Question

Compute the gradient of the following: a. $f(x,y)=x^2-y^2$ b. $f(x,y,z)=yz+xy+xz$. c. $f(x,y)={\tan }^{-1}\left(\frac{y}{x}\right)$. d. $f(x,y,z)=2y^x\cos z-5\sin z\cos y$.

Short answer

The gradients for the functions are: a. $(2x,-2y)$, b. $(y+z,x+z,x+y)$, c. $(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})$, d. $(2\cos (z)\ln (y)y^x,2x{y}^{x-1}\cos (z)+5\sin (z)\sin (y),2y^x(-\sin (z))-5\cos (z)\cos (y))$.

Step-by-step solution

Compute the gradient of $f(x,y)=x^2-y^2$

The gradient of a function can be found by taking the partial derivatives with respect to each variable. So, the gradient of this function is given by $(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$, which becomes $(\frac{\partial }{\partial x}(x^2-y^2),\frac{\partial }{\partial y}(x^2-y^2))=(2x,-2y)$

Compute the gradient of $f(x,y,z)=yz+xy+xz$

Using the same procedure as in step 1 but considering each variable $x,y,z$ in turn, the gradient of this function will be $(\frac{\partial }{\partial x}(yz+xy+xz),\frac{\partial }{\partial y}(yz+xy+xz),\frac{\partial }{\partial z}(yz+xy+xz))=(y+z,x+z,x+y)$

Compute the gradient of $f(x,y)={\tan }^{-1}(\frac{y}{x})$

Using the chain rule, which in this case states that $\frac{\partial }{\partial x}({\tan }^{-1}(g(x,y))=\frac{1}{1+(g(x,y){)}^2}\cdot \frac{\partial }{\partial x}(g(x,y))$, we find the gradient to be the derived pair $(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})$

Compute the gradient of $f(x,y,z)=2y^x\cos (z)-5\sin (z)\cos (y)$

This function involves a slightly more complex chain rule, due to functions of functions. After applying these rules, we find the gradient to be $(\frac{\partial }{\partial x}(2y^x\cos (z)-5\sin (z)\cos (y)),\frac{\partial }{\partial y}(2y^x\cos (z)-5\sin (z)\cos (y)),\frac{\partial }{\partial z}(2y^x\cos (z)-5\sin (z)\cos (y)))=(2\cos (z)\ln (y)y^x,2x{y}^{x-1}\cos (z)+5\sin (z)\sin (y),2y^x(-\sin (z))-5\cos (z)\cos (y))$

Key concepts

Partial Derivatives

Partial derivatives play a crucial role in the toolbox of multivariable calculus. They measure how a function changes as one of the variables is varied, holding all other variables constant. This is similar to how a regular derivative reflects change in single-variable functions.

When computing the gradient of a function such as
$f(x,y)=x^2-y^2$,
we find the partial derivatives with respect to each variable: $x$ and $y$. For $x$, we treat $y$ as a constant and differentiate $x^2-y^2$ with respect to $x$ to get $2x$. Similarly, for $y$, we differentiate with respect to $y$ considering $x$ as a constant to obtain $-2y$. These operations result in the gradient vector
$(2x,-2y)$.

The process isolates each variable's impact on the function's rate of change, which is why partial derivatives are foundational in fields involving multiple variables, such as physics and engineering.

Multivariable Calculus

Beyond partial derivatives, multivariable calculus encompasses the study of functions that depend on several variables. It deals with concepts such as gradients, multiple integrals, and vectors. One of the main objectives is to understand how these functions behave in a multidimensional space.

In our example of computing the gradient of $f(x,y,z)=yz+xy+xz$, we are entering the realm of three-dimensional space and the functions within it. With each partial derivative
$(y+z,x+z,x+y)$,
we gain information about the slope of the function in the direction of each coordinate axis. Understanding and visualizing these slopes are essential for applications that range from optimizing multivariable functions to solving complex engineering problems.

Chain Rule

The chain rule in multivariable calculus is a powerful tool used to differentiate compositions of functions. It's especially pertinent when dealing with functions within functions—an everyday occurrence in mathematics and related disciplines.

For instance, to find the gradient of
$f(x,y)={\tan }^{-1}(\frac{y}{x})$,
we must apply the chain rule. The function inside the arctangent, $g(x,y)=\frac{y}{x}$, needs to be differentiated as well with respect to $x$ and $y$. According to the chain rule, the resulting partial derivatives are
$(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2})$.

This highlights the function's sensitivity to changes in either variable and reflects how intertwined the variables are within the function's composition.

Vector Calculus

Gradients in Vector Calculus

Vector calculus is an extension of calculus that deals with vector fields and operations on vectors. A gradient is a perfect example of where vector calculus comes into play, as it is fundamentally a vector consisting of partial derivatives of a function with respect to its variables.

Gradients represent the direction and rate of fastest increase of a function. Particularly, the gradient of the function
$f(x,y,z)=2y^x\text{cos}(z)-5\text{sin}(z)\text{cos}(y)$,
which involves both exponential and trigonometric parts, shows how each component of the vector along $x$, $y$, and $z$ is affected by a subtle interplay of these functions. The resulting gradient
$(2\text{cos}(z)\text{ln}(y)y^x,2x{y}^{x-1}\text{cos}(z)+5\text{sin}(z)\text{sin}(y),2y^x(-\text{sin}(z))-5\text{cos}(z)\text{cos}(y))$
turns out to be a vector that compactly encodes how modifications in any of the three variables influence the output in all three dimensions simultaneously.

Understanding vector calculus, hence, is essential for accurately capturing the nuances of change in multidimensional functions and fields, reinforcing the analytical power of multivariable calculus.