TF. 5 - Calculus

Question

$$\begin{gathered} \mathrm{A} \mathrm{chain} \mathrm{rule} \mathrm{for} \mathrm{functions} \mathrm{of} \mathrm{two} \mathrm{variables}: \mathrm{Let} f(x, y)=x^2{y}^3, x=s \cos t, \mathrm{and} y=s \sin t. \\ \mathrm{Find} \frac{\partial f}{\partial s} \mathrm{and} \frac{\partial f}{\partial t} \mathrm{using} \mathrm{the} \mathrm{preceding} \mathrm{results}. \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \frac{\partial f}{\partial s}=5s^4 {\cos }^{2}t {\sin }^{3}t \\ \frac{\partial f}{\partial t}=-2s^5 \cos t \sin t+3s^5 {\cos }^{3}t {\sin }^{2}t \end{gathered}$$

Step 1. Given information

$f(x, y)=x^2{y}^3, x=s \cos t, \mathrm{and} y=s \sin t.$

Step 2. Finding partial derivatives

$$\begin{gathered} f(x, y)=x^2{y}^3, x=s \cos t, \mathrm{and} y=s \sin t. \\ \Rightarrow f(x, y)=s^5 {\cos }^{2}t {\sin }^{3}t \\ 1. \frac{\partial f}{\partial s}=5s^4 {\cos }^{2}t {\sin }^{3}t \\ 2. \frac{\partial f}{\partial t}=s^5 \left(2\cos t\right) \left(-\sin t\right)+s^5 {\cos }^{2}t \left(3{\sin }^{2}t\right)\left(\cos t\right) \\ \Rightarrow \frac{\partial f}{\partial t}=-2s^5 \cos t \sin t+3s^5 {\cos }^{3}t {\sin }^{2}t \end{gathered}$$