Question

Let z = ln(x 2 + y), x = ret . and y = ter . Use the Chain Rule to compute ∂z ∂r and ∂z ∂t at the point where (r, t) = (1, 2).\

Answer 1

By the chain rule,

$\dfrac{\partial z}{\partial u}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u}$

where $u\in\{r,t\}$.

We have component partial derivatives

$\dfrac{\partial z}{\partial x}=\dfrac{2x}{x^2+y}=\dfrac{2re^t}{r^2e^{2t}+te^r}$

$\dfrac{\partial z}{\partial y}=\dfrac1{x^2+y}=\dfrac1{r^2e^{2t}+te^r}$

$\dfrac{\partial x}{\partial r}=e^t$

$\dfrac{\partial x}{\partial t}=re^t$

$\dfrac{\partial y}{\partial r}=te^r$

$\dfrac{\partial y}{\partial t}=e^r$

Putting the appropriate pieces together and setting $(r,t)=(1,2)$, we get

$\dfrac{\partial z}{\partial r}(1,2)=\dfrac{2e^3+2}{e^3+2}$

$\dfrac{\partial z}{\partial t}(1,2)=\dfrac{2e^3+1}{e^3+2}$