Question

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\sqrt{t+4} \mathbf{i}+\frac{t}{t+1} \mathbf{j}-e^{-t^{2}} \mathbf{k}$$

Short answer

Question: Determine the second and third derivatives of the vector function \(\mathbf{r}(t) = \sqrt{t + 4} \mathbf{i} + \frac{t}{t + 1} \mathbf{j} - e^{-t^2} \mathbf{k}\). Answer: The second derivative of the vector function is \(\mathbf{r}^{\prime \prime}(t) = -\frac{1}{4(t+4)^{\frac{3}{2}}} \mathbf{i} -\frac{2}{(t+1)^3} \mathbf{j} + 2 e^{-t^2} (1 - 2t^2) \mathbf{k}\). The third derivative of the vector function is \(\mathbf{r}^{\prime \prime \prime}(t) = \frac{3}{8(t+4)^{\frac{5}{2}}} \mathbf{i} + \frac{6}{(t+1)^4} \mathbf{j} + 4t e^{-t^2} (4t^2 - 3) \mathbf{k}\).

Step-by-step solution

Compute the first derivatives of each component

We first compute the first derivative of each component: $$ \frac{d}{dt} \sqrt{t+4} = \frac{1}{2\sqrt{t+4}} $$ $$ \frac{d}{dt} \frac{t}{t+1} = \frac{1}{(t+1)^2} $$ $$ \frac{d}{dt} -e^{-t^2} = 2t e^{-t^2} $$ So the first derivative of the vector function \(\mathbf{r}(t)\) is: $$ \mathbf{r}^{\prime}(t) = \frac{1}{2\sqrt{t+4}} \mathbf{i} + \frac{1}{(t+1)^2} \mathbf{j} + 2t e^{-t^2} \mathbf{k} $$

Compute the second derivatives of each component

Now we compute the second derivative of each component: $$ \frac{d^2}{dt^2} \sqrt{t+4} = -\frac{1}{4(t+4)^{\frac{3}{2}}} $$ $$ \frac{d^2}{dt^2} \frac{t}{t+1} = -\frac{2}{(t+1)^3} $$ $$ \frac{d^2}{dt^2} -e^{-t^2} = 2 e^{-t^2} (1 - 2t^2) $$ So the second derivative of the vector function \(\mathbf{r}(t)\) is: $$ \mathbf{r}^{\prime \prime}(t) = -\frac{1}{4(t+4)^{\frac{3}{2}}} \mathbf{i} -\frac{2}{(t+1)^3} \mathbf{j} + 2 e^{-t^2} (1 - 2t^2) \mathbf{k} $$

Compute the third derivatives of each component

Finally, we compute the third derivative of each component: $$ \frac{d^3}{dt^3} \sqrt{t+4} = \frac{3}{8(t+4)^{\frac{5}{2}}} $$ $$ \frac{d^3}{dt^3} \frac{t}{t+1} = \frac{6}{(t+1)^4} $$ $$ \frac{d^3}{dt^3} -e^{-t^2} = 4t e^{-t^2} (4t^2 - 3) $$ So the third derivative of the vector function \(\mathbf{r}(t)\) is: $$ \mathbf{r}^{\prime \prime \prime}(t) = \frac{3}{8(t+4)^{\frac{5}{2}}} \mathbf{i} + \frac{6}{(t+1)^4} \mathbf{j} + 4t e^{-t^2} (4t^2 - 3) \mathbf{k} $$ Now we have computed both the second and third derivatives of the given vector function: $$ \mathbf{r}^{\prime \prime}(t) = -\frac{1}{4(t+4)^{\frac{3}{2}}} \mathbf{i} -\frac{2}{(t+1)^3} \mathbf{j} + 2 e^{-t^2} (1 - 2t^2) \mathbf{k} $$ $$ \mathbf{r}^{\prime \prime \prime}(t) = \frac{3}{8(t+4)^{\frac{5}{2}}} \mathbf{i} + \frac{6}{(t+1)^4} \mathbf{j} + 4t e^{-t^2} (4t^2 - 3) \mathbf{k} $$