Question

find the vectors t, n, and b at the given point. r(t) = <3 cos(t), 3 sin(t), 3 ln(cos(t))>, (3, 0, 0) t = n = b = resources read it

Explanation:

Step1: Find the value of \(t\)

Set \(r(t)=\langle3\cos(t),3\sin(t),3\ln(\cos(t))
angle=(3,0,0)\). We have \(3\cos(t) = 3\), \(3\sin(t)=0\) and \(3\ln(\cos(t)) = 0\). From \(3\cos(t)=3\), we get \(\cos(t)=1\), so \(t = 0\).

Step2: Calculate the derivative of \(r(t)\)

\(r'(t)=\langle - 3\sin(t),3\cos(t),-3\frac{\sin(t)}{\cos(t)}
angle\). When \(t = 0\), \(r'(0)=\langle0,3,0
angle\).

Step3: Calculate the unit - tangent vector \(T(t)\)

The formula for \(T(t)=\frac{r'(t)}{\vert r'(t)\vert}\). \(\vert r'(0)\vert=\sqrt{0 + 9+0}=3\), so \(T(0)=\frac{r'(0)}{\vert r'(0)\vert}=\langle0,1,0
angle\).

Step4: Calculate the derivative of \(T(t)\)

First, \(T(t)=\langle-\sin(t),\cos(t),-\tan(t)
angle\). Then \(T'(t)=\langle-\cos(t),-\sin(t),-\sec^{2}(t)
angle\). When \(t = 0\), \(T'(0)=\langle - 1,0,-1
angle\).

Step5: Calculate the unit - normal vector \(N(t)\)

The formula for \(N(t)=\frac{T'(t)}{\vert T'(t)\vert}\). \(\vert T'(0)\vert=\sqrt{1 + 0 + 1}=\sqrt{2}\), so \(N(0)=\langle-\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}}
angle\).

Step6: Calculate the binormal vector \(B(t)\)

The formula for \(B(t)=T(t)\times N(t)\). \(B(0)=\langle0,1,0
angle\times\langle-\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}}
angle=\langle-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}
angle\).

Answer:

\(T=\langle0,1,0
angle\)
\(N=\langle-\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}}
angle\)
\(B=\langle-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}
angle\)