Textbook Question
67–72. Derivatives Consider the following parametric curves.
a. Determine dy/dx in terms of t and evaluate it at 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
2:51 minutesProblem 12.1.80
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
Verified step by step guidance1
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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Video duration:
2mKey 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. 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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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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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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