Question

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\left\langle 3 t^{12}-t^{2}, t^{8}+t^{3}, t^{-4}-2\right\rangle$$

Short answer

Question: Find the second and third derivatives of the vector function \(\mathbf{r}(t) = \langle 3t^{12} - t^2, t^8 + t^3, t^{-4} - 2 \rangle\). Answer: The second and third derivatives of the given vector function are: $$\mathbf{r}^{\prime\prime}(t) = \left\langle 396t^{10} - 2, 56t^6 + 6t, 20t^{-6}\right\rangle$$ $$\mathbf{r}^{\prime\prime\prime}(t) = \left\langle 3960t^9, 336t^5 + 6, -120t^{-7}\right\rangle$$

Step-by-step solution

Find the first derivatives of the components

First, we need to find the first derivative of each component of the function \(\mathbf{r}(t)\) with respect to \(t\). For \(x(t) = 3t^{12} - t^2\), $$\frac{dx}{dt} = 36t^{11} - 2t$$ For \(y(t) = t^8 + t^3\), $$\frac{dy}{dt} = 8t^7 + 3t^2$$ For \(z(t) = t^{-4} - 2\), $$\frac{dz}{dt} = -4t^{-5}$$ Now, we can express the first derivative of the function \(\mathbf{r}(t)\) as: $$\mathbf{r}^\prime(t) = \left\langle 36t^{11} - 2t, 8t^7 + 3t^2, -4t^{-5}\right\rangle$$

Find the second derivatives of the components

Next, we need to find the second derivative of each component of the function \(\mathbf{r}(t)\) with respect to \(t\). For \(x(t)\), using \(\frac{dx}{dt} = 36t^{11} - 2t\), $$\frac{d^2x}{dt^2} = 396t^{10} - 2$$ For \(y(t)\), using \(\frac{dy}{dt} = 8t^7 + 3t^2\), $$\frac{d^2y}{dt^2} = 56t^6 + 6t$$ For \(z(t)\), using \(\frac{dz}{dt} = -4t^{-5}\), $$\frac{d^2z}{dt^2} = 20t^{-6}$$ Now, we can express the second derivative of the function \(\mathbf{r}(t)\) as: $$\mathbf{r}^{\prime\prime}(t) = \left\langle 396t^{10} - 2, 56t^6 + 6t, 20t^{-6}\right\rangle$$

Find the third derivatives of the components

Finally, we need to find the third derivative of each component of the function \(\mathbf{r}(t)\) with respect to \(t\). For \(x(t)\), using \(\frac{d^2x}{dt^2} = 396t^{10} - 2\), $$\frac{d^3x}{dt^3} = 3960t^9$$ For \(y(t)\), using \(\frac{d^2y}{dt^2} = 56t^6 + 6t\), $$\frac{d^3y}{dt^3} = 336t^5 + 6$$ For \(z(t)\), using \(\frac{d^2z}{dt^2} = 20t^{-6}\), $$\frac{d^3z}{dt^3} = -120t^{-7}$$ Now, we can express the third derivative of the function \(\mathbf{r}(t)\) as: $$\mathbf{r}^{\prime\prime\prime}(t) = \left\langle 3960t^9, 336t^5 + 6, -120t^{-7}\right\rangle$$ The second and third derivatives of the given vector function are: $$\mathbf{r}^{\prime\prime}(t) = \left\langle 396t^{10} - 2, 56t^6 + 6t, 20t^{-6}\right\rangle$$ $$\mathbf{r}^{\prime\prime\prime}(t) = \left\langle 3960t^9, 336t^5 + 6, -120t^{-7}\right\rangle$$