Use the chain rule to determine $\frac{d}{dt}g(\mathbf{r}(t))$.

Question

The functions $\mathbf{r}:\mathbb{R}\mapsto \mathbb{R}^2$ and $g:\mathbb{R}^2\mapsto \mathbb{R}$ are defined by \begin{equation*} \mathbf{r}(t) := \begin{bmatrix} \sinh{(t)} \\ t^2 \end{bmatrix} \quad \textit{and} \quad g(x,y) := x^2y. \end{equation*}
Use the chain rule to determine $\frac{d}{dt}g(\mathbf{r}(t))$.
Okay so $g(\mathbf{r}(t)) = t^2\sinh^2{(t)}$ so $\frac{d}{dt}g(\mathbf{r}(t)) = 2t\sinh^2{(t)}+2t^2\cosh{(t)}$. Is this correct? Thanks!

Answer 1

It is not correct and you also did not use Chain Rule. The second term is $2t^{2} \sinh t\cosh t$.

We have $(g(r(t))'=g_x\frac {dx} {dt}+g_y \frac {dy} {dt}=2xy\frac {dx} {dt}+x^{2} \frac {dy} {dt}$. Now verify that this gives the same answer.