Problem 31 - Calculus Early Transcendentals

Question

The curves \(\mathbf{r}_{1}(t)=\left\langle t, t^{2}, t^{3}\right\rangle\) and \(\mathbf{r}_{2}(t)=\langle\sin t, \sin 2 t, t\rangle\) intersect at the origin. Find their angle of intersection correct to the nearest degree.

Step-by-Step Solution

Answer

The angle of intersection is approximately 66 degrees.

Step 1: Identify Intersection Point

The curves intersect at the origin, given by the point \((0,0,0)\). To verify, substitute \(t=0\) into both parametric equations. For \(\mathbf{r}_1(t)\), substituting \(t=0\) gives \((0, 0^2, 0^3) = (0, 0, 0)\). For \(\mathbf{r}_2(t)\), substituting \(t=0\) gives \((\sin 0, \sin 0, 0) = (0, 0, 0)\). Both curve equations equal zero at \(t=0\), confirming the intersection at the origin.

Step 2: Find Tangent Vectors

Compute the tangent vector for each curve at the intersection point. The tangent vector can be found by differentiating the parametric equations. For \(\mathbf{r}_1(t)\), differentiate to get \(\mathbf{r'}_1(t) = \langle 1, 2t, 3t^2 \rangle\). At \(t=0\), the tangent vector is \(\mathbf{r'}_1(0) = \langle 1, 0, 0 \rangle\).For \(\mathbf{r}_2(t)\), differentiate to get \(\mathbf{r'}_2(t) = \langle \cos t, 2\cos 2t, 1 \rangle\). At \(t=0\), the tangent vector is \(\mathbf{r'}_2(0) = \langle 1, 2, 1 \rangle\).

Step 3: Use the Dot Product to Find the Angle

To find the angle \(\theta\) between the tangent vectors, use the dot product formula: \[ \mathbf{r'}_1(0) \cdot \mathbf{r'}_2(0) = \| \mathbf{r'}_1(0) \| \| \mathbf{r'}_2(0) \| \cos(\theta) \]Calculate the dot product: \[ \langle 1, 0, 0 \rangle \cdot \langle 1, 2, 1 \rangle = 1 \]Calculate the magnitudes: \[ \| \mathbf{r'}_1(0) \| = \sqrt{1^2+0^2+0^2} = 1 \]\[ \| \mathbf{r'}_2(0) \| = \sqrt{1^2+2^2+1^2} = \sqrt{6} \]Substitute these into the dot product formula: \[ 1 = 1 \times \sqrt{6} \times \cos(\theta) \]

Step 4: Calculate the Angle

Solve \(1 = \sqrt{6} \cos(\theta)\) for \(\theta\). Rearrange to find: \[ \cos(\theta) = \frac{1}{\sqrt{6}} \]Find \(\theta\) by taking the arccosine: \[ \theta = \cos^{-1}\left(\frac{1}{\sqrt{6}}\right) \approx 65.9052^\circ \]To the nearest degree, \(\theta \approx 66^\circ\).

Key Concepts

Tangent Vectors

Tangent vectors are crucial when examining the direction a curve takes at a particular point. Imagine you're strolling along a curvy path; the tangent vector would indicate the path's direction at a precise spot. It's essentially the "straight line" that touches the curve at one point and points in the direction you're heading at that location. To find the tangent vector for a curve defined by parametric equations, you differentiate each component of the vector function.For example, consider the curve given by \( \mathbf{r}_{1}(t)=\langle t, t^{2}, t^{3} \rangle \). By taking the derivative with respect to \( t \), we obtain the tangent vector \( \mathbf{r'}_1(t) = \langle 1, 2t, 3t^2 \rangle \). Evaluating it at the intersection point \( t = 0 \) yields \( \mathbf{r'}_1(0) = \langle 1, 0, 0 \rangle \).Similarly, for \( \mathbf{r}_{2}(t)=\langle \sin t, \sin 2t, t \rangle \), the derivative provides the tangent vector \( \mathbf{r'}_2(t) = \langle \cos t, 2\cos 2t, 1 \rangle \). Substituting \( t = 0 \) results in \( \mathbf{r'}_2(0) = \langle 1, 2, 1 \rangle \). These vectors give us the best linear approximation of each curve at the intersection point.

Dot Product

The dot product is an efficient way to discover the angle between two vectors. Imagine you're seeing how closely parallel or perpendicular two directions are. The dot product gives a scalar value that hints at this relationship. The formula for the dot product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]Using this product, you can derive the angle \( \theta \) between the vectors via the relation: \[ \mathbf{a} \cdot \mathbf{b} = \| \mathbf{a} \| \| \mathbf{b} \| \cos(\theta) \]Here, \( \| \mathbf{a} \| \) and \( \| \mathbf{b} \| \) are the magnitudes of the vectors, calculated as \( \sqrt{a_1^2 + a_2^2 + a_3^2} \) and \( \sqrt{b_1^2 + b_2^2 + b_3^2} \) respectively.For our problem, the tangent vectors are \( \mathbf{r'}_1(0) = \langle 1, 0, 0 \rangle \) and \( \mathbf{r'}_2(0) = \langle 1, 2, 1 \rangle \). Their dot product is:\[ 1 \cdot 1 + 0 \cdot 2 + 0 \cdot 1 = 1 \]By solving the equation \( \cos(\theta) = \frac{1}{\sqrt{6}} \), we determine the angle \( \theta \) that gives us the angle of intersection.

Parametric Equations

The beauty of parametric equations is in their flexibility to describe paths that traditional functions might struggle with, such as loops or spirals. Unlike standard equations that express one variable solely in terms of another, parametric equations use a third variable, called the parameter, to define each component of the vector.In the calculated example, the parameter \( t \) helps trace points on the curves \( \mathbf{r}_{1}(t)=\langle t, t^{2}, t^{3} \rangle \) and \( \mathbf{r}_{2}(t)=\langle \sin t, \sin 2t, t \rangle \) by presenting each coordinate position along the curve as a function of \( t \). As \( t \) changes, it draws out the path of the curve in three-dimensional space.Parametric equations are highly adaptable, as they can easily express movements that depend on time or other continuously changing variables. This flexibility is valuable in physics, engineering, and computer graphics, where understanding paths and trajectories is essential. The connection between parametric representation and derivatives make them highly useful in finding tangent vectors, as shown in this exercise when determining angles at intersection points.