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Magazine Chapter 3: Problem 39
Find the tangent line to the parametric curve \(x=\varphi_{1}(t), y=\varphi_{2}(t)\) at the point corresponding to the given value \(t_{0}\) of the parameter. $$ \varphi_{1}(t)=t+\cos (t), \varphi_{2}(t)=t+\sin (t) \quad t_{0}=\pi / 4 $$
Short Answer
Expert verified
The tangent line is \( y - (\frac{\pi}{4} + \frac{\sqrt{2}}{2}) = (3 + 2\sqrt{2})(x - (\frac{\pi}{4} + \frac{\sqrt{2}}{2})) \).
Step by step solution
01
Differentiate the Parametric Equations
First, we need to find the derivatives of the parametric equations with respect to \( t \). For \( x = t + \cos(t) \), the derivative is \( \frac{dx}{dt} = 1 - \sin(t) \).For \( y = t + \sin(t) \), the derivative is \( \frac{dy}{dt} = 1 + \cos(t) \).
02
Calculate the Slope of the Tangent
The slope of the tangent to the curve at a particular \( t = t_0 \) is given by the ratio \( \frac{dy/dt}{dx/dt} \). At \( t_0 = \frac{\pi}{4} \), we find:\[ \frac{dy}{dt} = 1 + \cos\left(\frac{\pi}{4}\right) = 1 + \frac{\sqrt{2}}{2} \]\[ \frac{dx}{dt} = 1 - \sin\left(\frac{\pi}{4}\right) = 1 - \frac{\sqrt{2}}{2} \]Therefore, the slope \( m \) is:\[ m = \frac{1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}} \]
03
Compute the Slope Value
Evaluate the expression to find the slope:\[ m = \frac{1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}} \]Simplify using conjugates:\[ m = \frac{1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}} \times \frac{1 + \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \frac{(1 + \frac{\sqrt{2}}{2})^2}{(1 - \frac{\sqrt{2}}{2})(1 + \frac{\sqrt{2}}{2})} \]\[ = \frac{(1 + \frac{\sqrt{2}}{2})^2}{1 - \left(\frac{\sqrt{2}}{2}\right)^2} = \frac{\left(1 + \frac{\sqrt{2}}{2}\right)^2}{1 - \frac{1}{2}}\]\[ = \frac{1 + \sqrt{2} + \frac{1}{2}}{\frac{1}{2}} = 3 + 2\sqrt{2} \]
04
Calculate the Point on the Curve
Evaluate \( x(t_0) \) and \( y(t_0) \) at \( t_0 = \frac{\pi}{4} \) to find the specific point:\[ x\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + \cos\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{\sqrt{2}}{2} \]\[ y\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + \sin\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{\sqrt{2}}{2} \]
05
Write the Equation of the Tangent Line
Use the point-slope form of the line equation: \( y - y_0 = m(x - x_0) \). Here, \( m = 3 + 2\sqrt{2} \), and the point is \( \left(\frac{\pi}{4} + \frac{\sqrt{2}}{2}, \frac{\pi}{4} + \frac{\sqrt{2}}{2} \right) \).Plug these into the formula:\[ y - \left(\frac{\pi}{4} + \frac{\sqrt{2}}{2}\right) = (3 + 2\sqrt{2})(x - \left(\frac{\pi}{4} + \frac{\sqrt{2}}{2}\right)) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Tangent Line
A tangent line to a curve is a line that touches the curve at a single point without crossing it. This contact point is called the point of tangency.
Imagine a tangent line as gently grazing the curve without diving in or away from it abruptly. It only sketches a path along the edge of the curve, perfectly matching the curve's direction at that particular spot.
An equation for a tangent line to a parametric curve can be found using the derivatives of the parametric equations. These derivatives give us the slope, which is pivotal for drawing the tangent line at a specific point. Understanding how a tangent line interacts with parametric curves is essential for grasping advanced calculus topics.
Imagine a tangent line as gently grazing the curve without diving in or away from it abruptly. It only sketches a path along the edge of the curve, perfectly matching the curve's direction at that particular spot.
An equation for a tangent line to a parametric curve can be found using the derivatives of the parametric equations. These derivatives give us the slope, which is pivotal for drawing the tangent line at a specific point. Understanding how a tangent line interacts with parametric curves is essential for grasping advanced calculus topics.
Parametric Differentiation
Parametric differentiation allows us to find the derivative of a curve when the curve is defined via parametric equations.
Parametric equations define a curve by expressing the x and y coordinates as separate functions of a parameter, often denoted as \(t\).
Parametric equations define a curve by expressing the x and y coordinates as separate functions of a parameter, often denoted as \(t\).
- This method of differentiation involves finding the derivative of x and y concerning t \(\left(\frac{dx}{dt}, \frac{dy}{dt}\right)\).
- It is particularly useful when dealing with curves that are not easy to express as explicit functions of x or y.
Slope Calculation
The slope of a tangent line at a given point on a parametric curve is found using the derivatives \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\).
To find the slope \(m\), take the ratio \(\frac{dy/dt}{dx/dt}\). This ratio gives a measure of how steep the curve is at a specific point determined by \(t\).
In practical terms, slope calculation tells us whether the tangent line rises (positive slope) or falls (negative slope) as we move along the x-axis.
This calculation is not only fundamental for drawing tangent lines but also influences understanding how two-dimensional motion changes direction. Slope calculations find application in countless areas like graphing functions and analyzing the movement of objects.
To find the slope \(m\), take the ratio \(\frac{dy/dt}{dx/dt}\). This ratio gives a measure of how steep the curve is at a specific point determined by \(t\).
In practical terms, slope calculation tells us whether the tangent line rises (positive slope) or falls (negative slope) as we move along the x-axis.
This calculation is not only fundamental for drawing tangent lines but also influences understanding how two-dimensional motion changes direction. Slope calculations find application in countless areas like graphing functions and analyzing the movement of objects.
Point-Slope Form
The point-slope form of a line offers a convenient way to write the equation for a line when you know the point it passes through and its slope. Given a point \((x_0, y_0)\) and a slope \(m\), the point-slope form is expressed as:
\[ y - y_0 = m(x - x_0) \]
This formula is very helpful because it translates the geometric concept of a line directly into an algebraic expression. It not only simplifies the process of writing equations for lines but also assists in visualizing how a change in slope or position affects the line.
Understanding point-slope form is a pillar in geometry and algebra, enabling students and professionals to solve practical problems in areas like engineering, physics, and computer graphics. When working with parametric curves, the point-slope form allows us to succinctly write the equation for a tangent line, which is essential for further analysis and application.
\[ y - y_0 = m(x - x_0) \]
This formula is very helpful because it translates the geometric concept of a line directly into an algebraic expression. It not only simplifies the process of writing equations for lines but also assists in visualizing how a change in slope or position affects the line.
Understanding point-slope form is a pillar in geometry and algebra, enabling students and professionals to solve practical problems in areas like engineering, physics, and computer graphics. When working with parametric curves, the point-slope form allows us to succinctly write the equation for a tangent line, which is essential for further analysis and application.
