Question

Find \(\mathbf{r}^{\prime}(t)\). $$ \mathbf{r}(t)=\frac{1}{t} \mathbf{i}+16 t \mathbf{j}+\frac{t^{2}}{2} \mathbf{k} $$

Short answer

\(\mathbf{r}^{\prime}(t) = -\frac{1}{t^2} \mathbf{i} + 16 \mathbf{j} + t \mathbf{k}\)

Step-by-step solution

Derivative of the First Component

To find the derivative of the first component \(\frac{1}{t}\), apply the power rule for derivatives, which states that the derivative of \(t^n\) is \(nt^{n-1}\), with \(n=-1\). So, the derivative of \(\frac{1}{t}\) is \(-\frac{1}{t^2}\).

Derivative of the Second Component

The second component of the vector function is \(16t\). The derivative of a constant times a function is the constant times the derivative of the function. Hence, the derivative of \(16t\) is \(16\).

Derivative of the Third Component

The third component of the vector function is \(\frac{t^{2}}{2}\). Applying the power rule for derivatives here gives us \(t\).

Combine the Derivatives

Combine these derivatives into a vector to conclude that \(\mathbf{r}^{\prime}(t) = -\frac{1}{t^2} \mathbf{i} + 16 \mathbf{j} + t \mathbf{k}\).