Question

Consider the curve \(\mathbf{r}(t)=\sin t \cos t \mathbf{i}+\sin ^{2} t \mathbf{j}+\cos t \mathbf{k}\), \(0 \leq t \leq 2 \pi\) (a) Show that the curve lies on a sphere centered at the origin. (b) Where does the tangent line at \(t=\pi / 6\) intersect the \(x y\) -plane?

Short answer

(a) The curve lies on a sphere of radius 1 centered at the origin. (b) The tangent line intersects the xy-plane at \( (\frac{1}{4} + \frac{\sqrt{3}}{2}, \frac{7}{4}, 0) \).\)

Step-by-step solution

Identify the Parametric Equation

The curve is given by the parametric vector function \( \mathbf{r}(t) = \sin t \cos t \mathbf{i} + \sin^2 t \mathbf{j} + \cos t \mathbf{k} \).

Check Sphere Condition

To show that the curve lies on a sphere centered at the origin, compute the magnitude of \( \mathbf{r}(t) \) and verify that it is constant:\[\left| \mathbf{r}(t) \right| = \sqrt{(\sin t \cos t)^2 + (\sin^2 t)^2 + (\cos t)^2}\]Simplify:\[ (\sin t \cos t)^2 = \sin^2 t \cos^2 t \ (\sin^2 t)^2 = \sin^4 t \ (\cos t)^2 = \cos^2 t \\Rightarrow \sin^2 t \cos^2 t + \sin^4 t + \cos^2 t = \sin^2 t (\cos^2 t + \sin^2 t) + \cos^2 t = \sin^2 t + \cos^2 t\]Since \( \sin^2 t + \cos^2 t = 1 \), we have \( \left| \mathbf{r}(t) \right| = 1 \), confirming the curve lies on a sphere of radius 1 centered at the origin.

Find the Tangent Vector

To find where the tangent line at \( t = \frac{\pi}{6} \) intersects the \( xy \)-plane, first compute the derivative \( \mathbf{r}'(t) \):\[\mathbf{r}'(t) = \frac{d}{dt}(\sin t \cos t) \mathbf{i} + \frac{d}{dt}(\sin^2 t) \mathbf{j} + \frac{d}{dt}(\cos t) \mathbf{k} \= (\cos^2 t - \sin^2 t) \mathbf{i} + 2\sin t \cos t \mathbf{j} - \sin t \mathbf{k}\]Evaluate \( \mathbf{r}'\left( \frac{\pi}{6} \right) \) to find the direction vector:\[ \cos^2\left( \frac{\pi}{6} \right) - \sin^2\left( \frac{\pi}{6} \right) = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} \ 2 \sin\left( \frac{\pi}{6} \right) \cos\left( \frac{\pi}{6} \right) = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \ \sin\left( \frac{\pi}{6} \right) = \frac{1}{2} \\Rightarrow \mathbf{r}'\left( \frac{\pi}{6} \right) = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} - \frac{1}{2} \mathbf{k}\]

Set Up Equation for Tangent Line

The tangent line at \( t = \frac{\pi}{6} \) is given by the parametric equations:\[\mathbf{r}\left( \frac{\pi}{6} \right) + \lambda \mathbf{r}'\left( \frac{\pi}{6} \right) \= (\frac{1}{4} \mathbf{i} + \frac{1}{4} \mathbf{j} + \frac{\sqrt{3}}{2} \mathbf{k}) + \lambda (\frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} - \frac{1}{2} \mathbf{k})\]\(= \left(\frac{1}{4} + \lambda \frac{1}{2}\right) \mathbf{i} + \left(\frac{1}{4} + \lambda \frac{\sqrt{3}}{2}\right) \mathbf{j} + \left(\frac{\sqrt{3}}{2} - \lambda \frac{1}{2}\right) \mathbf{k}\) To find the intersection with the \( xy \)-plane, set the \( z\)-component to 0 and solve for \( \lambda \):\[ \frac{\sqrt{3}}{2} - \lambda \frac{1}{2} = 0 \ \frac{\sqrt{3}}{2} = \lambda \frac{1}{2} \ \lambda = \sqrt{3}\]

Calculate Intersection Coordinates

Substitute \( \lambda = \sqrt{3} \) into the parametric equations of the tangent line to find the point of intersection:\[ x = \frac{1}{4} + \sqrt{3} \times \frac{1}{2} = \frac{1}{4} + \frac{\sqrt{3}}{2} \ y = \frac{1}{4} + \sqrt{3} \times \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{2} = \frac{7}{4}\]Thus, the coordinates of the intersection with the \( xy \)-plane are \( \left( \frac{1}{4} + \frac{\sqrt{3}}{2}, \frac{7}{4}, 0 \right) \).

Key concepts

Tangent Lines

Tangent lines are fundamental in calculus and geometry. They represent a line that just "touches" a curve at a given point and indicates its direction at that point.
Tangent lines are essential for understanding how a curve behaves in its vicinity. For any curve described by a parametric equation \(\mathbf{r}(t)\), the tangent at a specific point is determined by calculating the derivative \(\mathbf{r}'(t)\) at that point. This derivative, also known as the tangent vector, provides the direction of the tangent line.
Consider the curve described by \(\mathbf{r}(t) = \sin t \cos t \mathbf{i} + \sin^2 t \mathbf{j} + \cos t \mathbf{k}\). To find the tangent line at \(t = \frac{\pi}{6}\), we compute the derivative \(\mathbf{r}'(t)\) and evaluate it at \(t = \frac{\pi}{6}\) to obtain the direction of the tangent. The result, \(\mathbf{r}'\left( \frac{\pi}{6} \right) = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} - \frac{1}{2} \mathbf{k}\), gives the slope of the tangent line in three dimensions.
The tangent line is then tracked along this direction, and its equation is set up by using the point on the curve at \(t = \frac{\pi}{6}\) and adding the tangent vector scaled by a parameter \(\lambda\). This parameterizes all points along the tangent line.

Spheres

A sphere in three-dimensional space is a set of points that are equidistant from a fixed point, known as the center. It's similar to a circle in 2D but extended into 3D. The radius of the sphere is the constant distance from the center to any point on its surface.
To prove that a curve lies on a sphere, we need to show that the distance from each point on the curve to the center of the sphere is constant. In our example, the curve \(\mathbf{r}(t) = \sin t \cos t \mathbf{i} + \sin^2 t \mathbf{j} + \cos t \mathbf{k}\) is considered. The center of this sphere is at the origin (0, 0, 0) and its radius is intended to be 1.
We calculate the magnitude \(\left| \mathbf{r}(t) \right|\) and find it to be \(\sqrt{\left(\sin t \cos t\right)^2 + \left(\sin^2 t\right)^2 + \left(\cos t\right)^2}\). Upon simplifying, we see this results in \(\sin^2 t + \cos^2 t\), which equals 1 for any angle \(t\). This confirms that every point on the curve maintains a constant distance of 1 from the origin, thus demonstrating the curve indeed resides on the sphere.

Vector Calculus

Vector calculus is a field of mathematics concentrating on vector functions, which are essential for analyzing curves and surfaces in space. It extends differentiation and integration to vector-valued functions.
In our analysis, we use vector calculus to understand the curve described by the parametric equation \(\mathbf{r}(t) = \sin t \cos t \mathbf{i} + \sin^2 t \mathbf{j} + \cos t \mathbf{k}\). The derivative \(\mathbf{r}'(t)\) provides a vector indicating the curve's rate of change and direction at any point. This derivative plays a crucial role in finding tangent lines, as previously discussed.
Vector calculus also aids in examining how the curve interacts with surfaces, such as the \(xy\)-plane. By setting the \(z\)-component of the tangent line equation to zero, we can find where the tangent line intersects the plane. This intersection involves solving linear equations that arise from substituting parameters into the equations derived from the vector components.
Such calculations and analyses involve fundamental concepts from vector calculus, which are invaluable for working with geometrical shapes and understanding spatial relationships in mathematics.