Problem 33 - Thomas Calculus

Question

As mentioned in the text, the tangent line to a smooth curve \(\mathbf{r}(t)=f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}\) at \(t=t_{0}\) is the line that passes through the point \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right)\) parallel to \(\mathbf{v}\left(t_{0}\right),\) the curve's velocity vector at \(t_{0} .\) In Exercises \(33-36,\) find parametric equations for the line that is tangent to the given curve at the given parameter value \(t=t_{0}\) . $$ \mathbf{r}(t)=(\sin t) \mathbf{i}+\left(t^{2}-\cos t\right) \mathbf{j}+e^{t} \mathbf{k}, \quad t_{0}=0 $$

Step-by-Step Solution

Answer

The tangent line has parametric equations: \( x(t) = t \), \( y(t) = -1 \), \( z(t) = 1 + t \).

Step 1: Find the Point on the Curve

Evaluate the position vector \( \mathbf{r}(t) = (\sin t) \mathbf{i} + (t^2 - \cos t) \mathbf{j} + e^t \mathbf{k} \) at \( t_0 = 0 \). This gives us the point on the curve at \( t_0 \). \- Compute \( f(t_0) = \sin(0) = 0 \) \- Compute \( g(t_0) = 0^2 - \cos(0) = 0 - 1 = -1 \) \- Compute \( h(t_0) = e^0 = 1 \) Thus, the point on the curve is \((0, -1, 1)\).

Step 2: Find the Velocity Vector \( \mathbf{v}(t) \)

Calculate the velocity vector \( \mathbf{v}(t) \) by taking the derivative of \( \mathbf{r}(t) \) with respect to \( t \).\- Differentiate \( \sin t \) to get \( \cos t \)\- Differentiate \( t^2 - \cos t \) to get \( 2t + \sin t \)\- Differentiate \( e^t \) to get \( e^t \)Thus, \( \mathbf{v}(t) = (\cos t) \mathbf{i} + (2t + \sin t) \mathbf{j} + e^t \mathbf{k} \).

Step 3: Evaluate the Velocity Vector at \( t_0 = 0 \)

Substitute \( t_0 = 0 \) into the velocity vector \( \mathbf{v}(t) \).\- Compute \( \cos(0) = 1 \) \- Compute \( 2(0) + \sin(0) = 0 \)\- Compute \( e^0 = 1 \)Thus, the velocity vector at \( t_0 = 0 \) is \( (1, 0, 1) \).

Step 4: Write the Parametric Equations of the Tangent Line

The tangent line at \( t_0 = 0 \) is given by:\[\begin{align*}x(t) &= f(t_0) + v_1 \cdot t = 0 + 1 \cdot t = t, \y(t) &= g(t_0) + v_2 \cdot t = -1 + 0 \cdot t = -1, \z(t) &= h(t_0) + v_3 \cdot t = 1 + 1 \cdot t = 1+t. \\end{align*}\]Therefore, the parametric equations for the tangent line are \( x(t) = t \), \( y(t) = -1 \), and \( z(t) = 1 + t \).

Key Concepts

Velocity Vector

The concept of a velocity vector is pivotal when dealing with parametric equations and curves. It represents the rate of change of the position vector with respect to time, essentially indicating the direction and speed at which a point moves along a curve.

• For a curve given by the position vector \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \), the velocity vector \( \mathbf{v}(t) \) is derived by differentiating each component of the position vector with respect to \( t \).
• In our exercise, \( \mathbf{v}(t) = (\cos t) \mathbf{i} + (2t + \sin t) \mathbf{j} + e^t \mathbf{k} \). This shows how each part of the curve changes over time.

Calculating the velocity vector involves applying basic differentiation rules to functions like \( \sin t, t^2, \) and \( e^t \). Once we find \( \mathbf{v}(t) \), evaluating it at specific times gives us instantaneous rate of motion, like our evaluated \( \mathbf{v}(0) = (1, 0, 1) \).
Understanding the velocity vector helps in analyzing and predicting the trajectory of moving points on the curve.

Tangent Line

The tangent line is essential for understanding the curve's local behavior at a given point. It's a linear approximation that 'touches' the curve, accurately demonstrating the curve's flow at an instant.

• When you have the velocity vector \( \mathbf{v}(t_0) \), this directly becomes the direction vector for the tangent line at \( t_0 \).
• In our case, the tangent line at \( t_0 = 0 \) is determined using the velocity vector \( (1, 0, 1) \)

To find the parametric equations of the tangent line, start from the point on the curve at \( t_0\), which is \( (0, -1, 1) \) and move in the direction of the velocity vector.
The parametric equations are established as:
\( x(t) = t \), \( y(t) = -1 \), and \( z(t) = 1 + t \).
These equations represent a straight line mirroring the curve's immediate path at that point.

Position Vector

The position vector describes the exact spot on a curve at any given time \( t \). It is a composite of function components, each associated with a different spatial dimension.

• For example, \( \mathbf{r}(t) = (\sin t) \mathbf{i} + (t^2 - \cos t) \mathbf{j} + e^t \mathbf{k} \) shows how the position of a point varies in three-dimensional space.
• The functions describe movement along the X, Y, and Z axes.

Using the position vector allows us to pinpoint the exact location on a curve for any parameter \( t \).
Evaluating at \( t_0 = 0 \), as in our example, gives the point \( (0, -1, 1) \). This provides the foundation for both the velocity vector and tangent line analysis.

Differentiation

Differentiation is a key mathematical process for finding how functions change, and is crucial in calculating both velocity vectors and tangent lines.

• Through differentiation, we derive the rate of change, transforming our position vector into the velocity vector.
• For our involved functions, different rules apply: \( \sin t \) becomes \( \cos t \), \( t^2 \) turns into \( 2t \), and \( \cos t \) transforms into \( -\sin t \).

This procedure is vital for understanding both the instantaneous motion and slope of a curve. At \( t_0 = 0 \), differentiation reveals how the position vector \( \mathbf{r}(t) \) influences \( \mathbf{v}(t) \) and ensures the tangent line accurately reflects the curve's immediate path.
In summary, any analysis of parametric curves heavily relies on differentiation.