Find the slope of the parametric curve x=−2t ³ +1, y=3t ², for −∞\u003ct\u003c∞, at the point corresponding to t=2.

Ch.12 - Parametric and Polar Curves

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch.12 - Parametric and Polar Curves

Problem 12.1.7

Chapter 12, Problem 12.1.7

Find the slope of the parametric curve x=−2t ³ +1, y=3t ², for −∞<t<∞, at the point corresponding to t=2.

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Identify the parametric equations given: \(x = -2t^{3} + 1\) and \(y = 3t^{2}\).

Recall that the slope of the parametric curve at a given \(t\) is found by computing \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).

Find the derivative of \(x\) with respect to \(t\): \(\frac{dx}{dt} = \frac{d}{dt}(-2t^{3} + 1) = -6t^{2}\).

Find the derivative of \(y\) with respect to \(t\): \(\frac{dy}{dt} = \frac{d}{dt}(3t^{2}) = 6t\).

Evaluate \(\frac{dy}{dx}\) at \(t=2\) by substituting into the derivatives: \(\frac{dy}{dx} = \frac{6 \times 2}{-6 \times (2)^{2}}\).

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Key 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, usually denoted t. Instead of y as a function of x, both x and y depend on t, allowing the description of more complex curves and motions.

Derivative of Parametric Curves

To find the slope of a parametric curve at a given parameter t, compute dx/dt and dy/dt, then find dy/dx by dividing dy/dt by dx/dt. This gives the instantaneous rate of change of y with respect to x at that point.

Evaluating Derivatives at a Specific Parameter

After finding the general expression for dy/dx in terms of t, substitute the given value of t to find the slope at that specific point on the curve. This step provides the exact slope corresponding to the parameter value.

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