3.5.6.
Now that we’ve found the center of the circle tangent to the curve \(y=\sqrt{2x}\) at \((2,2)\text{,}\) use this to find the equation of the lines normal and tangent to the curve at that point. Plot the curve and these two lines on the same set of axes to see if they really are normal and tangent to the curve.
Problem3.5.7.
Use Descartes’ Method of Normals to find the slope of the line tangent to the curve \(y=\sqrt{x}\) at the point \((4,2)\text{.}\)
Problem3.5.8.
Use Descartes’ Method of Normals to find the slope of the line tangent to the curve \(y=x^2\) at the point \((3,9)\) and compare with Problem 3.4.10.
Hint.
You may want to use circles that are centered on the \(y\)–axis instead of the \(x\)–axis.
Problem3.5.9.
Apply Descartes’ Method of Normals to the line \(y-3x=2\) at the point \((1,5)\text{.}\) What equation did you find for the tangent line? Does this make sense to you?
Descartes’ Method, unlike Fermat’s, is free of the logical pitfalls. But it is algebraically cumbersome because it requires that we find the “double root” of a polynomial. Double roots of quadratic functions like the one in Example 3.5.2 are relatively easy to find. But for higher degree polynomials this can be a substantial problem. For transcendental functions like \(y=1-\cos(x)\) this problem is almost insurmountable without Calculus.
