Find Magnitudes of Acceleration Components: Tangential & Normal

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SUMMARY

The discussion focuses on calculating the magnitudes of tangential and normal acceleration components defined by the equations \(\vec{a_{T}} = \frac{4cos(2t)sin(2t)}{1 + 2sin^{2}(2t)} \cdot (\hat{x} - sin(2t)\hat{y} + sin(2t)\hat{z})\) and \(\vec{a_{N}} = - \frac{2cos(2t)}{1 + 2sin^{2}(2t)} \cdot (2sin(2t)\hat{x} + \hat{y} - \hat{z})\). The magnitudes of these components are derived as \(\vec{a_{T}} = \frac{4cos(2t)sin(2t)}{\sqrt{1 + 2sin^{2}(2t)}}\) and \(\vec{a_{N}} = \frac{2\sqrt{2}cos(2t)}{\sqrt{1 + 2sin^{2}(2t)}}\). The methodology for finding the magnitude of a vector is also discussed, illustrating the application of the Pythagorean theorem to vector components.

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Statement;
The tangential acceleration and normal acceleration is given as follows:
[tex]\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})[/tex] (#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,
[tex]\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)} }[/tex] (#1)

I mean if we had to solve for the magnitude of some vector [tex]a = 3\hat{x} - 33\hat{z},[/tex] then we say,
[tex]|a| = \sqrt{3^{2} + (-33)^{2}}[/tex]. How would we apply this to our equation (#0) to get equation (#1)?Thanks,JL
 
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[tex]|a\hat{x} + b\hat{y} + c\hat{z}| = \sqrt{a^2+b^x+c^2}[/tex]

that's all there is to it.
 

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