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Chapter 9: Problem 11

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. \(x=\cos t \sin (2 t), y=\sin t \sin (2 t)\) on \([0,2 \pi] ; \quad t=3 \pi / 4\)

Short Answer

Expert verified
To find \( \frac{dy}{dx} \), use \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). At \( t=\frac{3\pi}{4} \), tangent and normal lines are at \((0, 0)\); they ultimately coincide due to the parametric definitions.

Step by step solution

01

Differentiate x with respect to t

To find \( \frac{dy}{dx} \), we first need to find \( \frac{dx}{dt} \). Given \( x = \cos t \sin (2t) \), apply the product rule: \( \frac{dx}{dt} = \frac{d}{dt}(\cos t) \cdot \sin (2t) + \cos t \cdot \frac{d}{dt}(\sin (2t)) \). Compute derivatives: \( \frac{d}{dt}(\cos t) = -\sin t \) and \( \frac{d}{dt}(\sin (2t)) = 2 \cos (2t) \). Substituting these in, \( \frac{dx}{dt} = -\sin t \cdot \sin (2t) + \cos t \cdot 2 \cos (2t) \).
02

Differentiate y with respect to t

Now we find \( \frac{dy}{dt} \). Given \( y = \sin t \sin (2t) \), apply the product rule: \( \frac{dy}{dt} = \frac{d}{dt}(\sin t) \cdot \sin (2t) + \sin t \cdot \frac{d}{dt}(\sin (2t)) \). Compute derivatives: \( \frac{d}{dt}(\sin t) = \cos t \). Substituting these in, \( \frac{dy}{dt} = \cos t \cdot \sin (2t) + \sin t \cdot 2 \cos (2t) \).
03

Compute dy/dx

We know that \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). Substitute the expressions from Steps 1 and 2: \( \frac{dy}{dx} = \frac{\cos t \cdot \sin (2t) + \sin t \cdot 2 \cos (2t)}{-\sin t \cdot \sin (2t) + \cos t \cdot 2 \cos (2t)} \). Simplify this expression to find the slope of the tangent line at any \( t \).
04

Find the tangent and normal line at t=3蟺/4

First, calculate the specific \( x \) and \( y \) values at \( t = \frac{3 \pi}{4} \): \( x = \cos \left(\frac{3 \pi}{4}\right) \sin \left(\frac{3 \pi}{2}\right) = 0 \) and \( y = \sin \left(\frac{3 \pi}{4}\right) \sin \left(\frac{3 \pi}{2}\right) = 0 \). At this point, compute the slope of the tangent line \( m = \frac{dy}{dx} \) using the expression from Step 3 and the parameter \( t = \frac{3\pi}{4} \). The equation of the tangent line is \( y - 0 = m(x - 0) \). The normal line, perpendicular to the tangent, has a slope of \( -\frac{1}{m} \). Its equation is \( y = -\frac{1}{m}x \).
05

Sketch the parametric curve and lines

Plot the parametric curve using \( x = \cos t \sin (2t) \) and \( y = \sin t \sin (2t) \) for \( t \) from 0 to \( 2\pi \). At \( t = \frac{3 \pi}{4} \), draw the tangent line with the computed slope from Step 4 and passing through the calculated point \((0, 0)\). Similarly, draw the normal line with the perpendicular slope from Step 4 passing through the same point. This provides a visual understanding of the curve and its tangent and normal lines at the specified \( t \).

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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 is a straight line that touches a curve at a single point without crossing it. This means that at the point of tangency, the tangent line has the same direction as the curve. The slope of the tangent line is the derivative of the curve at that point. For parametric equations, this involves calculating the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), and then using these to find \(\frac{dy}{dx}\). This derivative represents the slope of the tangent to the curve at a specific parameter \(t\).

The equation of the tangent line at a point \((x_0, y_0)\) is given by\[ y - y_0 = m(x - x_0) \],where \(m\) is the slope of the tangent line.
  • Find the point of tangency using the specific parameter value of \(t\).
  • Calculate \(\frac{dy}{dx}\) to determine the slope of the tangent line.
  • Plug \(x_0\), \(y_0\), and \(m\) into the tangent line equation.
Normal Line
A normal line at a given point on a curve is a straight line that is perpendicular to the tangent line at that point. It plays an essential role, especially in problems that involve normals in physical contexts like optics and mechanics. To find the normal line, you first need the slope of the tangent line. Once you have the tangent slope, the slope of the normal line is the negative reciprocal of it, which is denoted as \(-\frac{1}{m}\).

The equation for the normal line at a point \((x_0, y_0)\) is:\[ y - y_0 = -\frac{1}{m} (x - x_0) \].
  • Identify the point where the normal line is needed, often the same as the tangent point.
  • Compute the slope of the normal as \(-\frac{1}{m}\) using the known tangent slope \(m\).
  • Use the point \((x_0, y_0)\) to find the equation of the normal line.
Differentiation
Differentiation is a fundamental concept in calculus that measures how a function changes as its input changes. It's essential for understanding rates of change and patterns in functions. In parametric equations, both \(x\) and \(y\) are expressed in terms of a third parameter, often \(t\). Thus, to find \(\frac{dy}{dx}\), we use the chain rule, which applies differentiation to parametric equations by finding \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), then forming the ratio of these derivatives: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \].
  • Differentiate both \(x\) and \(y\) with respect to \(t\).
  • Use the product rule for functions that are products of more than one element.
  • Simplify \(\frac{dy}{dx}\) to find the slope of the tangent line.
Sketching Curves
Sketching curves involves plotting points and drawing the path of the curve that connects these points. For parametric equations, this can be executed by calculating \(x\) and \(y\) for various values of \(t\), and plotting these coordinates. The curve can show significant features such as loops, intercepts, and symmetry.

To include tangent and normal lines, you need:
  • The point of tangency, where the tangent and normal lines touch the curve.
  • The equations of both the tangent and normal lines to accurately plot their paths.
  • A clear range for \(t\) ensuring all interesting features appear within the graph.
Include the main elements using software or a graphing calculator to produce an accurate visual representation of the curve along with its tangent and normal lines.

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Most popular questions from this chapter

Pick a integer value for \(n,\) where \(n \neq 2,3,\) and use technology to plot \(r=\sin \left(\frac{m}{n} \theta\right)\) for three different integer values of \(m\). Sketch these and determine a minimal interval on which the entire graph is shown.

Convert the rectangular equation to a polar equation. . \(y=5\)

Graph the polar function on the given interval. \(r=2+\sin \theta, \quad[0,2 \pi]\)

Find: (a) \(\frac{d y}{d x}\) (b) the equation of the tangent and normal lines to the curve at the indicated \(\theta\) -value. \(r=1-3 \cos \theta ; \quad \theta=3 \pi / 4\)

Find \(t=t_{0}\) where the graph of the given parametric equations is not smooth, then find \(\lim _{t \rightarrow t_{0}} \frac{d y}{d x}\). \(x=-t^{3}+7 t^{2}-16 t+13, \quad y=t^{3}-5 t^{2}+8 t-2\)
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