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Ch. 11 - Parametric Equations and Polar Coordinates
Hass - Thomas' Calculus 15th Edition
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Tutti i libri di testoHass 15th EditionCh. 11 - Parametric Equations and Polar CoordinatesProblema 11.2.6
Capitolo 11, Problema 11.2.6

Tangent Lines to Parametrized Curves


In Exercises 1−14, find an equation for the line tangent to the curve at the point defined by the given value of t. Also, find the value of d²y/dx² at this point.


x = sec² t − 1, y = tan t, t = −π/4

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First, find the derivatives of the parametric equations with respect to the parameter \(t\). Compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) from the given functions \(x = \sec^{2} t - 1\) and \(y = \tan t\).
Next, find the slope of the tangent line \(\frac{dy}{dx}\) at the given value of \(t = -\frac{\pi}{4}\) by using the chain rule for parametric curves: \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
Evaluate \(x\) and \(y\) at \(t = -\frac{\pi}{4}\) to find the specific point on the curve where the tangent line touches.
Use the point-slope form of a line, \(y - y_0 = m(x - x_0)\), where \(m\) is the slope found in step 2 and \((x_0, y_0)\) is the point found in step 3, to write the equation of the tangent line.
To find \(\frac{d^{2}y}{dx^{2}}\) at \(t = -\frac{\pi}{4}\), first compute \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\), then divide by \(\frac{dx}{dt}\) using the formula \(\frac{d^{2}y}{dx^{2}} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\), and finally evaluate at \(t = -\frac{\pi}{4}\).

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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 shape and behavior by varying t, rather than expressing y directly as a function of x.
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Parameterizing Equations

Finding the Tangent Line to a Parametric Curve

The tangent line at a point on a parametric curve is found using derivatives dx/dt and dy/dt. The slope of the tangent line is dy/dx = (dy/dt) / (dx/dt), evaluated at the given t. Using this slope and the point coordinates, the tangent line equation can be written in point-slope form.
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Differentiation of Parametric Curves

Second Derivative d²y/dx² for Parametric Curves

The second derivative d²y/dx² measures the curvature of the parametric curve and is found by differentiating dy/dx with respect to x. Using the chain rule, d²y/dx² = (d/dt(dy/dx)) / (dx/dt). Evaluating this at the given t provides the curve's concavity at that point.
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Higher Order Derivatives of Parametric Curves
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Centroids


Find the coordinates of the centroid of the curve x = cos t, y = t + sin t, 0 ≤ t ≤ π.

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Finding Cartesian from Parametric Equations


In Exercises 19–24, match the parametric equations with the parametric curves labeled A through F.


x = cos t, y = sin 3t


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Finding Lengths of Polar Curves


Find the lengths of the curves in Exercises 21–28.


The curve r = cos³(θ/3), 0 ≤ θ ≤ π/4

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Shifting Conic Sections


You may wish to review Section 1.2 before solving Exercises 39-56.


Exercises 53-56 give equations for hyperbolas and tell how many units up or down and to the right or left each hyperbola is to be shifted. Find an equation for the new hyperbola, and find the new center, foci, vertices, and asymptotes.


x²/4 − y²/5 = 1, right 2, up 2

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Hyperbolas


Exercises 27-34 give equations for hyperbolas. Put each equation in standard form and find the hyperbola's asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch.


8x² − 2y² = 16

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Polar to Cartesian Equations


Replace the polar equations in Exercises 27–52 with equivalent Cartesian equations. Then describe or identify the graph.


r = 3 cos θ

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