The calculator will find the tangential 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 , 2 t , 3 t 2 ⟩ \mathbf{\vec{r}^{\prime}\left(t\right)} = \left\langle 1, 2 t, 3 t^{2}\right\rangle r ′ ( t ) = ⟨ 1 , 2 t , 3 t 2 ⟩ derivative calculator ).
Find the magnitude of r ⃗ ′ ( t ) \mathbf{\vec{r}^{\prime}\left(t\right)} r ′ ( t ) ∣ r ⃗ ′ ( t ) ∣ = 9 t 4 + 4 t 2 + 1 \mathbf{\left\lvert \mathbf{\vec{r}^{\prime}\left(t\right)}\right\rvert} = \sqrt{9 t^{4} + 4 t^{2} + 1} ∣ r ′ ( t ) ∣ = 9 t 4 + 4 t 2 + 1 magnitude calculator ).
Find the derivative of r ⃗ ′ ( t ) \mathbf{\vec{r}^{\prime}\left(t\right)} r ′ ( t ) r ⃗ ′ ′ ( t ) = ⟨ 0 , 2 , 6 t ⟩ \mathbf{\vec{r}^{\prime\prime}\left(t\right)} = \left\langle 0, 2, 6 t\right\rangle r ′′ ( t ) = ⟨ 0 , 2 , 6 t ⟩ derivative calculator ).
Find the dot product: r ⃗ ′ ( t ) ⋅ r ⃗ ′ ′ ( t ) = 18 t 3 + 4 t \mathbf{\vec{r}^{\prime}\left(t\right)}\cdot \mathbf{\vec{r}^{\prime\prime}\left(t\right)} = 18 t^{3} + 4 t r ′ ( t ) ⋅ r ′′ ( t ) = 18 t 3 + 4 t dot product calculator ).
Finally, the tangential component of acceleration is a T ( t ) = r ⃗ ′ ( t ) ⋅ r ⃗ ′ ′ ( t ) ∣ r ⃗ ′ ( t ) ∣ = 18 t 3 + 4 t 9 t 4 + 4 t 2 + 1 . a_T\left(t\right) = \frac{\mathbf{\vec{r}^{\prime}\left(t\right)}\cdot \mathbf{\vec{r}^{\prime\prime}\left(t\right)}}{\mathbf{\left\lvert \mathbf{\vec{r}^{\prime}\left(t\right)}\right\rvert}} = \frac{18 t^{3} + 4 t}{\sqrt{9 t^{4} + 4 t^{2} + 1}}. a T ( t ) = ∣ r ′ ( t ) ∣ r ′ ( t ) ⋅ r ′′ ( t ) = 9 t 4 + 4 t 2 + 1 18 t 3 + 4 t .
Answer The tangential component of acceleration is a T ( t ) = 18 t 3 + 4 t 9 t 4 + 4 t 2 + 1 a_T\left(t\right) = \frac{18 t^{3} + 4 t}{\sqrt{9 t^{4} + 4 t^{2} + 1}} a T ( t ) = 9 t 4 + 4 t 2 + 1 18 t 3 + 4 t A .
