OhMyMarkov said:Hello everyone, I remember this from single-variable calculus: if f(x) = 2x, then df = 2dx, right?<br><br>What if $f(r, \theta) = r\cos \theta$, how can I express $\partial f$ in terms of $r$ and $\theta$.?<br><br>Thanks!
The formula for calculating partial derivatives is $\frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h,y)-f(x,y)}{h}$ for a function $f(x,y)$ with respect to the variable $x$. Similarly, for a function $f(x,y)$ with respect to the variable $y$, the formula is $\frac{\partial f}{\partial y} = \lim_{h \to 0} \frac{f(x,y+h)-f(x,y)}{h}$.
To calculate partial derivatives for a function with polar coordinates, we use the chain rule. For a function $f(r,\theta)$, the partial derivative with respect to $r$ is $\frac{\partial f}{\partial r} = \cos \theta$ and the partial derivative with respect to $\theta$ is $\frac{\partial f}{\partial \theta} = -r\sin \theta$.
No, the polar coordinate equations can only be used to calculate $\partial f$ for functions that are defined in polar coordinates, i.e. functions that have variables $r$ and $\theta$.
Calculating partial derivatives allows us to determine the rate of change of a function with respect to a specific variable. This is useful in many applications, such as optimization problems, where we want to find the maximum or minimum value of a function.
The values of $\partial f$ for a function with polar coordinates represent the rate of change of the function in the direction of the specified variable. For example, $\frac{\partial f}{\partial r}$ represents the rate of change of the function in the radial direction, while $\frac{\partial f}{\partial \theta}$ represents the rate of change in the angular direction.