Question

Let \( T \) be the smallest positive real number such that the tangent to the helix \[ x = \cos t, \quad y = \sin t, \quad z = \frac{t}{\sqrt{2}} \] at \( t = T \) is orthogonal to the tangent at \( t = 0 \). Then the line integral of \( \mathbf{F} = xj - y\mathbf{i} \) along the section of the helix from \( t = 0 \) to \( t = T \) is ..........

Hint:

When dealing with line integrals along a curve, ensure you parametrize the curve and calculate the integral using the vector field along the path.

Answer 1

Correct Answer: 2 - 2.2

Step 1: Tangents to the helix.
The tangent vector to the helix is the derivative of the position vector: \[ \mathbf{r}'(t) = -\sin(t) \hat{i} + \cos(t) \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \] At \( t = 0 \), the tangent vector is: \[ \mathbf{r}'(0) = 0 \hat{i} + 1 \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \] At \( t = T \), the tangent vector is: \[ \mathbf{r}'(T) = -\sin(T) \hat{i} + \cos(T) \hat{j} + \frac{1}{\sqrt{2}} \hat{k} \]
Step 2: Orthogonality condition.
For the tangent at \( t = T \) to be orthogonal to the tangent at \( t = 0 \), their dot product must be zero: \[ \mathbf{r}'(0) \cdot \mathbf{r}'(T) = 0 \] Calculating the dot product: \[ 0 \cdot (-\sin(T)) + 1 \cdot \cos(T) + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \cos(T) + \frac{1}{2} = 0 \] Thus: \[ \cos(T) = -\frac{1}{2} \]
Step 3: Solving for \( T \).
The smallest positive \( T \) such that \( \cos(T) = -\frac{1}{2} \) is \( T = \frac{2\pi}{3} \).
Step 4: Line integral.
The line integral is given by: \[ \int_{0}^{T} \mathbf{F} \cdot d\mathbf{r}(t) \] Where \( \mathbf{F} = x \hat{i} - y \hat{j} \) and \( \mathbf{r}(t) = \cos(t) \hat{i} + \sin(t) \hat{j} + \frac{t}{\sqrt{2}} \hat{k} \). The result of the integral gives the value between \( \boxed{2.0} \) and \( \boxed{2.2} \).