Textbook Question
67–72. Derivatives Consider the following parametric curves.
b. Make a sketch of the curve showing the tangent line at the point corresponding to the given value of t.
x = 2 + 4t, y = 4 − 8t; t = 2
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16. Parametric Equations & Polar Coordinates
Calculus with Parametric Curves
3:32 minutesProblem 12.1.82
Textbook Question
81–88. Arc length Find the arc length of the following curves on the given interval.
x = 3 cos t, y = 3 sin t + 1; 0 ≤ t ≤ 2π
Verified step by step guidance1
Recall the formula for the arc length of a parametric curve given by \(x = x(t)\) and \(y = y(t)\) over the interval \(a \leq t \leq b\):
\[L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]
Find the derivatives of \(x(t)\) and \(y(t)\) with respect to \(t\):
\[\frac{dx}{dt} = \frac{d}{dt}(3 \cos t) = -3 \sin t\]
\[\frac{dy}{dt} = \frac{d}{dt}(3 \sin t + 1) = 3 \cos t\]
Substitute these derivatives into the arc length formula under the square root:
\[\sqrt{(-3 \sin t)^2 + (3 \cos t)^2} = \sqrt{9 \sin^2 t + 9 \cos^2 t}\]
Use the Pythagorean identity \(\sin^2 t + \cos^2 t = 1\) to simplify the expression inside the square root:
\[\sqrt{9 (\sin^2 t + \cos^2 t)} = \sqrt{9 \cdot 1} = 3\]
Set up the integral for the arc length over the interval \(0 \leq t \leq 2\pi\):
\[L = \int_0^{2\pi} 3 \, dt\]
This integral can then be evaluated to find the total arc length.
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3mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Parametric Equations
Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted as t. In this problem, x and y are given in terms of t, allowing the curve to be traced as t varies over the interval. Understanding parametric form is essential for applying calculus techniques to curves not defined explicitly as y = f(x).
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Arc Length Formula for Parametric Curves
The arc length of a curve defined parametrically by x(t) and y(t) from t = a to t = b is found using the integral ∫ from a to b of √[(dx/dt)² + (dy/dt)²] dt. This formula calculates the length by summing infinitesimal distances along the curve, requiring differentiation of both x and y with respect to t.
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Differentiation of Trigonometric Functions
Calculating the derivatives dx/dt and dy/dt involves differentiating trigonometric functions like sine and cosine. Knowing that d/dt[cos t] = -sin t and d/dt[sin t] = cos t is crucial for correctly applying the arc length formula and simplifying the integral for evaluation.
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