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Magazine Chapter 10: Problem 33
Find the equation of the tangent line to the given curve at the given value of \(t\) without eliminating the parameter. Make a sketch. $$ x=2 \sec t, y=2 \tan t ; t=-\frac{\pi}{6} $$
Short Answer
Expert verified
Tangent line equation: \( y = -2x + \frac{4\sqrt{3}}{3} \).
Step by step solution
01
Identify the Equation of the Curve
The problem gives the parametric equations of the curve as follows:\[ x = 2 \sec t \] \[ y = 2 \tan t \] These equations define the curve in terms of the parameter \( t \).
02
Find the Coordinates of the Point on the Curve
To locate the point on the curve at \( t = -\frac{\pi}{6} \), we substitute this value into the parametric equations:\[ x = 2 \sec \left(-\frac{\pi}{6}\right) = 2 \times \frac{1}{\cos\left(-\frac{\pi}{6}\right)} = 2 \times \frac{2}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \] \[ y = 2 \tan \left(-\frac{\pi}{6}\right) = 2 \times \left(-\frac{1}{\sqrt{3}}\right) = -\frac{2\sqrt{3}}{3} \]The coordinates of the point on the curve are \( \left( \frac{4\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3} \right) \).
03
Differentiate Parametric Equations to Find Slope
Find the derivatives of \( x \) and \( y \) with respect to \( t \):\[ \frac{dx}{dt} = \frac{d}{dt}[2 \sec t] = 2 \sec t \tan t \] \[ \frac{dy}{dt} = \frac{d}{dt}[2 \tan t] = 2 \sec^2 t \]Now find the slope of the tangent line (\( \frac{dy}{dx} \)):\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2 \sec^2(t)}{2 \sec(t) \tan(t)} = \frac{\sec t}{\tan t} = \frac{1}{\sin t} \]
04
Calculate Slope at Specific Parameter Value
Substitute \( t = -\frac{\pi}{6} \) into the slope expression:\[ \frac{dy}{dx} = \frac{1}{\sin\left(-\frac{\pi}{6}\right)} = \frac{1}{-\frac{1}{2}} = -2 \]The slope of the tangent line at \( t = -\frac{\pi}{6} \) is \(-2\).
05
Write the Equation of the Tangent Line
Use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where the slope \( m = -2 \) and point \( (x_1, y_1) = \left( \frac{4\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3} \right) \):\[ y - \left(-\frac{2\sqrt{3}}{3}\right) = -2 \left(x - \frac{4\sqrt{3}}{3}\right) \]Simplify the equation:\[ y + \frac{2\sqrt{3}}{3} = -2x + \frac{8\sqrt{3}}{3} \] \[ y = -2x + 2\sqrt{3} - \frac{2\sqrt{3}}{3} \] \[ y = -2x + \frac{4\sqrt{3}}{3} \]This is the equation of the tangent line.
06
Make a Sketch of the Curve and Tangent Line
Draw the curve defined by the parametric equations \( x = 2 \sec t \) and \( y = 2 \tan t \). Plot the point \( \left( \frac{4\sqrt{3}}{3}, -\frac{2\sqrt{3}}{3} \right) \) on the curve. Draw the tangent line to the curve at this point, which has the equation \( y = -2x + \frac{4\sqrt{3}}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parametric Equations
Parametric equations are a powerful tool in mathematics that express the coordinates of points along a curve as functions of a variable, typically denoted as parameter \( t \). These equations are particularly useful when describing curves that are not easily represented by a single function \( y = f(x) \).
- Each parametric equation relates a separate function to \( t \): one for \( x \) and one for \( y \).
- Such a system defines a curve in a coordinate plane as \( t \) changes.
- \( x = 2 \sec t \)
- \( y = 2 \tan t \)
Differentiation
Differentiation is a mathematical process that allows us to find the rate of change of a function with respect to a variable. In the context of parametric equations, we find the derivatives of each equation with respect to the parameter \( t \).
- For the given parametric equations, differentiation helps us find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
- These derivatives tell us how the \( x \) and \( y \) coordinates change as \( t \) varies.
- \( \frac{dx}{dt} = 2 \sec t \tan t \)
- \( \frac{dy}{dt} = 2 \sec^2 t \)
Slope of Tangent Line
The slope of a tangent line gives us an insight into the direction and angle at which the tangent touches the curve at a particular point. In parametric equations, the slope of the tangent is calculated by dividing the derivative of \( y \) with respect to \( t \) by the derivative of \( x \) with respect to \( t \).
- This ratio, \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \), provides the slope of the tangent line at a given point on the curve.
- For the given parametric functions, substituting \( t = -\pi/6 \) yields the slope.
- \( \frac{dy}{dx} = \frac{1}{\sin(-\pi/6)} = -2 \)
