Problem 2 - Calculus Early Transcendentals

Question

(a) Make a large sketch of the curve described by the vector function \(\mathbf{r}(t)=\left\langle t^{2}, t\right\rangle, 0 \leqslant t \leqslant 2,\) and draw the vectors \(\mathbf{r}(1), \mathbf{r}(1.1),\) and \(\mathbf{r}(1.1)-\mathbf{r}(1)\) (b) Draw the vector \(\mathbf{r}^{\prime}(1)\) starting at \((1,1)\) and compare it with the vector $$ \frac{\mathbf{r}(1.1)-\mathbf{r}(1)}{0.1} $$ Explain why these vectors are so close to each other in length and direction.

Step-by-Step Solution

Answer

The vectors \( \mathbf{r}'(1) \) and \( \frac{\mathbf{r}(1.1)-\mathbf{r}(1)}{0.1} \) are close because the difference quotient approximates the derivative.

Step 1: Sketch the Curve

To sketch the curve described by \( \mathbf{r}(t) = \langle t^2, t \rangle \), we note that for each \( t \), the x-coordinate is \( t^2 \) and the y-coordinate is \( t \). As \( t \) ranges from 0 to 2, we calculate some key points: \( \mathbf{r}(0) = \langle 0, 0 \rangle \), \( \mathbf{r}(1) = \langle 1, 1 \rangle \), and \( \mathbf{r}(2) = \langle 4, 2 \rangle \). Drawing these points and connecting them smoothly gives a parabolic curve, opening to the right.

Step 2: Draw Vectors \( \mathbf{r}(1) \) and \( \mathbf{r}(1.1) \)

Calculate \( \mathbf{r}(1) = \langle 1, 1 \rangle \) and \( \mathbf{r}(1.1) = \langle (1.1)^2, 1.1 \rangle = \langle 1.21, 1.1 \rangle \). Plot these points on the sketch. The vector \( \mathbf{r}(1) \) starts from the origin and ends at \( \langle 1, 1 \rangle \), while \( \mathbf{r}(1.1) \) starts from the origin and ends at \( \langle 1.21, 1.1 \rangle \).

Step 3: Calculate and Draw the Vector \( \mathbf{r}(1.1) - \mathbf{r}(1) \)

The vector \( \mathbf{r}(1.1) - \mathbf{r}(1) = \langle 1.21, 1.1 \rangle - \langle 1, 1 \rangle = \langle 0.21, 0.1 \rangle \) represents the change from \( \mathbf{r}(1) \) to \( \mathbf{r}(1.1) \). Draw this vector starting at \( \langle 1, 1 \rangle \) to \( \langle 1.21, 1.1 \rangle \) on the curve.

Step 4: Differentiate \( \mathbf{r}(t) \) to find \( \mathbf{r}'(t) \)

Differentiate \( \mathbf{r}(t) = \langle t^2, t \rangle \) to find \( \mathbf{r}'(t) = \langle 2t, 1 \rangle \). Evaluating at \( t = 1 \) gives \( \mathbf{r}'(1) = \langle 2, 1 \rangle \). This is the instantaneous rate of change or tangent vector at \( \langle 1, 1 \rangle \).

Step 5: Draw and Compare Vectors \( \mathbf{r}'(1) \) and \( \frac{\mathbf{r}(1.1) - \mathbf{r}(1)}{0.1} \)

We calculate \( \frac{\mathbf{r}(1.1) - \mathbf{r}(1)}{0.1} = \frac{\langle 0.21, 0.1 \rangle}{0.1} = \langle 2.1, 1 \rangle \). Draw \( \mathbf{r}'(1) = \langle 2, 1 \rangle \) starting at the point \( \langle 1, 1 \rangle \). Both vectors \( \mathbf{r}'(1) \) and \( \frac{\mathbf{r}(1.1) - \mathbf{r}(1)}{0.1} \) have similar directions and lengths, reflecting that the smaller the interval over which we measure the change, the closer it tends to the derivative at a point.

Key Concepts

Parametric Curves

In vector calculus, parametric curves allow us to describe curves in a plane or space using parameters. Unlike standard functions where y is expressed directly in terms of x, parametric equations use a separate parameter, typically denoted by \(t\). This parameter runs along the curve, assigning coordinates on the plane to it.

For example, in the vector function \( \mathbf{r}(t)=\left\langle t^2, t \right\rangle \),

• The x-coordinate is represented by \(t^2\)
• The y-coordinate by \(t\).

As \(t\) varies, it traces out a curve. In this case, a parabolic curve is created as t runs from 0 to 2.

Using parametric curves is incredibly beneficial in cases where descriptions are natural in terms of movement or paths, such as in physics for particle trajectories.

Vector Functions

Vector functions play a crucial role in describing parametric curves. A vector function is a function that assigns a vector to each value of \(t\) in its domain. These functions are often represented as \(\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle \) for 3D space or simply \(\mathbf{r}(t) = \langle f(t), g(t) \rangle \) for 2D space.

They allow us to encapsulate complex movement in a system of equations, describing how a point moves over time in a space. In our given example,

• \(f(t) = t^2\) describes the movement in the horizontal direction
• \(g(t) = t\) in the vertical direction.

Thus, vector functions are powerful tools for describing and analyzing the geometry and motion of points and objects.

Derivative of Vector Functions

The derivative of vector functions is analogous to taking the derivative of a traditional function, but instead, it reveals how the curve changes with respect to the parameter \(t\). It indicates the velocity of the point as it moves along the curve.

To find the derivative of the vector function \(\mathbf{r}(t) = \langle t^2, t \rangle \), we differentiate each component with respect to \(t\).

• The x-component's derivative is \(\frac{d}{dt}(t^2) = 2t\)
• The y-component's derivative is \(\frac{d}{dt}(t) = 1\).

Thus, the derivative is \(\mathbf{r}'(t) = \langle 2t, 1 \rangle\).

At \(t = 1\), \(\mathbf{r}'(1) = \langle 2, 1 \rangle\), showing the instantaneous change at that point. Differentiating vector functions helps us identify tangent vectors and understand the dynamics of curves.

Tangents to Curves

Tangents to curves are pivotal in vector calculus because they give information about a curve's direction at a specific point. For a vector function \(\mathbf{r}(t)\), the tangent at a point \(t = a\) is given by the vector \(\mathbf{r}'(a)\).

The tangent vector \(\mathbf{r}'(1) = \langle 2, 1 \rangle\) represents the direction of the curve at \((1, 1)\). To visually interpret this, it can be drawn starting from the point on the curve, demonstrating the curve's slope and direction.

Comparing \(\mathbf{r}'(1)\) with the difference vector \(\frac{\mathbf{r}(1.1) - \mathbf{r}(1)}{0.1}\), students notice that they're quite similar.

• This is because as the interval for \(t\) gets smaller, the difference vector approximates the tangent vector.

This closeness shows how derivatives offer precise insights into the behavior of curves.