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FobelloE

Answered question

2020-11-02

Find the tangential and normal components of the acceleration vector

r (t) = (3 t - t^{3}) i + 3 t^{2} j

Answer & Explanation

SabadisO

Skilled2020-11-03Added 108 answers

Given
$r (t) = (3 t - t^{3}) i + 3 t^{2} j$
Differentiate
$r^{'} (t) = (3 t - 3 t^{2}) i + 3 t^{2} j$
Again Differentiate
$r^{″} (t) = - 6 t i + 6 j$
Note that
$| r^{'} (t) | = \sqrt{(3 - 3 t^{2})^{2} + 6 t^{2}}$
$= \sqrt{(9 + 9 t^{4} - 18 t^{2}) + 36 t^{2}}$
$= \sqrt{9 + 9 t^{4} + 18 t^{2}}$
$= 3 \sqrt{t^{4} + 2 t^{2} + 1}$
$= 3 \sqrt{(t^{2} + 1)^{2}}$
$= 3 t^{2} + 1$
The tangential component is given in Equation 9 as
$a_{T} = \frac{r^{'} (t) \cdot r^{″} (t)}{| r^{'} (t) |}$
$a_{T} = \frac{[(3 - 3 t^{2}) i + 6 t j] \cdot [- 6 t i + 6 j]}{3 (1 + t^{2})}$
$a_{T} = \frac{(- 18 t + 18 t^{3}) + 36 t}{3 (1 + t^{2})} = \frac{18 t + 18 t^{3}}{3 (1 + t^{2})} = \frac{18 t (1 + t^{2})}{3 (1 + t^{2})} = 6 t$
The normal component is provided by Equation 9.
$a_{N} = \frac{| r^{'} (t) * r^{″} (t) |}{| r^{'} (t) |}$
$r^{'} (t) \cdot r^{″} (t) = [\begin{matrix} i & j & k \\ (3 - 3 t^{2}) & 6 t & 0 \\ - 6 t & 6 & 0 \end{matrix}]$
The third column should be added to the determinant to get
$r^{'} (t) \cdot r^{″} (t) = k (6 (3 - 3 t^{2}) - (- 36 t^{2})) = (18 + 18 t^{2}) k$
Threrefore, $| r^{'} (t) \cdot r^{″} (t) | = 18 (1 + t^{2})$
And hence $a_{N} = \frac{18 (1 + t^{2})}{3 (1 + t^{2})} = 6$
Result $a_{T} = 6 t, a_{N} = 6$

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