Math Problem Statement

Find the curvature of the curve given by r(t) = ⟨−6sin(t), 2t, 6cos(t)⟩.

Solution

To find the curvature κ(t)\kappa(t) of the curve r⃗(t)=⟨−6sin⁡(t),2t,6cos⁡(t)⟩\vec{r}(t) = \langle -6\sin(t), 2t, 6\cos(t) \rangle, we use the formula:

κ(t)=∥r⃗′(t)×r⃗′′(t)∥∥r⃗′(t)∥3.\kappa(t) = \frac{\|\vec{r}'(t) \times \vec{r}''(t)\|}{\|\vec{r}'(t)\|^3}.

Step 1: Compute r⃗′(t)\vec{r}'(t) (the first derivative)

r⃗(t)=⟨−6sin⁡(t),2t,6cos⁡(t)⟩\vec{r}(t) = \langle -6\sin(t), 2t, 6\cos(t) \rangle r⃗′(t)=⟨−6cos⁡(t),2,−6sin⁡(t)⟩.\vec{r}'(t) = \langle -6\cos(t), 2, -6\sin(t) \rangle.

Step 2: Compute r⃗′′(t)\vec{r}''(t) (the second derivative)

r⃗′′(t)=⟨6sin⁡(t),0,−6cos⁡(t)⟩.\vec{r}''(t) = \langle 6\sin(t), 0, -6\cos(t) \rangle.

Step 3: Compute r⃗′(t)×r⃗′′(t)\vec{r}'(t) \times \vec{r}''(t) (the cross product)

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -6\cos(t) & 2 & -6\sin(t) \\ 6\sin(t) & 0 & -6\cos(t) \end{vmatrix}.$$ Expanding this determinant: $$\vec{r}'(t) \times \vec{r}''(t) = \mathbf{i} \begin{vmatrix} 2 & -6\sin(t) \\ 0 & -6\cos(t) \end{vmatrix} - \mathbf{j} \begin{vmatrix} -6\cos(t) & -6\sin(t) \\ 6\sin(t) & -6\cos(t) \end{vmatrix} + \mathbf{k} \begin{vmatrix} -6\cos(t) & 2 \\ 6\sin(t) & 0 \end{vmatrix}.$$ Simplify each component: - $$\mathbf{i}$$: $$(2)(-6\cos(t)) - (0)(-6\sin(t)) = -12\cos(t)$$. - $$\mathbf{j}$$: $$(-6\cos(t))(-6\cos(t)) - (-6\sin(t))(6\sin(t)) = 36\cos^2(t) + 36\sin^2(t) = 36$$. - $$\mathbf{k}$$: $$(-6\cos(t))(0) - (6\sin(t))(2) = -12\sin(t)$$. Thus: $$\vec{r}'(t) \times \vec{r}''(t) = \langle -12\cos(t), -36, -12\sin(t) \rangle.$$ ### Step 4: Compute $$\|\vec{r}'(t) \times \vec{r}''(t)\|$$ $$\|\vec{r}'(t) \times \vec{r}''(t)\| = \sqrt{(-12\cos(t))^2 + (-36)^2 + (-12\sin(t))^2}.$$ Simplify: $$\|\vec{r}'(t) \times \vec{r}''(t)\| = \sqrt{144\cos^2(t) + 1296 + 144\sin^2(t)}.$$ Using $$\cos^2(t) + \sin^2(t) = 1$$: $$\|\vec{r}'(t) \times \vec{r}''(t)\| = \sqrt{144 + 1296} = \sqrt{1440} = 12\sqrt{10}.$$ ### Step 5: Compute $$\|\vec{r}'(t)\|$$ $$\|\vec{r}'(t)\| = \sqrt{(-6\cos(t))^2 + 2^2 + (-6\sin(t))^2}.$$ $$\|\vec{r}'(t)\| = \sqrt{36\cos^2(t) + 4 + 36\sin^2(t)}.$$ Simplify: $$\|\vec{r}'(t)\| = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}.$$ ### Step 6: Compute $$\kappa(t)$$ $$\kappa(t) = \frac{\|\vec{r}'(t) \times \vec{r}''(t)\|}{\|\vec{r}'(t)\|^3}.$$ Substitute the values: $$\kappa(t) = \frac{12\sqrt{10}}{(2\sqrt{10})^3}.$$ Simplify: $$\kappa(t) = \frac{12\sqrt{10}}{8 \cdot 10 \cdot \sqrt{10}} = \frac{12\sqrt{10}}{80\sqrt{10}} = \frac{12}{80} = \frac{3}{20}.$$ ### Final Answer: $$\kappa(t) = \frac{3}{20}.$$ Would you like additional steps clarified or assistance with related concepts? Here are some related questions: 1. How is curvature used in physics or engineering? 2. What does the magnitude of curvature signify about a curve? 3. How do we compute the unit tangent vector for this curve? 4. Can curvature be used to compute the radius of a circle at a point on the curve? 5. How does the curvature formula change in 2D space? **Tip:** Always check if the curve's parameterization affects the simplifications when solving for curvature!

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Curvature
Cross Product
Derivatives
Vector Norm

Formulas

Curvature: κ(t) = ||r'(t) × r''(t)|| / ||r'(t)||^3
Cross Product determinant expansion for three vectors
Norm of a vector: ||v|| = sqrt(v1^2 + v2^2 + v3^2)

Theorems

Pythagorean identity: cos^2(t) + sin^2(t) = 1
Curvature formula in 3D space

Suitable Grade Level

Undergraduate Calculus

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