Problem 63 - Calculus Early Transcendentals

Question

Consider the following parametric curves. a. Determine \(d y / d x\) in terms of \(t\) and evaluate it at the given value of \(t.\) b. Make a sketch of the curve showing the tangent line at the point corresponding to the given value of \(t.\) $$x=t+1 / t, y=t-1 / t ; t=1$$

Step-by-Step Solution

Answer

Question: Determine the equation of the tangent line to the curve defined by the parametric equations \(x = t + \frac{1}{t}\) and \(y = t - \frac{1}{t}\) at the point corresponding to \(t = 1\), and provide a rough sketch of the curve and tangent line. Answer: The tangent line to the curve at the point corresponding to \(t = 1\) is vertical and passes through the point (2, 0). Since it is vertical, its equation is \(x = 2\).

Step 1: Finding \(dx/dt\) and \(dy/dt\)

First, we need to find the derivatives of \(x\) and \(y\) with respect to \(t\). Given the parametric equations: $$x = t + \frac{1}{t}$$ $$y = t - \frac{1}{t}$$ Now, differentiate both equations with respect to \(t\): $$\frac{dx}{dt} = 1 - \frac{1}{t^2}$$ $$\frac{dy}{dt} = 1 + \frac{1}{t^2}$$

Step 2: Finding \(\frac{dy}{dx}\)

Now, we will use the chain rule to find the derivative \(\frac{dy}{dx}\). Recall, that \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). Now we can substitute the expressions found in Step 1 into this equation: $$\frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}}$$

Step 3: Evaluating \(\frac{dy}{dx}\) at \(t=1\)

Now that we have found the expression for \(\frac{dy}{dx}\), we need to evaluate it at \(t=1\). So, we replace \(t\) with \(1\): $$\frac{dy}{dx}\Big|_{t=1} = \frac{1 + \frac{1}{1^2}}{1 - \frac{1}{1^2}} = \frac{1+1}{1-1}$$ However, this expression evaluates to an undefined value as the denominator becomes zero, which means the tangent is going to be vertical.

Step 4: Find the point corresponding to \(t=1\)

Now, we need to find the coordinate of the point on the graph corresponding to \(t=1\). Substitute \(t=1\) into the parametric equations to find the point \((x,y)\): $$x = 1 + \frac{1}{1} = 2$$ $$y = 1 - \frac{1}{1} = 0$$ So, our point is \((2, 0)\).

Step 5: Sketch the curve and the tangent line

Since the tangent line is vertical, the line will pass through the point \((2,0)\) with an undefined slope. It's going to be parallel to the \(y\)-axis. To sketch the curve, we can generate some points using different values of \(t\), and then plot the points on the Cartesian plane together with the tangent line passing through \((2,0)\).

Key Concepts

Derivative of Parametric Equations

In calculus, parametric equations allow us to describe a set of related quantities as functions of an independent variable, often represented by \( t \). To find the rate of change of the dependent variable \( y \) with respect to \( x \), one must first determine their derivatives with respect to \( t \). This involves differentiating each parametric equation separately.

• From the parametric equations \( x = t + \frac{1}{t} \) and \( y = t - \frac{1}{t} \), their derivatives with respect to \( t \) are \( \frac{dx}{dt} = 1 - \frac{1}{t^2} \) and \( \frac{dy}{dt} = 1 + \frac{1}{t^2} \).
• These derivatives express how \( x \) and \( y \) change with the parameter \( t \).

Finding the ratio \( \frac{dy}{dx} \) is crucial for understanding how the curve behaves as \( t \) varies.

Chain Rule in Calculus

The chain rule is a fundamental theorem in calculus used to determine the derivative of composite functions. In the context of parametric equations, it simplifies finding \( \frac{dy}{dx} \) by expressing it in terms of \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \).

• This is done through the relation \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).
• For our parametric curve, substituting the derivatives found earlier gives \( \frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \).

The chain rule here links the change in \( y \) with respect to \( x \) indirectly using \( t \), making complex problems much more tractable. Its usage is crucial especially when direct differentiation with respect to \( x \) is not feasible.

Tangent Line to Curve

A tangent line to a curve at a given point is a straight line that just barely "touches" the curve at that point. Mathematically, it's defined by the derivative \( \frac{dy}{dx} \) at a particular point, representing the slope of the tangent.

• In the parametric context, it describes the instantaneous direction of the curve.
• For \( t = 1 \), substituting into our expression for \( \frac{dy}{dx} \) results in an undefined slope due to division by zero, signifying a vertical tangent line.

Thus, in this case, the tangent line is aligned with the y-axis through the point \( (2, 0) \), indicating a vertical orientation.

Evaluating Derivatives

Evaluating derivatives means substituting specific values into our derivative expression to find the slope or different characteristics of the curve at those points. This helps us understand the curve’s shape and behavior locally.

• Substituting \( t = 1 \) into the parametric equations \( x = t + \frac{1}{t} \) and \( y = t - \frac{1}{t} \) yields the point \((2, 0)\).
• Plugging the same value into the derivative expression \( \frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \) leads to an undefined result.

This specific outcome shows that at \( t = 1 \), the slope is undefined, confirming a vertical tangent, and highlights how derivative evaluation can offer insights into the nature of curves.