Skip to main content
Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh.12 - Parametric and Polar CurvesProblem 12.1.78
Chapter 12, Problem 12.1.78

77–80. Slopes of tangent lines Find all points at which the following curves have the given slope.


x = 2 cos t, y = 8 sin t; slope = -1

Verified step by step guidance
1
Recall that the slope of the tangent line to a parametric curve given by \(x = x(t)\) and \(y = y(t)\) is found by computing \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
Differentiate both parametric equations with respect to \(t\): compute \(\frac{dx}{dt}\) for \(x = 2 \cos t\) and \(\frac{dy}{dt}\) for \(y = 8 \sin t\).
Write the expression for the slope \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) using the derivatives found in the previous step.
Set the slope equal to the given value \(-1\) and solve the resulting equation for \(t\).
Substitute the values of \(t\) found back into the parametric equations \(x = 2 \cos t\) and \(y = 8 \sin t\) to find the corresponding points on the curve.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Parametric Equations and Curves

Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted t. Understanding how x and y depend on t allows us to analyze the curve's behavior and find points corresponding to specific conditions, such as a given slope.
Recommended video:
Guided course
08:02
Parameterizing Equations

Derivative of Parametric Curves (dy/dx)

For parametric curves, the slope of the tangent line dy/dx is found by dividing the derivative of y with respect to t by the derivative of x with respect to t, i.e., (dy/dt) / (dx/dt). This ratio gives the instantaneous rate of change of y with respect to x at any parameter value t.
Recommended video:
06:49
Differentiation of Parametric Curves

Finding Points with a Given Slope

To find points where the curve has a specific slope, set the expression for dy/dx equal to the given slope and solve for the parameter t. Then, substitute t back into the parametric equations to find the corresponding (x, y) points on the curve.
Recommended video:
05:45
Understanding Slope Fields
Related Practice
Textbook Question

77–80. Slopes of tangent lines Find all points at which the following curves have the given slope.


x = 2 + √t, y = 2 - 4t; slope = -8

51
views
Textbook Question

57–64. Graphing polar curves Graph the following equations. Use a graphing utility to check your work and produce a final graph.


r = sin 3θ

51
views
Textbook Question

33–40. Areas of regions Make a sketch of the region and its bounding curves. Find the area of the region.


The region inside one leaf of r = cos 3θ

63
views
Textbook Question

Find the slope of the parametric curve x=−2t ³ +1, y=3t ², for −∞<t<∞, at the point corresponding to t=2. 

69
views
Textbook Question

93–94. Parametric equations of ellipses Find parametric equations (not unique) of the following ellipses (see Exercises 91–92). Graph the ellipse and find a description in terms of x and y.


An ellipse centered at (-2, -3) with major and minor axes of lengths 30 and 20, parallel to the x- and y-axes, respectively, generated counterclockwise (Hint: Shift the parametric equations.)

160
views
Textbook Question

37–52. Curves to parametric equations Find parametric equations for the following curves. Include an interval for the parameter values. Answers are not unique.


The horizontal line segment starting at P(8, 2) and ending at Q(−2, 2)

29
views