Skip to main content
Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
12장, 문제 12.R.7a

7–8. Parametric curves and tangent lines
a. Eliminate the parameter to obtain an equation in x and y.
x = 8cos t + 1, y = 8sin t + 2, for 0 ≤ t ≤ 2π; t = π/3

검증된 단계별 안내
1
Identify the given parametric equations: \(x = 8\cos t + 1\) and \(y = 8\sin t + 2\) with the parameter \(t\) in the interval \(0 \leq t \leq 2\pi\).
Recall the Pythagorean identity: \(\cos^2 t + \sin^2 t = 1\). This will help us eliminate the parameter \(t\) by expressing \(\cos t\) and \(\sin t\) in terms of \(x\) and \(y\).
Isolate \(\cos t\) and \(\sin t\) from the parametric equations: \(\cos t = \frac{x - 1}{8}\) and \(\sin t = \frac{y - 2}{8}\).
Substitute these expressions into the Pythagorean identity to get an equation involving only \(x\) and \(y\): \(\left(\frac{x - 1}{8}\right)^2 + \left(\frac{y - 2}{8}\right)^2 = 1\).
Simplify the equation by multiplying both sides by \(64\) (since \(8^2 = 64\)) to obtain the Cartesian equation of the curve without the parameter \(t\).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
2m
도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted as t. Instead of y as a function of x, both x and y depend on t, allowing the description of more complex curves like circles or ellipses.
추천 영상:
08:02
Parameterizing Equations

Eliminating the Parameter

Eliminating the parameter involves manipulating the parametric equations to remove t, resulting in a direct relationship between x and y. This often requires using trigonometric identities or algebraic techniques to rewrite the curve in Cartesian form.
추천 영상:
05:59
Eliminating the Parameter

Tangent Lines to Parametric Curves

The tangent line to a parametric curve at a given parameter value t is found by computing derivatives dx/dt and dy/dt, then using dy/dx = (dy/dt)/(dx/dt). This slope helps write the equation of the tangent line at the specified point on the curve.
추천 영상:
가이드 코스
06:49
Differentiation of Parametric Curves