Problem 3 - Calculus Early Transcendentals

Question

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. $$x=t^{4}+1, \quad y=t^{3}+t ; \quad t=-1$$

Step-by-Step Solution

Answer

The equation of the tangent is \(y = -x\).

Step 1: Understand the Problem

We need to find the equation of the tangent to the curve defined by parametric equations \(x = t^4 + 1\) and \(y = t^3 + t\) at the parameter \(t = -1\). This means we need to find the slope of the tangent at \(t = -1\) and the coordinates of the point on the curve.

Step 2: Find the Point on the Curve

Substitute \(t = -1\) into the parametric equations to find the coordinates of the point.\(x(-1) = (-1)^4 + 1 = 2\)\(y(-1) = (-1)^3 + (-1) = -2\)So, the point on the curve is \((2, -2)\).

Step 3: Differentiate the Parametric Equations

To find the slope of the tangent, we need \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\). Differentiate both equations with respect to \(t\):\(\frac{dx}{dt} = 4t^3\)\(\frac{dy}{dt} = 3t^2 + 1\).

Step 4: Compute the Slope of the Tangent

The slope \(m\) of the tangent line is given by \(\frac{dy}{dx}\). Use the derivatives:\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 + 1}{4t^3}\).Substitute \(t = -1\) to find the slope:\(m = \frac{3(-1)^2 + 1}{4(-1)^3} = \frac{4}{-4} = -1\).

Step 5: Write the Equation of the Tangent Line

We use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point on the curve and \(m\) is the slope.Using \((x_1, y_1) = (2, -2)\) and \(m = -1\):\(y - (-2) = -1(x - 2)\).Simplify to get the equation of the tangent:\(y + 2 = -x + 2\), or\(y = -x + 0\).

Step 6: Present the Final Answer

The equation of the tangent line to the curve at \(t = -1\) is \(y = -x\). This line has a slope of \(-1\) and passes through the point \((2, -2)\).

Key Concepts

Parametric Equations

Parametric equations express the coordinates of the points making up a geometric object, like a curve, as functions of a parameter. In this case, the equations are given as \(x = t^4 + 1\) and \(y = t^3 + t\). Here, \(t\) is the parameter. This parameter allows us to define both \(x\) and \(y\) in terms of \(t\), enabling a dynamic representation of the curve.

• Parameter \(t\): Represents a variable that helps to describe each point on the curve. It acts like a time variable, telling you at different "times" where the point will be on this curve.
• Curve Representation: Rather than expressing \(y\) directly as a function of \(x\), these equations allow complex curves that might not be functions in the usual sense (like circles or ellipses) to be expressed as the path traced by a moving point.
• Efficiency: Parametric equations simplify finding specific points (like our point \((2, -2)\) when \(t = -1\)) on a curve by offering a straightforward substitution method.

Understanding parametric equations is crucial because it transforms how we perceive and calculate curves and their points, enabling advanced calculus tasks like finding tangent lines, as seen in this exercise.

Differentiation in Calculus

Differentiation is a key tool in calculus used to determine rates of change and slopes of curves. For the given parametric equations \(x = t^4 + 1\) and \(y = t^3 + t\), differentiation allows us to find the slope of the tangent by computing \(\frac{dy}{dx}\).

• Partial Derivatives: We first calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), partial derivatives with respect to the parameter \(t\). This step gives us the rate at which \(x\) and \(y\) change concerning \(t\).
• Slope of the Tangent Line: After finding these derivatives, the slope of the curve in terms of \(x\) (not \(t\)) is determined using \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\). This ratio delivers the slope at any given \(t\).
• Example: By substituting \(t = -1\) into these derivative formulas, we calculated \(\frac{dy}{dx} = -1\), revealing that the tangent line at this point has a slope of \(-1\).

Differentiation helps us transform the abstract representation of motion provided by parametric forms into the concrete notion of slopes, making them integral for solving real-world geometric problems.

Point-Slope Form of a Line

The point-slope form of a line, \(y - y_1 = m(x - x_1)\), is a practical tool for writing the equation of a line. In this scenario, we identified the needed point \((2, -2)\) and slope \(-1\) from our calculations.

• Formula Breakdown: This line equation format is ideal when you know a point on the line \((x_1, y_1)\) and the slope \(m\). It lets us neatly encapsulate these into a linear equation.
• Application: Here, substituting \((x_1, y_1) = (2, -2)\) and \(m = -1\) into the formula provides us the equation \(y + 2 = -1(x - 2)\).
• Resulting Line Equation: Simplifying gives us \(y = -x\), which describes the tangent line at the point on our parametric curve.

Using the point-slope form is beneficial for quickly determining line equations directly from essential information about parameters, reflecting its significant utility in calculus problems, such as finding tangents to parametric curves.