Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the curve over the indicated domain for \(t\). Find \(\mathbf{v}, \mathbf{a}, \mathbf{T}\), and \(\kappa\) at the point where \(t=t_{1}\). \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} ; \quad 0 \leq t \leq 2 ; t_{1}=1\)

Step-by-Step Solution

Answer

The vectors are \( \mathbf{v}(1)=\mathbf{i}+2\mathbf{j} \), \( \mathbf{a}(1)=2\mathbf{j} \), \( \mathbf{T}(1)=\frac{1}{\sqrt{5}}(\mathbf{i}+2\mathbf{j}) \), and \( \kappa=\frac{2}{5\sqrt{5}} \).

Step 1: Parametric Equation Analysis

The parametric equation is given as \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \). This represents a curve in the plane with coordinates \( (x, y) = (t, t^2) \).

Step 2: Sketch the Curve

To sketch the curve, note that \( x = t \) and \( y = t^2 \). This shows that as \( t \) varies, \( y \) is simply the square of \( x \). The graph is a parabola opening upwards, starting from \( (0, 0) \) and ending at \( (2, 4) \).

Step 3: Find the Velocity Vector \( \mathbf{v} \)

The velocity vector is found by differentiating \( \mathbf{r}(t) \) with respect to \( t \). Thus, \( \mathbf{v}(t) = \frac{d}{dt}(t \mathbf{i} + t^2 \mathbf{j}) = \mathbf{i} + 2t \mathbf{j} \). At \( t = 1 \), \( \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} \).

Step 4: Find the Acceleration Vector \( \mathbf{a} \)

The acceleration vector is found by differentiating \( \mathbf{v}(t) \) with respect to \( t \). Hence, \( \mathbf{a}(t) = \frac{d}{dt}(\mathbf{i} + 2t \mathbf{j}) = 2 \mathbf{j} \). At \( t = 1 \), \( \mathbf{a}(1) = 2 \mathbf{j} \).

Step 5: Calculate the Magnitude of the Velocity Vector

The magnitude of the velocity vector \( \mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} \) is given by \( \| \mathbf{v}(t) \| = \sqrt{(1)^2 + (2t)^2} = \sqrt{1 + 4t^2} \). At \( t = 1 \), \( \| \mathbf{v}(1) \| = \sqrt{5} \).

Step 6: Find the Tangent Vector \( \mathbf{T} \)

The unit tangent vector \( \mathbf{T} \) is given by \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\| \mathbf{v}(t) \|} \). Thus, at \( t = 1 \), \( \mathbf{T}(1) = \frac{1}{\sqrt{5}} (\mathbf{i} + 2 \mathbf{j}) \).

Step 7: Find the Curvature \( \kappa \)

The curvature \( \kappa \) can be calculated using \( \kappa = \frac{\| \mathbf{v}(t) \times \mathbf{a}(t) \|}{\| \mathbf{v}(t) \|^3} \). Here, \( \mathbf{v}(1) = \mathbf{i} + 2 \mathbf{j} \) and \( \mathbf{a}(1) = 0 \mathbf{i} + 2 \mathbf{j} \). Thus, \( \mathbf{v}(1) \times \mathbf{a}(1) = 2 \). The magnitude of the velocity from step 5 is \( \sqrt{5} \), so the curvature \( \kappa = \frac{2}{(\sqrt{5})^3} = \frac{2}{5\sqrt{5}} \).

Step 8: Final Computations Summary

The velocity vector at \( t = 1 \) is \( \mathbf{i} + 2 \mathbf{j} \), the acceleration vector is \( 2 \mathbf{j} \), the tangent vector is \( \frac{1}{\sqrt{5}}(\mathbf{i} + 2 \mathbf{j}) \), and the curvature is \( \frac{2}{5\sqrt{5}} \).

Key Concepts

Curvature

Curvature reflects how sharply a curve bends at a given point. To determine curvature for a parametric curve, we often use vectors like the velocity and acceleration vectors. These vectors help depict the motion along the curve. The curvature can be calculated using the formula:\[\kappa(t) = \frac{\| \mathbf{v}(t) \times \mathbf{a}(t) \|}{\| \mathbf{v}(t) \|^3}\]where \(\mathbf{v}(t)\) is the velocity vector, and \(\mathbf{a}(t)\) is the acceleration vector. The cross product \( \mathbf{v}(t) \times \mathbf{a}(t) \) gives us a vector perpendicular to both velocity and acceleration, and its magnitude provides insight into the twisting of the curve at the point considered. In our example, curvature at \(t = 1\) was found to be \( \frac{2}{5\sqrt{5}} \), indicating a specific rate of change of direction of the curve, dependent on both speed and change of speed.

Tangent Vector

The tangent vector of a parametric curve gives a sense of direction at any point on the curve. For a parametric curve \(\mathbf{r}(t)\), the tangent vector \(\mathbf{T}(t)\) is crucial to understand how the curve is oriented at each point. To find it, we use the velocity vector, which is the derivative of \(\mathbf{r}(t)\):\[\mathbf{v}(t) = \frac{d\mathbf{r}}{dt}\]The unit tangent vector \(\mathbf{T}(t)\) is derived from normalizing the velocity vector:\[\mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|}\]This unit vector provides direction but not magnitude, making it useful for determining orientation across different points without regard to speed. For instance, at \(t=1\), the unit tangent vector helps define the precise path direction of our parabola while controlling for how fast the point moves along the curve.

Velocity and Acceleration Vectors

Velocity and acceleration vectors provide vital information about the motion of a point along a parametric curve.

• The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)\), representing the instantaneous rate of change of position. It gives both the speed and direction.
• Acceleration vector \(\mathbf{a}(t)\), the derivative of the velocity vector, gives the rate of change of velocity, showing how quickly something is speeding up or slowing down along the curve.

For our example, the velocity at \(t = 1\) was \(\mathbf{i} + 2\mathbf{j}\), meaning that the point is moving rightward and upward at equal rates. The acceleration \(2\mathbf{j}\) indicates that the motion primarily affects the vertical component, suggesting an upward acceleration along the parabola.

Parametric Curves

Parametric curves offer a flexible way to describe geometric figures and motion paths. These curves use parameters—often time \(t\)—to express coordinates in terms of equations like \( x(t) \) and \( y(t) \). By adjusting \(t\), the curve can depict complex paths beyond simple linear or angular motion.In our example, the parametric equations \(x(t) = t\) and \(y(t) = t^2\) describe a parabola. This parameterization allows us to explore the curve's properties, such as tangent and normal vectors, velocity and acceleration, all in relation to the chosen parameter \(t\). The power of parametric curves lies in their ability to represent intricate paths and facilitate calculations concerning motion and geometry on those paths.