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Chapter 3: Problem 73

Plot the given parametric curve \(x=\varphi_{1}\) \((t), y=\varphi_{2}(t)\) in a viewing rectangle that contains the given point \(P_{0}\). Find the equation of the tangent line at \(P_{0}\). Add the tangent line to your plot. $$ \begin{aligned} &\varphi_{1}(t)=8 \cos (t), \varphi_{2}(t)=8 \sin (t)(\sin (t / 2))^{2}\\\ &P_{0}=(4, \sqrt{(3)} \quad \text { (Tear Drop Curve) } \end{aligned} $$

Short Answer

Expert verified
The tangent line at \( P_0 = (4, \sqrt{3}) \) is \( y - \sqrt{3} = -\frac{\sqrt{3}}{6}(x - 4) \).

Step by step solution

01

Find the parameter t at Point P0

Given the coordinates of \( P_0 = (4, \sqrt{3}) \), we need to solve for \( t \) using the parametric equations. The x-coordinate equation gives \( 4 = 8 \cos(t) \), which simplifies to \( \cos(t) = \frac{1}{2} \). Solving this, we find \( t = \frac{\pi}{3} \).
02

Verify the y-coordinate for t

With \( t = \frac{\pi}{3} \), we check the y-coordinate to ensure it's correct. Substitute into \( \varphi_{2}(t) = 8 \sin(t)(\sin(t/2))^2 \):\[ y = 8 \sin\left(\frac{\pi}{3}\right)\left(\sin\left(\frac{\pi}{6}\right)\right)^2 = 8 \cdot \frac{\sqrt{3}}{2} \cdot \left(\frac{1}{2}\right)^2 = \sqrt{3} \]
03

Derive x'(t) and y'(t)

We need the derivatives \( x'(t) \) and \( y'(t) \) for the tangent line slope.\[ x'(t) = -8\sin(t) \]\[ y'(t) = 8 \left[ \cos(t) \left(\sin\left(\frac{t}{2}\right)\right)^2 + \sin(t) \cdot \sin\left(\frac{t}{2}\right) \cdot \cos\left(\frac{t}{2}\right) \right] \]
04

Evaluate derivatives at t = 蟺/3

Substitute \( t = \frac{\pi}{3} \) into the derivatives:\[ x'\left(\frac{\pi}{3}\right) = -8\sin\left(\frac{\pi}{3}\right) = -4\sqrt{3} \]\[ y'\left(\frac{\pi}{3}\right) = 8 \left[ \frac{1}{2} \cdot \left(\frac{1}{2}\right)^2 + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} \right] = 2 \]
05

Find the slope of the tangent line

The slope \( m \) of the tangent line is given by \( m = \frac{y'(t)}{x'(t)} \). At \( t = \frac{\pi}{3} \):\[ m = \frac{2}{-4\sqrt{3}} = -\frac{\sqrt{3}}{6} \]
06

Equation of the tangent line at P0

Using the point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), at \( P_0 = (4, \sqrt{3}) \):\[ y - \sqrt{3} = -\frac{\sqrt{3}}{6}(x - 4) \]Simplify to obtain the equation of the tangent line.
07

Plot the parametric curve and tangent line

On a coordinate plane, plot \( x = 8\cos(t) \) and \( y = 8\sin(t)(\sin(t/2))^2 \). The curve looks like a teardrop. Add the tangent line using its equation, and ensure \( P_0 \) is marked within the viewport.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Line Equation
A tangent line to a curve at a specific point gives the best linear approximation of the curve at that point. It "touches" the curve without crossing it. The equation of the tangent line can be derived using the point-slope form of a line, which is:
  • \( y - y_1 = m(x - x_1) \)
Here, \((x_1, y_1)\) is the point of tangency, and \(m\) is the slope of the tangent line. In the exercise, the point \(P_0 = (4, \sqrt{3})\) is given, and the slope was calculated as \(-\frac{\sqrt{3}}{6}\). So, the equation of the tangent line at \(P_0\) becomes:
  • \( y - \sqrt{3} = -\frac{\sqrt{3}}{6}(x - 4) \)
This line closely approximates the curve near \(P_0\) and intersects the curve at precisely that point.
Derivatives of Parametric Equations
Parametric equations are defined as a set of equations where both \(x\) and \(y\) are expressed in terms of a third variable, usually \(t\), called the parameter. To find the derivatives of these equations, we differentiate each component with respect to \(t\). Here鈥檚 how it鈥檚 done: First, compute \(x'(t)\) and \(y'(t)\) using differentiation rules:
  • \(x(t) = 8\cos(t) \Rightarrow x'(t) = -8\sin(t)\)
  • \(y(t) = 8\sin(t)(\sin(t/2))^2 \Rightarrow y'(t) \= 8 \left[ \cos(t) (\sin(t/2))^2 + \sin(t) \cdot \sin(t/2) \cdot \cos(t/2) \right]\)
These derivatives are crucial for calculating the slope of the tangent line at any point \((x(t),y(t))\). By evaluating \(x'(t)\) and \(y'(t)\) at a specific \(t\), one can find the rate of change of the parametric curve at that point.
Plotting Parametric Equations
Plotting parametric equations involves representing the relationship defined by the equations across a range of \(t\) values. Here, \(x(t) = 8\cos(t)\) and \(y(t) = 8\sin(t)(\sin(t/2))^2\) form a teardrop-shaped curve. To create the full plot:
  • Select a suitable range for \(t\). Common ranges for trigonometric functions often extend from \(0\) to \(2\pi\).
  • Generate pairs \((x(t), y(t))\) for evenly spaced values of \(t\) within this interval.
  • Plot the points on the coordinate plane. By connecting these points, you reveal the curve's shape.
  • Add the tangent line at \(P_0\) to this plot for a visual representation of its interaction with the curve.
A complete plot showcases the curve's journey through the axes and helps identify critical points like the tangent point \(P_0\).
Slope of Tangent Line
The slope of a tangent line to a parametric curve at a particular point is the key to determining the curve's direction at that point. For parametric equations, the slope \(m\) at a point is given by the rate of change of \(y(t)\) with respect to \(x(t)\), or \(\frac{dy}{dx}\). This is evaluated via:
  • \(m = \frac{y'(t)}{x'(t)}\)
In the given problem where \(t = \frac{\pi}{3}\), the slope is:
  • \(x'\left(\frac{\pi}{3}\right) = -4\sqrt{3}\)
  • \(y'\left(\frac{\pi}{3}\right) = 2\)
Thus, the slope \(m\) is calculated as \(-\frac{\sqrt{3}}{6}\). This value indicates the rate at which \(y\) increases or decreases relative to \(x\) around the point \(P_0 = (4, \sqrt{3})\). Such analysis of slope is vital for understanding how the curve behaves in its local neighborhood.

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Most popular questions from this chapter

Differentiate the given expression with respect to \(x\). $$ \operatorname{arcsec}(1 / x) $$

Show that \(\cos (2 \arccos (x))\) is the restriction to the interval [-1,1] of a polynomial in \(x\).

Calculate the linearization \(L(x)=f(c)+\) \(f^{\prime}(c), \cdot(x-c)\) for the given function \(f\) at the given value \(c\) $$ f(x)=x \ln (x), c=e $$

An initial value problem is given, along with its exact solution. (Read the instructions for Exercises \(47-50\) for terminology.) Verify that the given solution is correct by substituting it into the given differential equation and the initial value condition. Calculate the Euler's Method approximation \(y_{1}=y_{0}+F\left(x_{0}, y_{0}\right) \Delta x\) of \(y\left(x_{1}\right)\) where \(\Delta x=x_{1}-x_{0} .\) Let \(m_{1}=\left(F\left(x_{0}, y_{0}\right)+F\left(x_{1}, y_{1}\right)\right) / 2\) and \(z_{1}=y_{0}+\) \(m_{1} \Delta x .\) This is the Improved Euler Method approximation of \(y\left(x_{1}\right) .\) Calculate \(z_{1} .\) By evaluating \(y\left(x_{1}\right),\) determine which of the two approximations, \(y_{1}\) or \(z_{1},\) is more accurate. $$ \begin{array}{l} d y / d x=1+y / x, y(2)=1 / 2, x_{1}=3 / 2 ; \text { Exact solution: } y(x)= \\\ x \ln (x)+x(1 / 4-\ln (2)) \end{array} $$

Explain how the linearizations of the differentiable functions \(f\) and \(g\) at \(c\) may be used to discover the product rule for \((f \cdot g)^{\prime}(c)\).
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