Question

Find the radius of curvature at the indicated value. \(\vec{r}(t)=\langle\cos t, \sin (3 t)\rangle,\) at \(t=0\)

Short answer

The radius of curvature at \(t=0\) is 9.

Step-by-step solution

Find the first derivative of the vector function

Given the vector function \(\vec{r}(t) = \langle \cos t, \sin(3t) \rangle\), find the first derivative \(\vec{r}'(t)\). Calculate the derivatives of the component functions:- \(\frac{d}{dt}(\cos t) = -\sin t\)- \(\frac{d}{dt}(\sin(3t)) = 3\cos(3t)\)Therefore, \(\vec{r}'(t) = \langle -\sin t, 3\cos(3t) \rangle\).

Find the second derivative of the vector function

Now find the second derivative \(\vec{r}''(t)\).Calculate the derivatives of the first derivative components:- \(\frac{d}{dt}(-\sin t) = -\cos t\)- \(\frac{d}{dt}(3\cos(3t)) = -9\sin(3t)\)Thus, \(\vec{r}''(t) = \langle -\cos t, -9\sin(3t) \rangle\).

Calculate the magnitude of the first derivative

Find the magnitude of \(\vec{r}'(t)\):\[ \|\vec{r}'(t)\| = \sqrt{(-\sin t)^2 + (3\cos(3t))^2} = \sqrt{\sin^2 t + 9\cos^2(3t)} \].At \(t = 0\), this becomes:\[ \|\vec{r}'(0)\| = \sqrt{\sin^2(0) + 9\cos^2(0)} = \sqrt{0 + 9(1)} = 3 \].

Compute the cross product of derivatives

For 2D vector functions, the "cross product" simplifies to the determinant involving derivatives:\[ \vec{r}'(t) \times \vec{r}''(t) = -\sin t \, (-9\sin(3t)) - (3\cos(3t))(-\cos t) = 9\sin t \sin(3t) + 3\cos(3t)\cos t \].At \(t = 0\),\[ \vec{r}'(0) \times \vec{r}''(0) = 9(0)\sin(0) + 3(1)(1) = 3 \].

Calculate the radius of curvature

The formula for the radius of curvature \(\rho\) is given by\[ \rho = \frac{\|\vec{r}'(t)\|^3}{|\vec{r}'(t) \times \vec{r}''(t)|} \].Substitute \(t = 0\):\[ \rho = \frac{3^3}{|3|} = \frac{27}{3} = 9 \].

Conclusion: Final radius of curvature

Hence, the radius of curvature of the curve at \(t = 0\) is \(9\).

Key concepts

Understanding Vector Calculus

Vector calculus is a fundamental branch of mathematics concerned with differentiation and integration of vector fields. It combines the core concepts of calculus with vector algebra, enabling the exploration of a multitude of directions. In problems like finding the radius of curvature, vector calculus allows us to understand how a curve behaves and changes within a given space.

• Vectors bring direction and magnitude into play, essential for modeling real-world phenomena.
• Through calculus, these vectors can be differentiated or integrated for deeper insight.

By differentiating vector functions, we delve into how vectors change over time or along a path, laying the groundwork for analyzing complex motions and shapes.

The Role of the First Derivative

When working with vector functions like \(ackslashvec{r}(t) = \langle \cos t, \sin(3t) \rangle\), the first derivative provides crucial information about the tangent to the curve. Differentiating each component of the vector helps you understand how fast and in what direction the curve is speeding or slowing at any point.

• The first derivative of a function is its rate of change or slope.
• In vector calculus, it represents the tangent vector to the curve at point \( t \).

For instance, the first derivative \(\vec{r}'(t) = \langle -\sin t, 3\cos(3t) \rangle\) tells us how the \(ackslashcos t\) and \(\sin(3t)\) parts of the curve are altering, providing a basis on which further calculations, like curvatures, depend.

Exploring the Second Derivative

The second derivative extends the journey into understanding the curvature of a path described by a vector function. It tells us how the rates themselves are changing, essentially addressing the curve's concavity or convexity. Calculating the second derivative of a vector function allows us to determine the acceleration or the 'bending' of the curve.

• The second derivative provides information about the change in the rate of slope or curvature.
• It is vital in computing the radius of curvature, which measures how sharply a curve bends at a specific point.

In our working example, \(\vec{r}''(t) = \langle -\cos t, -9\sin(3t) \rangle\) presents the acceleration of the curve — a key factor in assessing how the curvature evolves over time.

Magnitude of a Vector and Its Importance

In vector calculus, the magnitude of a vector is crucial as it denotes the length or size of a vector. It simplifies evaluating the stretch or compression a vector undergoes. The magnitude of a vector function's derivative also helps determine the velocity along the curve.

• Magnitude is equivalent to the length of a vector in Euclidean space.
• Understanding magnitude aids in assessing the speed of change of the vector components.

Calculating the magnitude of the first derivative \(ackslash|\vec{r}'(t)\|\) connects directly to how 'fast' the curve is changing its position. For example, evaluating \(ackslash|\vec{r}'(0)\| = 3\) tells us about the velocity at \( t = 0 \), a vital element in computing the radius of curvature precisely.