Q. 34 - Calculus

Question

Find the first-order partial derivatives for the functions in Exercises 27–36.  

$g\left(r,\theta \right)=r\cos \theta$

Step-by-Step Solution

Answer

The first-order partial derivatives are $\frac{\partial g}{\partial r}=\cos \theta \text{ and} \frac{\partial g}{\partial \theta }=-r\sin \theta$

Step 1: Given

The given function is: $g\left(r,\theta \right)=r\cos \theta$

Step 2: To find

We have to find the first-order partial derivatives.  

Step 3: Calculation

$$\begin{gathered} g\left(r,\theta \right)=r\cos \theta \\ \frac{\partial g}{\partial r}=\frac{\partial }{\partial r}\left(r\cos \theta \right) \\ \frac{\partial g}{\partial r}=\cos \theta \\ g\left(r,\theta \right)=r\cos \theta \\ \frac{\partial g}{\partial \theta }=\frac{\partial }{\partial \theta }\left(r\cos \theta \right) \\ \frac{\partial g}{\partial \theta }=r\left(-\sin \theta \right) \\ \frac{\partial g}{\partial \theta }=-r\sin \theta \\ \text{Hence}, \frac{\partial g}{\partial r}=\cos \theta \text{ and} \frac{\partial g}{\partial \theta }=-r\sin \theta \end{gathered}$$