Problem 66 - Calculus Early Transcendentals

Question

Find an equation of the line tangent to the curve at the point corresponding to the given value of \(t.\) $$x=\sin t, y=\cos t ; t=\pi / 4$$

Step-by-Step Solution

Answer

Question: Find the equation of the tangent line to the curve defined by parametric equations \(x=\sin t\) and \(y=\cos t\) at the point corresponding to \(t=\pi / 4\). Answer: The equation of the tangent line to the curve at the point corresponding to \(t=\pi/4\) is \(y = -x + \sqrt{2}\).

Step 1: Determine the parametric equations of the curve

The given parametric equations of the curve are \(x=\sin t\) and \(y=\cos t\).

Step 2: Find the values of x and y at the given value of \(t\)

To find the coordinates of the point on the curve where \(t=\pi / 4\), plug the given value into the parametric equations: $$x = \sin (\pi/4) = \frac{1}{\sqrt{2}}$$ $$y = \cos (\pi/4) = \frac{1}{\sqrt{2}}$$ So the point on the curve is \((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\).

Step 3: Find the derivatives of x and y with respect to \(t\)

Differentiate the parametric equations with respect to \(t\). This gives us: $$\frac{dx}{dt} = \cos t$$ $$\frac{dy}{dt} = -\sin t$$

Step 4: Evaluate the derivatives at the given value of \(t\) to find the slope of the tangent line

Now, we evaluate the derivatives at \(t=\pi / 4\): $$\frac{dx}{dt}(\pi/4) = \cos (\pi/4) = \frac{1}{\sqrt{2}}$$ $$\frac{dy}{dt}(\pi/4) = -\sin (\pi/4) = -\frac{1}{\sqrt{2}}$$ The slope of the tangent line, \(m\), can be calculated as: $$m = \frac{dy/dt}{dx/dt} = \frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = -1$$

Step 5: Use the point-slope form of a line to write the equation of the tangent line

The point-slope form of a line is given by the equation: $$ y - y_1 = m(x - x_1) $$ Plugging in the point \((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\) and the slope \(m=-1\), we get: $$ y - \frac{1}{\sqrt{2}} = -1\left(x - \frac{1}{\sqrt{2}}\right) $$ Now, to find the equation of the tangent line, we need to simplify this equation. Expand and rearrange the terms: $$ y = -x + \frac{1}{\sqrt{2}}(1 + 1)$$ $$ y = -x + \sqrt{2} $$ So, the equation of the tangent line to the curve at the point corresponding to \(t=\pi/4\) is: $$ y = -x + \sqrt{2} $$

Key Concepts

Parametric Equations

Understanding parametric equations means recognizing that they express a set of quantities as explicit functions of a number of independent parameters. In the problem at hand, we have two parametric equations:

• The equation for the x-coordinate: \( x = \sin t \),
• And the equation for the y-coordinate: \( y = \cos t \).

These equations allow us to find the position on a curve as the parameter \( t \) varies. They offer a different view compared to traditional Cartesian equations because the change in position is related directly to changes in \( t \). In our exercise, the parameter \( t \) represents the angle in radians which helps find specific points such as \( t = \pi/4 \). At this value, substituting into the parametric equations gives precise points on the curve, which are essential for other calculations such as determining the tangent line.

Derivatives

Derivatives are our way of measuring how a function changes as its input changes. In this exercise, we specifically look at derivatives of parametric equations with respect to the parameter \( t \). For our parametric equations:

• The derivative of \( x \) with respect to \( t \) is \( \frac{dx}{dt} = \cos t \),
• and for \( y \) it is \( \frac{dy}{dt} = -\sin t \).

These derivatives provide the rate of change of the coordinates; hence they describe the direction and steepness of the curve at any point \( t \). Evaluating these derivatives at a specific \( t \), like \( \pi/4 \), allows us to find the slope of the tangent line to the curve at that point. Having this slope is crucial for writing the equation of the tangent line using point-slope form.

Point-Slope Form of a Line

To find the equation of the tangent line, we utilize the point-slope form, which is a way to write the equation of a line given a point on the line and its slope. The formula for point-slope form is:\[ y - y_1 = m(x - x_1) \]where \( (x_1, y_1) \) is a point on the line and \( m \) is the slope. In our exercise, the point derived from the parametric equations for \( t = \pi/4 \) is \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \), and the slope calculated is \( m = -1 \). Plugging these into the point-slope form results in:\[ y - \frac{1}{\sqrt{2}} = -1(x - \frac{1}{\sqrt{2}}) \]Simplifying this yields the tangent line's equation: \( y = -x + \sqrt{2} \). This equation gives the relationship between \( x \) and \( y \) on the tangent, describing the line that "just touches" the curve at that particular point and follows its direction.