73–76. Tangent lines Find an equation of the line tangent to the curve at the point corresponding to the given value of t.

x=t ²−1, y=t ³ +t; t=2

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Identify the parametric equations given: \(x = t^{2} - 1\) and \(y = t^{3} + t\), and the value of the parameter \(t = 2\) at which we want the tangent line.

Find the derivatives of \(x\) and \(y\) with respect to \(t\): compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). For \(x = t^{2} - 1\), \(\frac{dx}{dt} = 2t\). For \(y = t^{3} + t\), \(\frac{dy}{dt} = 3t^{2} + 1\).

Calculate the slope of the tangent line \(\frac{dy}{dx}\) at \(t=2\) using the chain rule for parametric equations: \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) evaluated at \(t=2\).

Find the coordinates of the point on the curve at \(t=2\) by substituting \(t=2\) into the parametric equations: \(x(2)\) and \(y(2)\).

Use the point-slope form of the line equation with the point \((x(2), y(2))\) and slope \(\frac{dy}{dx}\) to write the equation of the tangent line: \(y - y(2) = m (x - x(2))\), where \(m\) is the slope found in the previous step.

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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, often denoted as t. Here, x and y are given in terms of t, allowing us to analyze the curve's behavior by varying t.

Derivative of Parametric Curves

To find the slope of the tangent line to a parametric curve, we compute dy/dx as (dy/dt) divided by (dx/dt). This ratio gives the instantaneous rate of change of y with respect to x at a specific parameter value.

Equation of a Tangent Line

Once the slope of the tangent line and the point of tangency are known, the tangent line's equation can be written using the point-slope form: y - y₀ = m(x - x₀), where m is the slope and (x₀, y₀) is the point on the curve.

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