Question

Find the equation of the line tangent to the curve at the indicated \(t\) -value using the unit tangent vector. $$ \vec{r}(t)=\langle t, \cos t\rangle, \quad t=\pi / 4 $$

Short answer

The tangent line equation is \( x(t) = \pi/4 + \frac{2}{\sqrt{6}}t \) and \( y(t) = \frac{\sqrt{2}}{2} - \frac{1}{\sqrt{6}}t \).

Step-by-step solution

Differentiate the Parametric Equations

To find the tangent vector, we need to differentiate the parametric equations. Given the vector function \( \vec{r}(t) = \langle t, \cos t \rangle \), the derivative is \( \vec{r}'(t) = \langle 1, -\sin t \rangle \). This provides the components of the tangent vector at any \( t \).

Evaluate the Derivative at \( t = \pi/4 \)

Substitute \( t = \pi/4 \) into the derivative \( \vec{r}'(t) = \langle 1, -\sin t \rangle \) to find the tangent vector at this point. This gives us \( \vec{r}'(\pi/4) = \langle 1, -\sin(\pi/4) \rangle = \langle 1, -\frac{\sqrt{2}}{2} \rangle \).

Determine the Magnitude of the Tangent Vector

Calculate the magnitude of the tangent vector \( \vec{r}'(\pi/4) = \langle 1, -\frac{\sqrt{2}}{2} \rangle \). The magnitude is given by \( \sqrt{1^2 + (-\frac{\sqrt{2}}{2})^2} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}} \).

Find the Unit Tangent Vector

Divide each component of \( \vec{r}'(\pi/4) \) by its magnitude to find the unit tangent vector. Thus, the unit tangent vector is \( \left\langle \frac{2}{\sqrt{6}}, -\frac{\sqrt{2}/2}{\sqrt{3/2}} \right\rangle = \left\langle \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right\rangle \).

Find the Point on the Curve at \( t = \pi/4 \)

Evaluate the original vector function at \( t = \pi/4 \) to find the point of tangency. We have \( \vec{r}(\pi/4) = \langle \pi/4, \cos(\pi/4) \rangle = \langle \pi/4, \frac{\sqrt{2}}{2} \rangle \).

Write the Tangent Line Equation

Use the point found in Step 5 and the unit tangent vector from Step 4 to write the line equation in parametric form: \[ \langle x(t), y(t) \rangle = \langle \pi/4, \frac{\sqrt{2}}{2} \rangle + t \cdot \langle \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \rangle \]. This gives the equations: \( x(t) = \pi/4 + \frac{2}{\sqrt{6}}t \) and \( y(t) = \frac{\sqrt{2}}{2} - \frac{1}{\sqrt{6}}t \).

Key concepts

Parametric Equations

Parametric equations are a way of expressing mathematical relationships where each variable is expressed separately as a function of a common parameter, often denoted as "t".
These are especially useful when describing curves in a plane, as they allow us to specify both the x and y coordinates independently. In our exercise, we have a vector function, \( \vec{r}(t) = \langle t, \cos t \rangle \).
Here, \( t \) acts as the parameter that determines the specific point on the curve.These two functions, \( x = t \) and \( y = \cos t \), represent the coordinates of a point on the curve for a specific value of \( t \).

• "x = t" means that the x-coordinate of the curve is exactly equal to \( t \).
• "y = \cos t" determines the y-coordinate from the cosine function value at \( t \).

Parametric equations are powerful tools for working with curves where an explicit function \( y = f(x) \) doesn't neatly exist.

Unit Tangent Vector

The unit tangent vector is a normalized vector pointing in the direction of the curve's tangent at a particular point. It is derived from the tangent vector, providing the direction but with a magnitude of 1.
The process of finding this vector starts with the tangent vector, which is acquired by differentiating the parametric equations, as done in our example where \( \vec{r}'(t) = \langle 1, -\sin t \rangle \).
At the specific interest point, \( t = \pi/4 \), this derivative becomes \( \vec{r}'(\pi/4) = \langle 1, -\frac{\sqrt{2}}{2} \rangle \).To convert this into a unit tangent vector, we calculate the magnitude of the vector and divide each component by this magnitude. This ensures the vector maintains its direction but has a magnitude of exactly 1. The computations yield:

• The magnitude, calculated as \( \sqrt{1 + \left(-\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{3}{2}} \).
• The unit tangent vector is then \( \left\langle \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right\rangle \).

This vector perfectly suits linear approximations of the curve near \( t = \pi/4 \).

Derivative

In calculus, the derivative represents the rate at which a function is changing at any point. For parametric curves, you need to find derivatives for both x and y with respect to the parameter \( t \).
For the function \( \vec{r}(t) = \langle t, \cos t \rangle \), calculating the derivative gives us the tangent vector, \( \vec{r}'(t) = \langle 1, -\sin t \rangle \).
Here, the derivative \( 1 \) corresponds to a constant rate of change for "x = t", and \( -\sin t \) corresponds to the rate of change for the y-coordinates related to \( \cos t \). Derivatives describe how curves proceed through space,
and by evaluating these derivatives at specific \( t \) values, like \( \pi/4 \), it identifies the exact rate of change at that point.In our example,

• \( \vec{r}'(\pi/4) = \langle 1, -\frac{\sqrt{2}}{2} \rangle \)

This derivative is critical in constructing the tangent line's direction.

Magnitude

The magnitude of a vector measures its length and is crucial in determining the normalized, or unit, vector.
Mathematically, it's calculated using the Pythagorean theorem extended for higher dimensions:

• \( \sqrt{x^2 + y^2} \)

In our exercise, for the vector \( \vec{r}'(\pi/4) = \langle 1, -\frac{\sqrt{2}}{2} \rangle \), we find the length to be:
\[ \sqrt{1^2 + \left(-\frac{\sqrt{2}}{2}\right)^2} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}} \].Knowing a vector's magnitude helps in forming the unit tangent vector by dividing each element of the original vector by this length value. This results in:

• A unit vector of \( \left\langle \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right\rangle \)

Hence, magnitude is essential for converting any tangent vector to a standardized length of 1, providing a consistent basis for orientations.