Domanda del libro di testo
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
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Ch.12 - Parametric and Polar Curves
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 12, Problema 12.1.80
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
Guida verificata passo dopo passo1
Identify the given parametric equations: \(x = 2 + \sqrt{t}\) and \(y = 2 - 4t\).
Recall that the slope of the tangent line to a parametric curve is given by \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
Compute the derivatives with respect to \(t\): \(\frac{dx}{dt} = \frac{d}{dt}(2 + \sqrt{t})\) and \(\frac{dy}{dt} = \frac{d}{dt}(2 - 4t)\).
Set the slope equal to the given value: \(\frac{dy}{dx} = -8\), which means \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = -8\).
Solve the resulting equation for \(t\), then substitute back into the original parametric equations to find the corresponding points \((x, y)\).
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Parametric Equations
Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted as t. Here, x and y are given in terms of t, allowing us to analyze the curve's behavior by studying these functions.
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08:02
Parameterizing Equations
Derivative of Parametric Curves
The slope of the tangent line to a parametric curve is found by computing dy/dx = (dy/dt) / (dx/dt). This requires differentiating both x(t) and y(t) with respect to t and then dividing the results to find the instantaneous rate of change.
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Percorso guidato
06:49Differentiation of Parametric Curves
Finding Points with a Given Slope
To find points where the curve has a specific slope, set the derivative dy/dx equal to the given slope and solve for the parameter t. Substituting t back into x(t) and y(t) gives the coordinates of the points with that slope.
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05:45Understanding Slope Fields
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