Question

Find \(\mathbf{r}^{\prime}(t)\). $$ \mathbf{r}(t)=a \cos ^{3} t \mathbf{i}+a \sin ^{3} t \mathbf{j}+\mathbf{k} $$

Short answer

\(\mathbf{r}^{\prime}(t)=-3a\sin(t)\cos^2(t)\mathbf{i} + 3a\sin^2(t)\cos(t)\mathbf{j} + 0\mathbf{k}\)

Step-by-step solution

Differentiate the \(\mathbf{i}\) Component

Firstly, differentiate the component alongside the \(\mathbf{i}\) vector. This involves evaluating the derivative of \(a \cos ^{3} t\) with respect to \(t\). Use the chain rule here, which states that the derivative of a composition of functions is the product of the derivative of the inner function and the derivative of the outer function. The derivative is \(-3a\sin(t)\cos^2(t)\).

Differentiate the \(\mathbf{j}\) Component

Now, differentiate the component alongside the \(\mathbf{j}\) vector. Once again, applying the chain rule to \(a \sin ^{3} t\), we will get the derivative as \(3a\sin^2(t)\cos(t)\).

Differentiate the \(\mathbf{k}\) Component

Lastly, differentiate the component alongside the \(\mathbf{k}\) vector. As this is a constant term, its derivative is zero.