Question

Can someon help me with this?

Answer 1

(a)

`\vec T(2)= \Big < (4)/(9), \ (8)/(9), \ (1)/(9) \Big >`

(b)

`\vec N(2) = \Big < (2√(2) )/(3), \ (√(2) )/(6), \ -(√(2) )/(6) \Big >`

(c)

`\vec B(2) = \big < 0, \ 0, \ 0\big >`

Explanation:

To find the vectors T, N, and B at the given point (4, 16/3, 2), we can use the following formulas for the unit tangent vector (T), principal normal vector (N), and binormal vector (B):

Unit Tangent Vector (T):

The unit tangent vector T is the derivative of the position vector r(t) with respect to t, normalized to have a magnitude of 1. It gives us the direction of the curve at the given point.

`\rightarrow \vec T(t)= ( r'(t))/(\big|\big| r'(t)\big|\big|) = (v(t))/(\big|\big|v(t)\big|\big|)`

Principal Normal Vector (N):

The principal normal vector N is the derivative of the unit tangent vector T with respect to t, normalized to have a magnitude of 1. It is the vector that points towards the center of curvature.

`\rightarrow \vec N (t) = ((d\vec T(t))/(dt) )/( \big|\big|(d\vec T(t))/(dt)\big|\big| )`

Binormal Vector (B):

The binormal vector B is the cross product of the unit tangent vector T and the principal normal vector N. It points perpendicular to the osculating plane (the plane containing the curve and the principal normal) and indicates the twisting of the curve.

`\rightarrow \vec B(t) = \vec T(t) * \vec N(t)`

`\hrulefill`

Given:

`r(t)= \big < t^2, \ (2)/(3)t^3, \ t \big > ; \ P(4, \ (16)/(3), \ 2 )`

(a) Finding the unit tangent vector:

To compute r'(t), we take the derivative of r(t) with respect to t:

`\Longrightarrow r(t)= \big < t^2, \ (2)/(3)t^3, \ t \big > \\\\\\\\\therefore r'(t)= \big < 2t, \ 2t^2, \ 1 \big >`

Next, we calculate the magnitude of r'(t):

`\Longrightarrow \big|\big|r'(t)\big|\big|=√((2t)^2+(2t^2)^2+(1)^2) \\\\\\\\\therefore \big|\big|r'(t)\big|\big|=2t^2+1`

Now, we can find the unit tangent vector, T, at t = 2:

`\Longrightarrow \vec T(2)= ( \big < 2(2), \ 2(2)^2, \ 1 \big > )/(2(2)^2+1)\\\\\\\\\Longrightarrow \vec T(2)= ( \big < 4, \ 8, \ 1 \big > )/(9)\\\\\\\\\therefore \boxed{ \vec T(2)= \Big < (4)/(9), \ (8)/(9), \ (1)/(9) \Big > }`

(b) Finding the principle normal vector:

To compute dT/dt, we take the derivative of T(t) with respect to t:

`\Longrightarrow \vec T(t)= ( \big < 2t, \ 2t^2, \ 1 \big > )/(2t^2+1)\\\\\\\\\Longrightarrow \vec T(t)= \Big < (2t)/(2t^2+1), \ (2t^2)/(2t^2+1), \ \frac{1}\\{2t^2+1} \Big > \\\\\\\\\Longrightarrow \vec T'(t)= \Big < ((2t^2+1)(2)-(2t)(4t))/((2t^2+1)^2), \ ((2t^2+1)(4t)-(2t^2)(4t))/((2t^2+1)^2), \ ((2t^2+1)(0)-(1)(4t))/((2t^2+1)^2) \Big > \\\\\\\\`

`\Longrightarrow \vec T'(t)= \Big < ((4t^2+4)-(8t))/((2t^2+1)^2), \ ((8t^3+4t)-(8t^3))/((2t^2+1)^2), \ (-4t)/((2t^2+1)^2) \Big > \\\\\\\\\therefore \vec T'(t)= \Big < (4t^2-8t+4)/((2t^2+1)^2), \ (4t)/((2t^2+1)^2), \ (-4t)/((2t^2+1)^2) \Big >`

At this point finding the magnitude of the above vector with be time consuming, just plug in t=2 and find the magnitude from there:

`\Longrightarrow \vec T'(2)= \Big < (4(2)^2-8(2)+4)/((2(2)^2+1)^2), \ (4(2))/((2(2)^2+1)^2), \ (-4(2))/((2(2)^2+1)^2) \Big > \\\\\\\\\therefore \vec T'(2)= \Big < (32)/(81), \ (8)/(81), \ (-8)/(81) \Big >`

Now finding the magnitude of T(2):

`\Longrightarrow \big|\big| \vec T'(2)\big|\big|=\sqrt{\Big((32)/(81)\Big)^2+\Big((8)/(81)\Big)^2+\Big((-8)/(81)\Big)^2}\\\\\\\\\therefore \big|\big| \vec T'(2)\big|\big|=(8√(2) )/(27)`

Now, we can find the principle normal vector, N, at t = 4:

`\Longrightarrow \vec N(2)=(\Big < (32)/(81), \ (8)/(81), \ (-8)/(81) \Big > )/((8√(2) )/(27))\\\\\\\\\therefore \boxed{ \vec N(2) = \Big < (2√(2) )/(3), \ (√(2) )/(6), \ -(√(2) )/(6) \Big > }`

(b) Finding the binormal vector:

Take the cross of the unit tangent vector and the principle normal vector:

`\Longrightarrow \vec B(2)= \vec T(2)* \vec N(2) = \left|\begin{array}{ccc}\hat i & \hat j & \hat k\\\\ (32)/(81)& (8)/(81)& (-8)/(81)\\\\(2√(2) )/(3)& (√(2) )/(6)& -(√(2) )/(6) \end{array}\right|\\\\\\\\`

`\Longrightarrow \vec B(2) = [((8)/(81))((-√(2))/(6) )-((-8)/(81))((√(2))/(6) )]\hat i-[((32)/(81))((-√(2))/(6) )-((-8)/(81))((2√(2))/(3) )]\hat j+[((32)/(81))((√(2) )/(6))-((8)/(81))((2√(2) )/(3))]\hat k`

`\Longrightarrow \vec B(2) [(-4√(2) )/(243)+(4√(2) )/(243) ] \hat i - [(-16√(2) )/(242)+(16√(2) )/(242) ]\hat j + [(16√(2) )/(243) -(16√(2) )/(243)]\hat k\\\\\\\\\therefore \boxed{\vec B(2) = \big < 0, \ 0, \ 0\big > }`

Thus, all parts have been solved.