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

Verified step by step guidance

1. Identify the parametric equations given: \(x = -2t^{3} + 1\) and \(y = 3t^{2}\).
2. Recall that the slope of the parametric curve at a given \(t\) is found by computing \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
3. Find the derivative of \(x\) with respect to \(t\): \(\frac{dx}{dt} = \frac{d}{dt}(-2t^{3} + 1) = -6t^{2}\).
4. Find the derivative of \(y\) with respect to \(t\): \(\frac{dy}{dt} = \frac{d}{dt}(3t^{2}) = 6t\).
5. Evaluate \(\frac{dy}{dx}\) at \(t=2\) by substituting into the derivatives: \(\frac{dy}{dx} = \frac{6 \times 2}{-6 \times (2)^{2}}\).

Key Concepts

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.