Question

Parametric equations for a curve are given. (a) Find \(\frac{d y}{d x}\). (b) Find the equations of the tangent and normal line(s) at the point(s) given. (c) Sketch the graph of the parametric functions along with the found tangent and normal lines. \(x=\sec t, y=\tan t\) on \((-\pi / 2, \pi / 2) ; \quad t=\pi / 4\)

Short answer

(a) \(\frac{dy}{dx} = \csc t\). (b) Tangent: \(y = \sqrt{2}x - \sqrt{2} + 1\), Normal: \(y = -\frac{1}{\sqrt{2}}x + \sqrt{2} + 1\). (c) Curve, tangent, and normal lines intersect at \((\sqrt{2}, 1)\).

Step-by-step solution

Find dx/dt and dy/dt

For the parametric equations \(x = \sec t\) and \(y = \tan t\), find the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). The derivative of \(x\) with respect to \(t\) is \(\frac{d}{dt} (\sec t) = \sec t \tan t\). The derivative of \(y\) with respect to \(t\) is \(\frac{d}{dt} (\tan t) = \sec^2 t\).

Find dy/dx

Use the chain rule to find \(\frac{dy}{dx}\), which is \(\frac{dy}{dt} / \frac{dx}{dt}\). Substitute the derivatives found: \(\frac{dy}{dx} = \frac{\sec^2 t}{\sec t \tan t} = \frac{\sec t}{\tan t} = \csc t\).

Find the coordinates at t = π/4

Substitute \(t = \pi/4\) into the parametric equations to find the coordinates. \(x = \sec(\pi/4) = \sqrt{2}\) and \(y = \tan(\pi/4) = 1\). So the coordinates at \(t = \pi/4\) are \((\sqrt{2}, 1)\).

Slope of the tangent at t = π/4

Substitute \(t = \pi/4\) into \(\frac{dy}{dx} = \csc t\) to find the slope of the tangent. \(\csc(\pi/4) = \sqrt{2}\).

Equation of the tangent line

Use the point-slope form of the line equation: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point. For \(m = \sqrt{2}\) and the point \((\sqrt{2}, 1)\), the equation is \(y - 1 = \sqrt{2}(x - \sqrt{2})\). Simplifying, \(y = \sqrt{2}x - \sqrt{2} + 1\).

Slope of the normal line

The slope of the normal line is the negative reciprocal of the tangent slope. So \(-\frac{1}{\sqrt{2}}\).

Equation of the normal line

Use the point-slope form for the normal line: \(y - y_1 = m(x - x_1)\). For the slope \(-\frac{1}{\sqrt{2}}\) and point \((\sqrt{2}, 1)\), the equation is \(y - 1 = -\frac{1}{\sqrt{2}}(x - \sqrt{2})\). Simplifying, \(y = -\frac{1}{\sqrt{2}}x + \sqrt{2} + 1\).

Sketch the graph

Plot the parametric curve \(x = \sec t\) and \(y = \tan t\) for \(t \in (-\pi/2, \pi/2)\) on a graph. Also include the tangent line \(y = \sqrt{2}x - \sqrt{2} + 1\) and the normal line \(y = -\frac{1}{\sqrt{2}}x + \sqrt{2} + 1\) passing through point \((\sqrt{2}, 1)\). Ensure the tangent is exactly touching the curve and the normal is perpendicular at the point.

Key concepts

Derivative dy/dx

When dealing with parametric equations, finding the derivative \( \frac{dy}{dx} \) requires the application of the chain rule. For our given set of equations, where \( x = \sec t \) and \( y = \tan t \), we first determine each derivative with respect to the parameter \( t \). The derivative \( \frac{dx}{dt} \) is \( \sec t \tan t \), and \( \frac{dy}{dt} \) is \( \sec^2 t \). The chain rule states that \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Plugging in the derived expressions, we get \( \frac{dy}{dx} = \frac{\sec^2 t}{\sec t \tan t} \). Simplifying this, it results in \( \csc t \).

For practical purposes, understanding the formula \( \frac{dy}{dx} \) helps determine how the curve behaves as \( t \) changes. This derivative offers insights into the direction and steepness of the curve.

Tangent Line

The tangent line to a parametric curve at a given point provides a linear approximation of the curve near that point. To find the equation of this line, we first calculate the slope using \( \frac{dy}{dx} \). At \( t = \pi/4 \), substituting this into \( \csc t \) gives the slope \( \sqrt{2} \).

The equation for the tangent line in point-slope form is \( y - y_1 = m(x - x_1) \). Using the point \((\sqrt{2}, 1)\) and slope \(\sqrt{2}\), we find:

• \( y - 1 = \sqrt{2}(x - \sqrt{2}) \)
• This simplifies to: \( y = \sqrt{2}x - \sqrt{2} + 1 \)

The tangent line serves as an instantaneous direction indicator for the curve at specifically \( t = \pi/4 \). It captures the curve's immediate trend and can be used to understand local behavior.

Normal Line

A normal line to a curve at a point is perpendicular to the tangent at that same point. Finding its equation involves calculating the negative reciprocal of the tangent's slope. In this case, the normal line slope is \(-\frac{1}{\sqrt{2}}\).

Using the point-slope formula \( y - y_1 = m(x - x_1) \) again, with point \((\sqrt{2}, 1)\) and slope \(-\frac{1}{\sqrt{2}}\), we derive the equation:

• \( y - 1 = -\frac{1}{\sqrt{2}}(x - \sqrt{2}) \)
• This simplifies to: \( y = -\frac{1}{\sqrt{2}}x + \sqrt{2} + 1 \)

The normal line provides a way to analyze orthogonal behavior on the curve. It's essential for understanding more geometric properties such as curvature or intersections with other lines.

Graph Sketching

Plotting the graph of a parametric equation involves converting the relationship between \(x\) and \(y\) as defined by parameter \(t\). For the curve defined by \(x = \sec t\) and \(y = \tan t\), plot points for a range of \(t\) values in the interval \((-\pi/2, \pi/2)\). Include key points and observe trends such as asymptotes or symmetry.

Once the curve is plotted, add both the tangent and normal lines. Ensure the tangent line \(y = \sqrt{2}x - \sqrt{2} + 1\) touches the curve exactly at \((\sqrt{2}, 1)\), and the normal line \(y = -\frac{1}{\sqrt{2}}x + \sqrt{2} + 1\) crosses perpendicularly at the same point.
Sketching helps visualize intersections and curvature, providing a graphical understanding of the analytical outcomes.