Find T, N, and B for the given space curve.
-$$r ( t ) = ( \ln ( \cos t ) + 1 ) \mathbf { i } + 1 \mathbf { j } + ( 10 + t ) \mathbf { k } , - \pi / 2

Question

Find T, N, and B for the given space curve.
-$$r ( t ) = ( \ln ( \cos t ) + 1 ) \mathbf { i } + 1 \mathbf { j } + ( 10 + t ) \mathbf { k } , - \pi / 2 < t < \pi / 2$$
A) T = (sin t)i + (cos t)k; N = (cos t)i - (sin t)k; B = -j
B) T = (-sin t)i + (cos t)k; N = (-cos t)i - (sin t)k; B = -j
C) T = (-sin t)i + (cos t)k; N = (-cos t)i - (sin t)k; B = j
D) T = (sin t)i + (cos t)k; N = (cos t)i - (sin t)k; B = j

Quizplus · Accepted Answer

To find the unit tangent vector $T$, normal vector $N$, and binormal vector $B$ for the given space curve $r(t) = (\ln(\cos t) + 1)\mathbf{i} + 1\mathbf{j} + (10 + t)\mathbf{k}$, we first compute the derivative of $r(t)$ with respect to $t$, which gives the velocity vector $v(t)$. The derivative of $r(t)$ is $r'(t) = \frac{d}{dt}[\ln(\cos t) + 1]\mathbf{i} + 0\mathbf{j} + 1\mathbf{k} = \left(-\tan(t)\right)\mathbf{i} + 0\mathbf{j} + \mathbf{k}$. Using the trigonometric identity $\tan(t) = \frac{\sin(t)}{\cos(t)}$, we can rewrite this as $(- \sin(t))\mathbf{i} + \mathbf{k}$. The unit tangent vector $T$ is obtained by normalizing $v(t)$, which gives $T = \frac{v(t)}{\|v(t)\|}$. Given the form of $v(t)$, normalization directly leads to $T = (-\sin t)\mathbf{i} + \mathbf{k}$, since the magnitude of $v(t)$ would be $\sqrt{\sin^2(t) + 1}$, and the normalization process would adjust the coefficients accordingly.The normal vector $N$ is found by differentiating $T$ with respect to $t$ and then normalizing. Differentiating $T$ gives components that, when normalized, result in $N = (-\cos t)\mathbf{i} - (\sin t)\mathbf{k}$.The binormal vector $B$ is the cross product of $T$ and $N$, $B = T \times N$. Given the vectors $T$ and $N$, the cross product in this context simplifies to a vector in the direction of $\mathbf{j}$, specifically $-\mathbf{j}$, due to the right-hand rule and the orientation of $T$ and $N$.Therefore, the correct answer is B, which correctly identifies the unit tangent vector $T$, the normal vector $N$, and the binormal vector $B$ as $(-sin t)\mathbf{i} + \mathbf{k}$, $(-cos t)\mathbf{i} - (sin t)\mathbf{k}$, and $-\mathbf{j}$, respectively.

Find T, N, and B for the Given Space Curve r(t)=(ln⁡(cos⁡t)+1)i+1j+(10+t)k,−π/2<t<π/2r ( t ) = ( \ln ( \cos t ) + 1 ) \mathbf { i } + 1 \mathbf { j } + ( 10 + t ) \mathbf { k } , - \pi / 2 < t < \pi / 2r(t)=(ln(cost)+1)i+1j+(10+t)k,−π/2<t<π/2

Find T, N, and B for the Given Space Curve $r ( t ) = ( \ln ( \cos t ) + 1 ) \mathbf { i } + 1 \mathbf { j } + ( 10 + t ) \mathbf { k } , - \pi / 2 < t < \pi / 2$