Question

Can someone help me out with this?

Answer 1

Not very sure but here’s my attempt at part a. Tell me if you still need further help with b.

Answer 2

(a)

`T(t) = \Big < (\sqrt2)/(2\cosh(2t)), \ (e^(2t))/(2\cosh(2t)), \ -(e^(-2t))/(2\cosh(2t)) \Big >`

`N(t) = \Big < -\tanh(2t), \ (√(2))/(2\cosh(2t)), \ (√(2))/(2\cosh(2t)) \Big >`

(b)

`\kappa (t) = (√(2) )/(4\cosh^2(2t))`

Explanation:

Part (a): Finding the unit tangent and unit normal vectors T(t) and N(t):

To find the unit tangent vector, T(t), we first need to compute the derivative of the vector function r(t) with respect to t:

`r(t) = \big < 2t√(2), \ e^(2t), \ e^(-2t) \big > \\\\\\\\\Longrightarrow r'(t) = \big < (d)/(dt)[2t√(2)], \ (d)/(dt)[e^(2t)], \ (d)/(dt)[e^(-2t)]\big > \\\\\\\\\Longrightarrow \boxed{r'(t)= \big < 2√(2), \ 2e^(2t), \ -2e^(-2t)\big > }`

Next, we find the magnitude (or length) of the vector r'(t) by taking the square root of the sum of the squares of its components:

`\Longrightarrow \big|\big| r'(t)\big|\big| = \sqrt{(2√(2))^2+(2e^(2t))^2+(-2e^(-2t))^2} \\\\\\\\\Longrightarrow \big|\big| r'(t)\big|\big| = \sqrt{8+4e^(4t)+4e^(-4t)} \\\\\\\\\Longrightarrow \big|\big| r'(t)\big|\big| = √(8+8\cosh(4t)) \\\\\\\\\Longrightarrow \boxed\big`

Now, we can calculate the unit tangent vector T(t) by dividing the vector r'(t) by its magnitude:

`T(t) = (r'(t))/(\big|\big|r'(t)\big|\big|) \\\\\\\\\Longrightarrow T(t) = (\big < 2√(2), \ 2e^(2t), \ -2e^(-2t)\big > )/(4\cosh(2t))\\\\\\\\\Longrightarrow T(t) = \Big < (2√(2))/(4\cosh(2t)), \ (2e^(2t))/(4\cosh(2t)), \ (-2e^(-2t))/(4\cosh(2t))\Big > \\\\\\\\\Longrightarrow T(t) = \Big < (√(2))/(2\cosh(2t)), \ (e^(2t))/(2\cosh(2t)), \ -(e^(-2t))/(2\cosh(2t))\Big > \\\\\\\\`

`\therefore \boxed{\boxed{T(t) = \Big < (\sqrt2)/(2\cosh(2t)), \ (e^(2t))/(2\cosh(2t)), \ -(e^(-2t))/(2\cosh(2t)) \Big > }} \\\\\\`

To find the unit normal vector N(t), we differentiate the unit tangent vector T(t) with respect to t:

`T(t) = \Big < (\sqrt2)/(2\cosh(2t)), \ (e^(2t))/(2\cosh(2t)), \ -(e^(-2t))/(2\cosh(2t)) \\\\\\\\\Longrightarrow T'(t) = \Big < (d)/(dt) \Big[(√(2))/(2\cosh(2t))\Big], \ (d)/(dt)\Big[(e^(2t))/(2\cosh(2t))\Big], \ (d)/(dt) \Big[-(e^(-2t))/(2\cosh(2t))\Big]\Big >`

Finding T'_x(t):

`\Longrightarrow T_x'(t) = ((2\cosh(2t))(0)-(√(2) )(4\sinh(2t)))/((2\cosh(2t))^2)\\\\\\\\\Longrightarrow T_x'(t) = (-4√(2)\sinh(2t))/(4\cosh^2(2t))\\\\\\\\\Longrightarrow \boxed{T_x'(t) = -(√(2)\sinh(2t))/(\cosh^2(2t))}`

Finding T'_y(t):

`\Longrightarrow T_y'(t) = ((2\cosh(2t))(2e^(2t))-(e^(2t) )(4\sinh(2t)))/((2\cosh(2t))^2)\\\\\\\\\Longrightarrow T_y'(t) = (4e^(2t)\cosh(2t)- 4e^(2t)\sinh(2t))/(4\cosh^2(2t))\\\\\\\\\Longrightarrow T_y'(t) = (e^(2t)\cosh(2t)- e^(2t)\sinh(2t))/(\cosh^2(2t))\\\\\\\\\Longrightarrow \boxed{T_y'(t)=(1)/(\cosh^2(2t)) }`

Finding T'_z(t):

`\Longrightarrow T_z'(t) = -((2\cosh(2t))(-2e^(-2t))-(e^(-2t) )(4\sinh(2t)))/((2\cosh(2t))^2)\\\\\\\\\Longrightarrow T_z'(t) = (4e^(-2t)\cosh(2t)+4e^(-2t)\sinh(2t))/(4\cosh^2(2t))\\\\\\\\\Longrightarrow T_z'(t) = (e^(-2t)\cosh(2t)+e^(-2t)\sinh(2t))/(\cosh^2(2t))\\\\\\\\\Longrightarrow \boxed{T_z'(t)=(1)/(\cosh^2(2t)) }`

`\text{Thus}, \ \boxed{T'(t) = \Big < -(√(2)\sinh(2t))/(\cosh^2(2t)), \ (1)/(\cosh^2(2t)), \ (1)/(\cosh^2(2t)) \Big > }`

Again, we calculate the magnitude of the vector T'(t):

`\Longrightarrow \big|\big| T'(t)\big|\big| = \sqrt{\Big(\frac{√(2)\sinh(2t)}{\cosh^2{2t}} \Big)^2+\Big((1)/(\cosh^2(2t)) \Big)^2+\Big((1)/(\cosh^2(2t))\Big)^2}\\\\\\\\\Longrightarrow \boxed = (√(2))/(\cosh(2t))`

Finally, the unit normal vector N(t) is given by dividing T'(t) by its magnitude:

`N(t) = (T'(t))/(\big|\big|T'(t)\big|\big|) \\\\\\\\\Longrightarrow N(t) = ( \Big < -(√(2)\sinh(2t))/(\cosh^2(2t)), \ (1)/(\cosh^2(2t)), \ (1)/(\cosh^2(2t)) \Big > )/((√(2))/(\cosh(2t)) )`

`\therefore \boxed{\boxed{N(t) = \Big < -\tanh(2t), \ (√(2))/(2\cosh(2t)), \ (√(2))/(2\cosh(2t)) \Big > }}`

`\hrulefill`

Part (b): Finding the curvature using the formula κ(t) = ||T'(t)|| / ||r'(t)||:

The curvature of a vector function r(t) is given by the formula:

`\kappa (t) = (\big|\big|T'(t)\big|\big|)/(\big|\big|r'(t)\big|\big|)`

We have already calculated ||T'(t)|| and ||r'(t)|| in part (a):

`\bullet \ \big|\big| r'(t)\big|\big| = 4\cosh(2t)\\\\\\\bullet \ \big|\big| T'(t)\big|\big| = (√(2))/(\cosh(2t))`

Now, we can plug these values into the curvature formula:

`\kappa (t) = ( (√(2))/(\cosh(2t)))/(4\cosh(2t))\\\\\\\\\therefore \boxed{\boxed{\kappa (t) = (√(2) )/(4\cosh^2(2t)) }}`