Problem 32 - Essential Calculus

Question

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=4 \cos ^{2} \theta, \quad y=2 \sin \theta $$

Step-by-Step Solution

Answer

The points of horizontal tangency are at \((4,0)\), \((-4,0)\) every integer multiple of \(\pi\) and points of vertical tangency are at \((0,\pm 2)\) for \(\theta = \frac{(2n+1)\pi}{2}\) where \(n\) is an integer.

Step 1: Define Expression for Derivative

A tangent line to a curve defined by parametric equations has a slope given by \(dy/dx\), which can be obtained using the chain rule as \((dy/d\theta) / (dx/d\theta)\). Here, by differentiating the given equations with respect to \(\theta\), the derivative \(dy/dx\) has the form \(dy/dx = (dy/d\theta) / (dx/d\theta)\).

Step 2: Calculate Derivative

After differentiating, compute \((dy/d\theta) = 2 \cos(\theta)\) and \((dx/d\theta) = -8\cos(\theta)\sin(\theta)\). Then, divide \(dy/d\theta\) by \(dx/d\theta\) to obtain the derivative \(dy/dx = -\frac{1}{4\sin(\theta)}\).

Step 3: Points of Horizontal Tangency

Horizontal tangency occurs when \(dy/dx = 0\), which yields \(\sin(\theta) = 0\). Solve for \(\theta\) to get \(\theta= n\pi\), where \(n\) is an integer. By substituting these `\(\theta\)` values into the original equations will yield the points with horizontal tangency which are \((4,0)\), \((-4,0)\) for \(n = 0, \pm 1, \pm 2, \pm 3, . . .\).

Step 4: Points of Vertical Tangency

Vertical tangency occurs when \(dx/d\theta = 0\), which yields \(-8\cos(\theta)\sin(\theta)=0\), or \(\cos(\theta) = 0\) and \(\sin(\theta) = 0\). The first condition gives \(\theta = \frac{(2n+1)\pi}{2}\), where \(n\) is an integer. By substituting these `\(\theta\)` values into the original equations, we find that the vertical tangents lie on the \(y\)-axis, at points \((0,\pm 2)\) for \(n = 0, \pm 1, \pm 2, \pm 3, . . .\).

Step 5: Confirm with Graphing Utility

Use a graphing utility to graph the curve in the x-y plane using the parametric equations. The points of horizontal and vertical tangency should match the coordinates calculated earlier.