I was given that $$p(t)=(1+2\cos t)\mathbf i + 2(1+\sin t)\mathbf j + (9+4\cos t+8 \sin t)\mathbf k$$ and that I needed to find the tangent, normal, and binormal vectors. The curvature and the osculating and normal planes at $P(1,0,1)$.
The thing is that what I got for the tangent vectors was a HUGE messy answer. Please help and explain your answer so I can understand it.
Thanks!
The standard way of doing this involves a lot of work that shouldn't be necessary. To get the unit normal vector, we don't have to differentiate $\frac{T^\prime}{\left|T^\prime\right|}$. The first two derivatives of your original function have all of the information you need.
You've got this space curve, $p(t)$. Your first step is going to be taking two derivatives anyway, so we obtain $p^\prime(t)$ and $p^{\prime\prime}(t)$. If they're parallel, then there's no curvature at that point, no normal vector, no binormal vector. Thus, we assume they're not parallel.
Now, $T$ is just $p^\prime$ normalized so that it's of unit length. That usually makes it harder to work with, so let's not do that, yet. The independent vectors $p^\prime$ and $p^{\prime\prime}$ are in the osculating plane, so our binormal vector, being normal to that plane, is parallel to their cross product. Indeed, $B$ is nothing more than $\frac{p^\prime \times p^{\prime\prime}}{\left|p^\prime \times p^{\prime\prime}\right|}$, but let's not do that step yet, because $p^\prime \times p^{\prime\prime}$ alone is probably easier to work with. We have $p^\prime \times p^{\prime\prime}=k_1B$, and $p^{\prime}=k_2T$, where $k_1,k_2$ are positive constants. Their cross-product, $\left(p^\prime \times p^{\prime\prime}\right)\times p^\prime$ is therefore equal to $k_1k_2N$.
Finally, having obtained vectors that are positive scalar multiples of $T, N$, and $B$, we can make them all unit length, and then write down equations for the three planes they define at each point, etc. Come to think of it, write down those plane equations before dividing all of the numbers by the square root of blah-blah-blah.
I guess I didn't mention curvature, but it comes from the same pieces, equaling $\frac{\left|p^\prime \times p^{\prime\prime}\right|}{\left|p^\prime\right|^3}$. Its reciprocal is the radius $r_{osc}$ of the osculating circle, and if you take a particular point $p(t_0)$, calculate this radius for that point, as well as $N$, and add $p(t_0)+r_{osc}N$, you get the center $c_{osc}$ of the osculating circle. The actual circle is given by $c_{osc}+r_{osc}\left[\cos(t)N+\sin(t)T\right]$, or you can swap $\sin$ and $\cos$, and/or multiply either or both by $-1$, depending on how you roll ;)
The really nice thing about this method is that you don't do any calculus after finding $p^{\prime\prime}$, so if you're only interested in this analysis at one particular point, you can plug in that $t$-value immediately after taking that second derivative, and then just work with static vectors instead of a bunch of functions.
$$\text{Tangent: }-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}}\\ \text{Unit tangent: }T=\dfrac{-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}}}{|-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}}|}=\dfrac{-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}}}{2\sqrt{1+4(2\cos t-\sin t)^2}}\\ \text{Binormal: }\left(T\times\dfrac{T^{\prime}}{|T^{\prime}|}\right)$$ I'll leave the rest to you.
Here's the equation for $T^\prime$: $$T^\prime=\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}}}{2\sqrt{1+4(2\cos t-\sin t)^2}}\right)=\\ {\left(-2\sqrt{1+4(2\cos t-\sin t)^2}(2\cos t\mathbf{\hat{i}}+2\sin t \mathbf{\hat{j}}+(4\cos t+8\sin t)\mathbf{\hat{k}})-(-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}})\left(-\dfrac{(2\cos t-\sin t)(2\sin t+\cos t)}{8\sqrt{1+4(2\cos t-\sin t)^2}}\right)\right)}\div\left({4(1+4(2\cos t-\sin t)^2})\right)$$
