Find the principal unit normal vector N for the curve r(t).
-$$\mathbf { r } ( t ) = ( 10 + t ) \mathbf { i } + ( 9 + \ln ( \cos t ) ) \mathbf { k } , - \pi / 2

Question

Find the principal unit normal vector N for the curve r(t).
-$$\mathbf { r } ( t ) = ( 10 + t ) \mathbf { i } + ( 9 + \ln ( \cos t ) ) \mathbf { k } , - \pi / 2 < t < \pi / 2$$
A) N = (-cos t)i - (ln(cos t))k
B) N = (cos t)i + (sin t)k
C) N = (-sin t)i - (cos t)k
D) N = (-cos t)i - (tan t)k

Quizplus · Accepted Answer

To find the principal unit normal vector, we need to first find the unit tangent vector by taking the derivative of r(t) and dividing by its magnitude:

$$\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{\left\Vert\mathbf{r}'(t)\right\Vert}=\frac{\mathbf{i}-\sin t\mathbf{k}}{\sqrt{(\cos t)^2+(\sin t)^2}}=\frac{\mathbf{i}-\sin t\mathbf{k}}{\cos t}=\frac{\mathbf{i}}{\cos t}-\sin t\mathbf{k}$$

Then we take the derivative of the unit tangent vector and divide by its magnitude to get the principal unit normal vector:

$$\mathbf{N}(t)=\frac{\mathbf{T}'(t)}{\left\Vert\mathbf{T}'(t)\right\Vert}=-\frac{\sin t\mathbf{i}+\cos t\mathbf{k}}{\sqrt{(\cos t)^2+(\sin t)^2}}=-\sin t\mathbf{i}-\cos t\mathbf{k}=\boxed{\textbf{(C)}~N = (-\sin t)\mathbf{i} - (\cos t)\mathbf{k}}$$

Find the Principal Unit Normal Vector N for the Curve r(t)=(10+t)i+(9+ln⁡(cos⁡t))k,−π/2<t<π/2\mathbf { r } ( t ) = ( 10 + t ) \mathbf { i } + ( 9 + \ln ( \cos t ) ) \mathbf { k } , - \pi / 2 < t < \pi / 2r(t)=(10+t)i+(9+ln(cost))k,−π/2<t<π/2

Find the Principal Unit Normal Vector N for the Curve $\mathbf { r } ( t ) = ( 10 + t ) \mathbf { i } + ( 9 + \ln ( \cos t ) ) \mathbf { k } , - \pi / 2 < t < \pi / 2$