Problem 33 - Essential Calculus

Question

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=\sec \theta, \quad y=\tan \theta $$

Step-by-Step Solution

Answer

The point of horizontal tangency to the curve is at (1,0). There are no points of vertical tangency to the curve.

Step 1: Convert parametric to rectangular

Write the expression for \( x \) in terms of \( y \). Using the trigonometric identity \( \sec^{2}(\theta)=1+\tan^{2}(\theta) \), replace \( \sec \theta \) in the equation \( x=\sec \theta \) by \( x=\sqrt{1+y^{2}} \). Thus, the curve is described by the equation \( x=\sqrt{1+y^{2}} \).

Step 2: Find the derivative

Next, calculate the derivative of \( x \) in terms of \( y \). The derivative of \( x \) with respect to \( y \) is \( dx/dy=y/(y^{2}+1) \). Keep this expression for the subsequent steps.

Step 3: Find points of horizontal tangency

Horizontal tangency points occur where \( dx/dy=0 \). Hence, solve the equation \( y/(y^{2}+1)=0 \). The solution for \( y \) is \( y=0 \). After substituting \( y=0 \) in the rectangular equation \( x=\sqrt{1+y^{2}} \), the corresponding \( x \) value is \( x=1 \). Hence, the point of horizontal tangency is (1,0).

Step 4: Find points of vertical tangency

Vertical tangency occurs where the derivative is undefined. Hence, solve the equation \( y^{2}+1=0 \). However, there is no solution to this equation in the real number system. Hence, there are no points of vertical tangency to the curve.

Step 5: Validate with graphing utility

After plotting the curve \( x=\sqrt{1+y^{2}} \) using a graphing utility, it can be seen that the curve indeed has a flat point at (1,0), confirming our above analysis.