Problem 58 - Essential Calculus Early Transcendentals

Question

At what point do the curves \(\mathbf{r}_{1}(t)=\left\langle t, 1-t, 3+t^{2}\right\rangle\) and \(\mathbf{r}_{2}(s)=\left\langle 3-s, s-2, s^{2}\right\rangle\) intersect? Find their angle of intersection correct to the nearest degree.

Step-by-Step Solution

Answer

The curves intersect at \((2, -1, 7)\) with an angle of \(54\) degrees.

Step 1: Equate the Curves

To find the intersection, set the parametric equations equal: \( \langle t, 1-t, 3+t^2 \rangle = \langle 3-s, s-2, s^2 \rangle \). Solve for \( t \) and \( s \) so that the equations hold simultaneously.

Step 2: Solve for t and s

Equate the components: 1. \( t = 3-s \)2. \( 1-t = s-2 \)3. \( 3+t^2 = s^2 \)Solve Two of the simpler equations to find \( t \) and \( s \). From \( t = 3-s \), substitute into the second equation and solve to find \( t = 2 \) and \( s = 1 \).

Step 3: Find the Intersection Point

Substitute \( t = 2 \) into \( \mathbf{r}_1(t) \):\[ \mathbf{r}_1(2) = \langle 2, 1-2, 3+2^2 \rangle = \langle 2, -1, 7 \rangle \].This is the intersection point.

Step 4: Differentiate to Find Tangents

Find the derivative of \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(s) \):- \( \mathbf{r}_1'(t) = \langle \frac{d}{dt}(t), \frac{d}{dt}(1-t), \frac{d}{dt}(3+t^2) \rangle = \langle 1, -1, 2t \rangle \)- \( \mathbf{r}_2'(s) = \langle \frac{d}{ds}(3-s), \frac{d}{ds}(s-2), \frac{d}{ds}(s^2) \rangle = \langle -1, 1, 2s \rangle \)

Step 5: Evaluate Tangents at Intersection

Evaluate the tangents at \( t=2 \) and \( s=1 \):- \( \mathbf{r}_1'(2) = \langle 1, -1, 4 \rangle \)- \( \mathbf{r}_2'(1) = \langle -1, 1, 2 \rangle \)

Step 6: Calculate Dot Product and Magnitudes

Find the dot product of \( \mathbf{r}_1'(2) \) and \( \mathbf{r}_2'(1) \):\[ \mathbf{r}_1'(2) \cdot \mathbf{r}_2'(1) = (1)(-1) + (-1)(1) + (4)(2) = -1 - 1 + 8 = 6 \]Calculate magnitudes:- \( \|\mathbf{r}_1'(2)\| = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{18} \)- \( \|\mathbf{r}_2'(1)\| = \sqrt{(-1)^2 + 1^2 + 2^2} = \sqrt{6} \)

Step 7: Compute the Angle

Use the dot product formula to find the cosine of the angle:\[ \cos \theta = \frac{6}{\sqrt{18} \times \sqrt{6}} = \frac{6}{\sqrt{108}} = \frac{6}{6\sqrt{3}} = \frac{1}{\sqrt{3}} \]Find \( \theta \) using inverse cosine and convert to degrees, \( \theta \approx 54 \) degrees.

Key Concepts

Understanding Parametric Equations

In mathematics, parametric equations define a set of related quantities, expressed as functions of one or more independent variables, known as parameters. These equations are commonly used in scenarios where defining the relationship through a single equation might be complex. For instance, curves in space are often defined using parametric equations.For the given problem, two curves are represented by their parametric equations: \( \mathbf{r}_1(t) = \langle t, 1-t, 3+t^2 \rangle \) and \( \mathbf{r}_2(s) = \langle 3-s, s-2, s^2 \rangle \). Here, \( t \) and \( s \) are the parameters for their respective curves.

• The parametrization \( \mathbf{r}_1(t) \) specifies a trajectory in three-dimensional space by defining \( x, y, \) and \( z \) as functions of the parameter \( t \).
• Similarly, \( \mathbf{r}_2(s) \) defines another trajectory with \( s \) determining the path of the curve.

Setting these parametric equations equal helps in finding where the curves intersect, by solving for \( t \) and \( s \) such that the coordinates match.

Finding and Understanding the Angle of Intersection

The angle of intersection of two curves is the angle at which two tangent vectors meet at the point of intersection. It is an important measure because it quantitatively describes how two curves relate to one another at their intersection.In this exercise, after finding the intersection point where both curves meet, the next task is to determine their angle of intersection. To do this, you need to find the tangent vectors, which are the derivatives of the parametric equations.

• The tangent vector for \( \mathbf{r}_1(t) \) is \( \mathbf{r}_1'(t) = \langle 1, -1, 2t \rangle \).
• The tangent vector for \( \mathbf{r}_2(s) \) is \( \mathbf{r}_2'(s) = \langle -1, 1, 2s \rangle \).

By evaluating these derivatives at \( t=2 \) and \( s=1 \), you obtain the specific tangent vectors at the intersection point. These vectors are then used to calculate the angle of intersection.

Applying the Dot Product for Geometric Analysis

The dot product, also known as scalar product, is a way to quantitatively assess the relationship between two vectors. It is especially useful in finding the angle between vectors.For vectors \( \mathbf{a} \) and \( \mathbf{b} \), the dot product is given by:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]To find the angle between \( \mathbf{r}_1'(2) = \langle 1, -1, 4 \rangle \) and \( \mathbf{r}_2'(1) = \langle -1, 1, 2 \rangle \), compute their dot product. This is followed by finding the magnitudes of each vector:\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]In the exercise, the dot product is calculated as 6, and using the magnitudes, the formula helps determine \( \theta \), the angle between the vectors.

Differentiating Parametric Equations

Differentiation plays a key role in vector calculus, especially when working with parametric equations. The derivative of a parametric function provides the tangent vector, which is essential for understanding the behavior of the curve at any given point.The derivative of a vector function \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) is computed component-wise:

• \( \mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle \)

In the problem, the derivative of each curve is computed as:

• For \( \mathbf{r}_1(t) = \langle t, 1-t, 3+t^2 \rangle \), the derivative is \( \mathbf{r}_1'(t) = \langle 1, -1, 2t \rangle \).
• For \( \mathbf{r}_2(s) = \langle 3-s, s-2, s^2 \rangle \), the derivative is \( \mathbf{r}_2'(s) = \langle -1, 1, 2s \rangle \).

This step is vital for analyzing the tangent vectors and further calculating the angle of intersection.