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!
Best Answer
$$\text{Tangent: }-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}}\\ \text{Unit tangent: }T=\dfrac{-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}}}{|-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}}|}=\dfrac{-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}}}{2\sqrt{1+4(2\cos t-\sin t)^2}}\\ \text{Binormal: }\left(T\times\dfrac{T^{\prime}}{|T^{\prime}|}\right)$$ I'll leave the rest to you.
Here's the equation for $T^\prime$: $$T^\prime=\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}}}{2\sqrt{1+4(2\cos t-\sin t)^2}}\right)=\\ {\left(-2\sqrt{1+4(2\cos t-\sin t)^2}(2\cos t\mathbf{\hat{i}}+2\sin t \mathbf{\hat{j}}+(4\cos t+8\sin t)\mathbf{\hat{k}})-(-2\sin t\mathbf{\hat{i}}+2\cos t \mathbf{\hat{j}}+(-4\sin t+8\cos t)\mathbf{\hat{k}})\left(-\dfrac{(2\cos t-\sin t)(2\sin t+\cos t)}{8\sqrt{1+4(2\cos t-\sin t)^2}}\right)\right)}\div\left({4(1+4(2\cos t-\sin t)^2})\right)$$