Question

Consider the curve $$ \mathbf{r}(t)=2 t \mathbf{i}+\sqrt{7 t} \mathbf{j}+\sqrt{9-7 t-4 t^{2}} \mathbf{k}, 0 \leq t \leq \frac{1}{2} $$ (a) Show that the curve lies on a sphere centered at the origin. (b) Where does the tangent line at \(t=\frac{1}{4}\) intersect the \(x z\) -plane?

Short answer

The curve lies on a sphere with radius 3 centered at the origin. The tangent line at \(t=\frac{1}{4}\) intersects the \(xz\)-plane at \((1, 0, -\frac{7}{4})\).

Step-by-step solution

Find the equation of the sphere

To show the curve lies on a sphere, we need to confirm that the position vector \(\mathbf{r}(t)\) satisfies \(\mathbf{r}(t) \cdot \mathbf{r}(t) = R^2\) for some constant radius \(R\). First, calculate the magnitude squared of \(\mathbf{r}(t)\):\[|\mathbf{r}(t)|^2 = (2t)^2 + (\sqrt{7t})^2 + (\sqrt{9-7t-4t^2})^2\]Simplify this expression:\[|\mathbf{r}(t)|^2 = 4t^2 + 7t + (9-7t-4t^2) = 9\]This implies that \(|\mathbf{r}(t)|^2 = 9\), i.e., \(|\mathbf{r}(t)| = 3\), which means every point on the curve is 3 units away from the origin. Therefore, the curve lies on a sphere with radius 3 centered at the origin.

Compute the derivative of the curve

To find the tangent line at \(t = \frac{1}{4}\), compute the derivative of the position vector \(\mathbf{r}(t)\) with respect to \(t\):\[\mathbf{r}'(t) = \frac{d}{dt}(2t)\mathbf{i} + \frac{d}{dt}(\sqrt{7t})\mathbf{j} + \frac{d}{dt}(\sqrt{9-7t-4t^2})\mathbf{k}\]Calculate each component:- First component: \(\frac{d}{dt}(2t) = 2\)- Second component: \(\frac{d}{dt}(\sqrt{7t}) = \frac{1}{2}\cdot\frac{7}{\sqrt{7t}} = \frac{7}{2\sqrt{7t}}\)- Third component: \(\frac{d}{dt}(\sqrt{9-7t-4t^2}) = \frac{1}{2}\cdot\frac{-7-8t}{\sqrt{9-7t-4t^2}}\)Thus, \(\mathbf{r}'(t) = 2\mathbf{i} + \frac{7}{2\sqrt{7t}}\mathbf{j} - \frac{7+8t}{2\sqrt{9-7t-4t^2}}\mathbf{k}\).

Key concepts

Curve on a Sphere

A curve on a sphere represents a path traced by a point that maintains a constant distance from the center of the sphere. In our specific case, the curve is defined by the position vector \( \mathbf{r}(t) = 2 t \mathbf{i} + \sqrt{7 t} \mathbf{j} + \sqrt{9 - 7 t - 4 t^{2}} \mathbf{k} \), and it lies on a sphere centered at the origin. To determine this, we need to ensure that the distance of any point on the curve from the origin remains constant. This is done by verifying that the square of the magnitude of \( \mathbf{r}(t) \) is a constant.Our calculations show that \(|\mathbf{r}(t)|^2 = 4t^2 + 7t + (9-7t-4t^2) = 9\), which implies that the radius \( R \) of the sphere is 3. This means every point on our curve is exactly 3 units away from the origin, confirming that the curve is indeed on the surface of a sphere with radius 3.

Tangent Line

The tangent line to a curve at a given point is a straight line that touches the curve at the point without crossing it. It can be visualized as the curve's direction at that point. For the curve defined by \( \mathbf{r}(t) \), the tangent line at \( t = \frac{1}{4} \) is found by computing the derivative \( \mathbf{r}'(t) \). This gives us the velocity of a particle moving along the curve.The derivative \( \mathbf{r}'(t) = 2\mathbf{i} + \frac{7}{2\sqrt{7t}}\mathbf{j} - \frac{7+8t}{2\sqrt{9-7t-4t^2}}\mathbf{k} \) at \( t = \frac{1}{4} \) will yield the direction vectors for the tangent line. These derivative components help in setting up the equation of the tangent line, which can be extended to find where it intersects the xy-plane or any other required plane.

Position Vector

A position vector describes the position of a point in space relative to an origin. In vector calculus, it provides essential information about the location and movement along a curve. In our exercise, the position vector is \( \mathbf{r}(t) = 2 t \mathbf{i} + \sqrt{7 t} \mathbf{j} + \sqrt{9 - 7 t - 4 t^{2}} \mathbf{k} \).This vector tells us exactly where the point \( P \) is located on the curve for a particular value of \( t \). Each term in the position vector specifies how far along the x, y, and z directions the point is located from the origin. Understanding the position vector helps visualize how the curve moves through three-dimensional space, providing insight into the nature and shape of the curve.

Magnitude of a Vector

The magnitude of a vector represents the length or size of the vector. It is crucial in understanding how far a point is from the origin when described by a position vector. For any vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \), its magnitude is given by \( \|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2} \).For the curve \( \mathbf{r}(t) = 2 t \mathbf{i} + \sqrt{7 t} \mathbf{j} + \sqrt{9-7 t -4t^2} \mathbf{k} \), the magnitude is calculated as the square root of \((2t)^2 + (\sqrt{7t})^2 + (\sqrt{9-7t-4t^2})^2\). This simplifies to a constant value of 9, which means that \( \|\mathbf{r}(t)\| = 3 \). A constant magnitude across all values of \( t \) confirms that the entire curve remains on a sphere with a consistent radius throughout its length.