The calculator will find the normal component of acceleration for the object, described by the vector-valued function, at the given point, with steps shown.
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Solution Find the derivative of r ⃗ ( t ) \mathbf{\vec{r}\left(t\right)} r ( t ) r ⃗ ′ ( t ) = ⟨ 1 , 3 , 2 t ⟩ \mathbf{\vec{r}^{\prime}\left(t\right)} = \left\langle 1, 3, 2 t\right\rangle r ′ ( t ) = ⟨ 1 , 3 , 2 t ⟩ derivative calculator ).
Find the magnitude of r ⃗ ′ ( t ) \mathbf{\vec{r}^{\prime}\left(t\right)} r ′ ( t ) ∣ r ⃗ ′ ( t ) ∣ = 4 t 2 + 10 \mathbf{\left\lvert \mathbf{\vec{r}^{\prime}\left(t\right)}\right\rvert} = \sqrt{4 t^{2} + 10} ∣ r ′ ( t ) ∣ = 4 t 2 + 10 magnitude calculator ).
Find the derivative of r ⃗ ′ ( t ) \mathbf{\vec{r}^{\prime}\left(t\right)} r ′ ( t ) r ⃗ ′ ′ ( t ) = ⟨ 0 , 0 , 2 ⟩ \mathbf{\vec{r}^{\prime\prime}\left(t\right)} = \left\langle 0, 0, 2\right\rangle r ′′ ( t ) = ⟨ 0 , 0 , 2 ⟩ derivative calculator ).
Find the cross product: r ⃗ ′ ( t ) × r ⃗ ′ ′ ( t ) = ⟨ 6 , − 2 , 0 ⟩ \mathbf{\vec{r}^{\prime}\left(t\right)}\times \mathbf{\vec{r}^{\prime\prime}\left(t\right)} = \left\langle 6, -2, 0\right\rangle r ′ ( t ) × r ′′ ( t ) = ⟨ 6 , − 2 , 0 ⟩ cross product calculator ).
Find the magnitude of r ⃗ ′ ( t ) × r ⃗ ′ ′ ( t ) \mathbf{\vec{r}^{\prime}\left(t\right)}\times \mathbf{\vec{r}^{\prime\prime}\left(t\right)} r ′ ( t ) × r ′′ ( t ) ∣ r ⃗ ′ ( t ) × r ⃗ ′ ′ ( t ) ∣ = 2 10 \mathbf{\left\lvert \mathbf{\vec{r}^{\prime}\left(t\right)}\times \mathbf{\vec{r}^{\prime\prime}\left(t\right)}\right\rvert} = 2 \sqrt{10} ∣ r ′ ( t ) × r ′′ ( t ) ∣ = 2 10 magnitude calculator ).
Finally, the normal component of acceleration is a N ( t ) = ∣ r ⃗ ′ ( t ) × r ⃗ ′ ′ ( t ) ∣ ∣ r ⃗ ′ ( t ) ∣ = 2 5 2 t 2 + 5 . a_N\left(t\right) = \frac{\mathbf{\left\lvert \mathbf{\vec{r}^{\prime}\left(t\right)}\times \mathbf{\vec{r}^{\prime\prime}\left(t\right)}\right\rvert}}{\mathbf{\left\lvert \mathbf{\vec{r}^{\prime}\left(t\right)}\right\rvert}} = \frac{2 \sqrt{5}}{\sqrt{2 t^{2} + 5}}. a N ( t ) = ∣ r ′ ( t ) ∣ ∣ r ′ ( t ) × r ′′ ( t ) ∣ = 2 t 2 + 5 2 5 .
Answer The normal component of acceleration is a N ( t ) = 2 5 2 t 2 + 5 a_N\left(t\right) = \frac{2 \sqrt{5}}{\sqrt{2 t^{2} + 5}} a N ( t ) = 2 t 2 + 5 2 5 A .
