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MATH 119: Calculus 2
Multivariable functions
!!! definition - A multivariable function accepts more than one independent variable, e.g., \(f(x, y)\).
The signature of multivariable functions is indicated in the form [identifier]: [input type] → [return type]. Where \(n\) is the number of inputs:
\[f: \mathbb R^n \to \mathbb R\]
!!! example The following function is in the form \(f: \mathbb R^2\to\mathbb R\) and maps two variables into one called \(z\) via function \(f\).
$$(x,y)\longmapsto z=f(x,y)$$Sketching multivariable functions
!!! definition - In a scalar field, each point in space is assigned a number. For example, topography or altitude maps are scalar fields. - A level curve is a slice of a three-dimensional graph by setting to a general variable \(f(x, y)=k\). It is effectively a series of contour plots set in a three-dimensional plane. - A contour plot is a graph obtained by substituting a constant for \(k\) in a level curve.
Please see level set and contour line for example images.
In order to create a sketch for a multivariable function, this site does not have enough pictures so you should watch a YouTube video.
!!! example For the function \(z=x^2+y^2\):
For each $x, y, z$:
- Set $k$ equal to the variable and substitute it into the equation
- Sketch a two-dimensional graph with constant values of $k$ (e.g., $k=-2, -1, 0, 1, 2$) using the other two variables as axes
Combine the three **contour plots** in a three-dimensional plane to form the full sketch.A hyperbola is formed when the difference between two points is constant. Where \(r\) is the x-intercept:
\[x^2-y^2=r^2\]
If \(r^2\) is negative, the hyperbola is is bounded by functions of \(x\), instead.
Limits of two-variable functions
A function is continuous at \((x, y)\) if and only if all possible lines through \((x, y)\) have the same limit. Or, where \(L\) is a constant:
\[\text{continuous}\iff \lim_{(x, y)\to(x_0, y_0)}f(x, y) = L\]
In practice, this means that if any two paths result in different limits, the limit is undefined. Substituting \(x|y=0\) or \(y=mx\) or \(x=my\) are common solutions.
!!! example For the function \(\lim_{(x, y)\to (0,0)}\frac{x^2}{x^2+y^2}\):
Along $y=0$:
$$\lim_{(x,0)\to(0, 0)} ... = 1$$
Along $x=0$:
$$\lim_{(0, y)\to(0, 0)} ... = 0$$
Therefore the limit does not exist.Partial derivatives
Partial derivatives have multiple different symbols that all mean the same thing:
\[\frac{\partial f}{\partial x}=\partial_x f=f_x\]
For two-input-variable equations, setting one of the input variables to a constant will return the derivative of the slice at that constant.
By definition, the partial derivative of \(f\) with respect to \(x\) (in the x-direction) at point \((a, B)\) is:
\[\frac{\partial f}{\partial x}(a, B)=\lim_{h\to 0}\frac{f(a+h, B)-f(a, B)}{h}\]
Effectively:
- if finding \(f_x\), \(y\) should be treated as constant.
- if finding \(f_y\), \(x\) should be treated as constant.
!!! example With the function \(f(x,y)=x^2\sqrt{y}+\cos\pi y\):
\begin{align*}
f_x(1,1)&=\lim_{h\to 0}\frac{f(1+h,1)-f(1,1)} h \\
\tag*{$f(1,1)=1+\cos\pi=0$}&=\lim_{h\to 0}\frac{(1+h)^2-1} h \\
&=\lim_{h\to 0}\frac{h^2+2h} h \\
&= 2 \\
\end{align*}Higher order derivatives
!!! definition - wrt. is short for “with respect to”.
\[\frac{\partial^2f}{\partial x^2}=\partial_{xx}f=f_{xx}\]
Derivatives of different variables can be combined:
\[f_{xy}=\frac{\partial}{\partial y}\frac{\partial f}{\partial x}=\frac{\partial^2 f}{\partial xy}\]
The order of the variables matter: \(f_{xy}\) is the derivative of f wrt. x and then wrt. y.
Clairaut’s theorem states that if \(f_x, f_y\), and \(f_{xy}\) all exist near \((a, b)\) and \(f_{yx}\) is continuous at \((a,b)\), \(f_{yx}(a,b)=f_{x,y}(a,b)\) and exists.
!!! warning In multivariable calculus, differentiability does not imply continuity.
Linear approximations
A tangent plane represents all possible partial derivatives at a point of a function.
For two-dimensional functions, the differential could be used to extrapolate points ahead or behind a point on a curve.
\[ \Delta f=f'(a)\Delta d \\ \boxed{y=f(a)+f'(a)(x-a)} \]
The equations of the two unit direction vectors in \(x\) and \(y\) can be used to find the normal of the tangent plane:
\[ \vec n=\vec d_1\times\vec d_2 \\ \begin{bmatrix}-f_x(a,b) \\ -f_y(a,b) \\ 1\end{bmatrix} = \begin{bmatrix}1\\0\\f_x(a,b)\end{bmatrix} \begin{bmatrix}0\\1\\f_y(a,b)\end{bmatrix} \]
Therefore, the general expression of a plane is equivalent to:
\[ z=C+A(x-a)+B(x-b) \\ \boxed{z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)} \]
??? tip “Proof” The general formula for a plane is \(c_1(x-a)+c_2(y-b)+c_3(z-c)=0\).
If $y$ is constant such that $y=b$:
$$z=C+A(x-a)$$
which must represent in the x-direction as an equation in the form $y=b+mx$. It follows that $A=f_x(a,b)$. A similar concept exists for $f_y(a,b)$.
If both $x=a$ and $y=b$ are constant:
$$z=C$$
where $C$ must be the $z$-point.Usually, functions can be approximated via the tangent at \(x=a\).
\[f(x)\simeq L(x)\]
!!! warning Approximations are less accurate the stronger the curve and the farther the point is away from \(f(a,b)\). A greater \(|f''(a)|\) indicates a stronger curve.
!!! example Given the function \(f(x,y)=\ln(\sqrt[3]{x}+\sqrt[4]{y}-1)\), \(f(1.03, 0.98)\) can be linearly approximated.
$$
L(x=1.03, y=0.98)=f(1,1)=f_x(1,1)(x-1)+f_y(1,1)(y-1) \\
f(1.03,0.98)\simeq L(1.03,0.98)=0.005
$$