Question

Give the equation of the line tangent to the graph of \(\vec{r}(t)\) at the given \(t\) value. $$ \vec{r}(t)=\left\langle t^{2}+t, t^{2}-t\right\rangle \text { at } t=1 $$

Short answer

The tangent line equation is \( y = \frac{1}{3}x - \frac{2}{3} \).

Step-by-step solution

Differentiate the Vector Function

First, we need to find the derivative of the vector function \( \vec{r}(t) = \langle t^2 + t, t^2 - t \rangle \). Differentiating each component with respect to \( t \) gives: \( \vec{r}'(t) = \langle 2t + 1, 2t - 1 \rangle \).

Evaluate the Derivative at the Given Point

Next, evaluate the derivative \( \vec{r}'(t) \) at \( t = 1 \): \( \vec{r}'(1) = \langle 2(1) + 1, 2(1) - 1 \rangle = \langle 3, 1 \rangle \). This vector gives the direction of the tangent line.

Find the Point on the Curve

Calculate the coordinates of the point on the curve at \( t = 1 \) using \( \vec{r}(t) \): \( \vec{r}(1) = \langle 1^2 + 1, 1^2 - 1 \rangle = \langle 2, 0 \rangle \). This is the point of tangency.

Write the Parametric Equations for the Tangent Line

The parametric equations for a line with direction vector \( \vec{d} = \langle a, b \rangle \) and passing through the point \( (x_0, y_0) \) are: \( x = x_0 + at \) and \( y = y_0 + bt \). Substitute \( (2, 0) \) and \( \langle 3, 1 \rangle \) into these equations: \( x = 2 + 3t \), \( y = 0 + 1t \).

Convert Parametric to Cartesian Equation

To express the line in Cartesian form, solve \( x = 2 + 3t \) for \( t \): \( t = \frac{x - 2}{3} \). Substitute \( t \) into \( y = t \) to get: \( y = \frac{x - 2}{3} \). The Cartesian equation of the tangent line is \( y = \frac{1}{3}x - \frac{2}{3} \).

Key concepts

Differentiation of Vector Functions

When dealing with vector functions, differentiation is applied to each individual component of the function. Vector functions are often presented as \( \vec{r}(t) = \langle f(t), g(t) \rangle \), consisting of multiple scalar functions in a vector form. To differentiate a vector function, simply compute the derivative of each scalar function independently. In this exercise, the vector function \( \vec{r}(t) = \langle t^2 + t, t^2 - t \rangle \) is differentiated by finding the derivative of each component:

• The derivative of \( f(t) = t^2 + t \) with respect to \( t \) is \( 2t + 1 \).
• The derivative of \( g(t) = t^2 - t \) with respect to \( t \) is \( 2t - 1 \).

Thus, the derivative of the vector function is \( \vec{r}'(t) = \langle 2t + 1, 2t - 1 \rangle \). This result is essential as it provides the direction of the tangent line at any point \( t \).

Parametric Equations

Parametric equations express a set of related quantities as explicit functions of an independent parameter, usually time \( t \). In reference to our example, the tangent line to the vector function \( \vec{r}(t) \) at \( t = 1 \) can be expressed in parametric form. Here's how the equations are formulated:

• The direction vector \( \vec{d} = \langle 3, 1 \rangle \), derived from \( \vec{r}'(1) \), tells us how the line "moves" in space.
• The point of tangency is \( \vec{r}(1) = \langle 2, 0 \rangle \), which is the starting point of the line.

Thus, using the line equation \( x = x_0 + at \) and \( y = y_0 + bt \), we substitute:\
\( x = 2 + 3t \)
\( y = 0 + t \).
Parametric equations are powerful when modeling scenarios involving movement where position changes with time.

Cartesian Equations

To understand tangent lines in a more traditional format, we transform parametric equations into Cartesian form. This involves eliminating the parameter \( t \) to express one variable solely in terms of the other. Starting with the parametric equations \( x = 2 + 3t \) and \( y = t \), we can solve for \( t \) in terms of \( x \) as follows:

• From \( x = 2 + 3t \), isolate \( t \): \( t = \frac{x - 2}{3} \).

Substitute this expression for \( t \) into the equation for \( y \):
\( y = \frac{x - 2}{3} \).
Thus, the Cartesian form of the tangent line equation is \( y = \frac{1}{3}x - \frac{2}{3} \). This form clearly shows the relationship between \( x \) and \( y \) and is often more intuitive for graphing on a standard coordinate plane.

Vector Calculus

Vector calculus extends basic calculus concepts to vector-valued functions, which are often used to describe physical motions in space. In our exercise, vector calculus enables us to work with paths defined by vector functions like \( \vec{r}(t) \). Key aspects include:

• Differentiation: As illustrated, differentiating a vector function gives us the rate of change of its components, helpful for finding velocities or tangent directions.
• Parametric representations: Vector calculus frequently employs parametric equations, which describe how quantities change concurrently.

Through these foundations, vector calculus allows us to bridge theoretical mathematics with practical applications, such as finding tangent lines or analyzing curves in physics and engineering contexts.