jeff1evesque
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Statement;
The tangential acceleration and normal acceleration is given as follows:
\vec{a_{T}} = \frac{4cos(2t)sin(2t)}{1 + 2sin^{2}(2t)} \cdot (\hat{x} - sin(2t)\hat{y} + sin(2t)\hat{z}), <br /> \vec{a_{N}} = - \frac{2cos(2t)}{1 + 2sin^{2}(2t)} \cdot (2sin(2t)\hat{x} + \hat{y} - \hat{z}) (#0)
Find the magnitude of the tangential and normal acceleration components (given above).
Solution/Attempt:
The solution for the tangential and normal magnitudes are given by the following,
\vec{a_{T}} = \frac{4cos(2t)sin(2t) }{\sqrt{1 + 2sin^{2}(2t)}}, <br /> \vec{a_{N}} = \frac{2\sqrt{2}cos(2t)}{\sqrt{1 + 2sin^{2}(2t)} } (#1)
I mean if we had to solve for the magnitude of some vector a = 3\hat{x} - 33\hat{z}, then we say,
|a| = \sqrt{3^{2} + (-33)^{2}}. How would we apply this to our equation (#0) to get equation (#1)?Thanks,JL
The tangential acceleration and normal acceleration is given as follows:
\vec{a_{T}} = \frac{4cos(2t)sin(2t)}{1 + 2sin^{2}(2t)} \cdot (\hat{x} - sin(2t)\hat{y} + sin(2t)\hat{z}), <br /> \vec{a_{N}} = - \frac{2cos(2t)}{1 + 2sin^{2}(2t)} \cdot (2sin(2t)\hat{x} + \hat{y} - \hat{z}) (#0)
Find the magnitude of the tangential and normal acceleration components (given above).
Solution/Attempt:
The solution for the tangential and normal magnitudes are given by the following,
\vec{a_{T}} = \frac{4cos(2t)sin(2t) }{\sqrt{1 + 2sin^{2}(2t)}}, <br /> \vec{a_{N}} = \frac{2\sqrt{2}cos(2t)}{\sqrt{1 + 2sin^{2}(2t)} } (#1)
I mean if we had to solve for the magnitude of some vector a = 3\hat{x} - 33\hat{z}, then we say,
|a| = \sqrt{3^{2} + (-33)^{2}}. How would we apply this to our equation (#0) to get equation (#1)?Thanks,JL
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