Functions of several variables

1.1 Sketching simple surfaces in $\mathbb{R}^3$

The graph of a function $f$ of one variable, given by $y=f(x)$, gives rise to a curve in $\mathbb{R}^2$. The graph of a function $F$ of two variables, given by $z=F(x,y)$, gives rise to a surface in $\mathbb{R}^3$.

Sketching a graph in $\mathbb{R}^3$ can be challenging because the sketch must be represented in $\mathbb{R}^2$. The conventional orientation is as follows:

This is however difficult to properly represent for plots that cannot be visualised clearly from a fixed perspective, and also cannot be easily drawn by hand.

Topographic maps solve this problem by using contour lines to represent the height (above sea level) of the surface of the earth at various points. This idea can be adapted to sketching a surface $z=F(x,y)$ described by a function $F$. Here, $z$ represents the height of the surface above the $xy$-plane, and the contours of the surface can be defined as follows:

For each level curve, the corresponding value of $C$ gives the height of the curve above the $xy$-plane.

We can use a profile obtained from a plot of:

• $z$ versus $y$ (with $x=\text{constant}$) or
• $z$ versus $x$ (with $y=\text{constant}$).
  These can be thought of as vertical cuts through the function surface.

Basically level curves are top-down view; profile's are views from the side of the plane when seen in conventional orientation.

Example Sketch level curves for the function $F:\mathbb{R}^2 \rightarrow \mathbb{R}$, where $F(x,y)=x^2+y^2$.
Since the level curves are of the form $x^2+y^2=C$, where $C$ is nonnegative (as there is no solution for $C<0$), you can interpret this as circles with varying radii.

These are the level curves given by $x^2+y^2=\{0,1,2,3,4\}$.

Example Sketch the surface in $\mathbb{R}^3$ described by the equation $z=x^2+y^2$.
From the previous example, we know that the surface increases in radius on the $xy$-plane; the level curves for the surface are circles. When $x=0$, we have $z=y^2$. On the $yz$-plane, this is simply a parabola. The $yz$-profile and level curves help produce a sketch of the surface.

• (You can also use the $xz$-profile but since they are both the same you can just use one)
  
  This is the conventional orientation for the $x$-, $y$- and $z$-axes.

Example A surface in $\mathbb{R}^3$ is described by the equation $x^2 + y^2 − z^2 = 1$. Sketch some level curves and hence sketch the surface in $\mathbb{R}^3$ .

First obtain the level curve by setting $z$ equal to $C$, for some constant $C$.

For the profiles, if $x= 0$ you get $y^2-z^2=1$, which is a hyperbola:

Combining this, you get a hyperboloid of one sheet

Example. Sketch some level curves and profiles for the surface $z=x^2-y^2$, and hence describe the surface.
The level curves are of the form $x^2-y^2=C$:

• If $C=0$, then $y^2=x^2$, i.e. the pair of straight lines $y=x$ and $y=-x$.
• If $C>0$, the level curves are hyperbolas opening left-and-right, crossing the $x$-axis at $x=\pm\sqrt{C}$.
• If $C<0$, the level curves are hyperbolas opening up-and-down, crossing the $y$-axis at $y=\pm\sqrt{-C}$.
  For the profiles, setting $y=0$ gives $z=x^2$ (an upward parabola on the $xz$-plane), whilst setting $x=0$ gives $z=-y^2$ (a downward parabola on the $yz$-plane). So the surface rises in the $x$-direction but falls in the $y$-direction; this is a saddle (formally, a hyperbolic paraboloid). Notice how the two profiles are genuinely different here, so unlike the paraboloid you actually need both of them.

The surface $z=xy$ is the same saddle shape rotated $45^\circ$ (rotating $z=x^2-y^2$ by $45^\circ$ gives exactly $z=2xy$, so $z=xy$ is that saddle rotated and vertically scaled by $\frac{1}{2}$); its level curves $xy=C$ are hyperbolas with the $x$- and $y$-axes as asymptotes, and along the line $y=x$ the profile is the upward parabola $z=x^2$ whilst along $y=-x$ it is the downward parabola $z=-x^2$.

Example. The surfaces $z=x^2+y^2$ and $z=\sqrt{x^2+y^2}$ both have circles as their level curves. Explain how a sketch can still tell them apart.
For the paraboloid $z=x^2+y^2$, the level curve at height $C$ is a circle of radius $\sqrt{C}$, so the radii for $C=1,2,3,4$ are

$$1, \sqrt{2}, \sqrt{3}, 2;$$

the circles bunch closer and closer together, meaning the surface gets steeper as you move outwards. For $z=\sqrt{x^2+y^2}$, the level curve at height $C$ is a circle of radius exactly $C$, so the radii are $1,2,3,4$; evenly spaced circles, meaning constant steepness. The profiles confirm this: setting $x=0$ gives $z=y^2$ for the first surface but $z=|y|$ (a V shape) for the second, so the second surface is a cone with a sharp tip at the origin.
Two different surfaces can have identical-looking level curves; the spacing of the level curves and the profiles are what actually pin the surface down.

1.2 Partial differentiation

To introduce the notion of a partial derivative, consider the function $F$ given by

$$F(x,y) = x^2+y^2.$$

The goal is to quantify the rate of change of $F(x,y)$ at the point $(1,2)$.

Since there are two variables, there are two places where the rate of change is changing: the $x$- and the $y$-direction. This means we can differentiate with respect to one of the variables while holding the other variables fixed. Since we are considering $(1,2)$, we have $x=1$ and

$$z=F(x,y)=F(1,y)=1+y^2,$$

since $x$ is constant but $y$ is not. This can also be thought of as the plane $x=1$ intersecting with the plot of $F(x,y)$. The gradient in the $y$-direction can be calculated in the usual way:

$$\text{gradient } = \frac{d}{dy}(1+y^2) \bigg\rvert_{y=2}=2y\bigg\rvert_{y=2}=4.$$

Thus the rate of change of $F(x,y)$ at $(1,2)$ in the $y$-direction is 4.

A faster way of doing this is simply just treating $x$ as a constant:

$$F_y(x,y)=2y$$

and then evaluating

$$F_y(1,2)=2\times2=4.$$

The function $F_y$ is called the partial derivative of $F$ with respect to $y$.
Using the same method you can find $F_x(1,2)$ and obtain $2$.

The partial derivatives of a function $F$ may be defined formally by using limits.

Example. Using the limit definition, find $F_x$ where $F(x,y)=x^2y$.

$$\begin{align*} F_x(x,y)&=\lim_{h\rightarrow0}\frac{F(x+h,y)-F(x,y)}{h} \\ &=\lim_{h\rightarrow0}\frac{(x+h)^2y-x^2y}{h} \\ &=\lim_{h\rightarrow0}\frac{x^2y+2xhy+h^2y-x^2y}{h} \\ &=\lim_{h\rightarrow0}\frac{2xhy+h^2y}{h} \\ &=\lim_{h\rightarrow0}(2xy+hy) \\ &=2xy. \end{align*}$$

This agrees with the fast method of just treating $y$ as a constant and differentiating with respect to $x$.

Example. Let $F(x,y)=\sqrt{x^2+y^2}$ (the cone from 1.1). Show that $F_x(0,0)$ does not exist.
Blindly differentiating gives $F_x=\dfrac{x}{\sqrt{x^2+y^2}}$, which is $\frac{0}{0}$ at the origin, so we fall back to the limit definition:

$$\begin{align*} F_x(0,0)&=\lim_{h\rightarrow0}\frac{F(h,0)-F(0,0)}{h} \\ &=\lim_{h\rightarrow0}\frac{\sqrt{h^2}-0}{h} \\ &=\lim_{h\rightarrow0}\frac{|h|}{h}, \end{align*}$$

which is $1$ from the right but $-1$ from the left, so the limit does not exist. This makes sense geometrically; the cone has a sharp tip at the origin, so there is no well defined slope there (and, looking ahead to 1.3, no tangent plane either). If a question asks about a partial derivative at a point where the usual differentiation rules break down, use the limit definition.

Notation: Regarding $F_x$ and $F_y$, they can also be denoted by

$$\frac{\partial F}{\partial x} \quad \text{ or } \quad D_1F,$$

for $F_x$, while $F_y$ may be denoted by

$$\frac{\partial F}{\partial y} \quad \text{ or } \quad D_2F.$$

It is important to note that the notation involving the 'curly d' is a bit ambiguous. The notation $D_1F(y,x)$ simply means differentiate the first variable in the parenthesis and then evaluate the derivative at the point $(y,x)$. To do the same with the 'curly d', one would write $\dfrac{\partial F}{\partial x}(y,x)$. Here the $x$ represents the first variable of the function whilst the $x$ in parenthesis represents the second ordinate of the point $(y,x).$

Geometrical interpretations

1. $\dfrac{\partial F}{\partial x}$ is the slope of the surface $z=F(x,y)$ in the $x$ direction.
2. $F_x(a,b)$ is the gradient of the tangent to the cross section at $(a,b)$ when the surface $z=F(x,y)$ is intersected with the plane $y=b$.
   Similar definitions for $\dfrac{\partial F}{\partial y}$ and $F_y(a,b)$.

Example. Suppose that $F(x,y)=3e^{xy^3}\sin{y}$. Find $\dfrac{\partial F}{\partial x}$ and $\dfrac{\partial F}{\partial y}.$
For $\dfrac{\partial F}{\partial x}$, we treat $y$ as a constant to yield

$$3y^3e^{xy^3}\sin{y}.$$

For $\dfrac{\partial F}{\partial y}$, we treat $x$ as a constant and use the product rule to yield

$$\begin{align*} (3xy^2 \times 3e^{xy^3}\times\sin{y}) + (3e^{xy^3} \times \cos{y}) &= \\ 3e^{xy^3}(3xy^2\sin{y}+\cos{y}). \end{align*}$$

Notation p2:
For second order partial derivatives, the notation is as follows:

$$\begin{align*} \frac{\partial^2 F}{\partial x^2} &\text{ means } \frac{\partial F}{\partial x} (\frac{\partial F}{\partial x}); \\ \frac{\partial^2 F}{\partial y^2} &\text{ means } \frac{\partial F}{\partial y} (\frac{\partial F}{\partial y}); \\ \frac{\partial^2 F}{\partial x \partial y} &\text{ means } \frac{\partial F}{\partial x} (\frac{\partial F}{\partial y}); \\ \frac{\partial^2 F}{\partial y \partial x} &\text{ means } \frac{\partial F}{\partial y} (\frac{\partial F}{\partial x}). \end{align*}$$

Since the definition of continuous is sort of ambiguous for two variables, a simple way to determine if it's continuous is simply to see if it's made up of many compositions of single variable functions. This means for a function $F(x,y)=3xy^4+e^x\sin{y}$, you can rewrite as
$F(x,y)=f(x)g(y)+h(x)k(y)$.

Most functions of two variables given in this course are continuous on their domains.

To prove the theorem, you use the definition of the partial derivative.

Example. Compute all second order partial derivatives of $F$, where $F(x,y)=3xy^4+e^x\sin{y}$, and verify the mixed derivative theorem.
The first order partial derivatives are

$$\begin{align*} F_x(x,y)&=3y^4+e^x\sin{y}, \\ F_y(x,y)&=12xy^3+e^x\cos{y}. \end{align*}$$

Differentiating each of these again,

$$\begin{align*} \frac{\partial^2 F}{\partial x^2}&=e^x\sin{y}, \\ \frac{\partial^2 F}{\partial y^2}&=36xy^2-e^x\sin{y}, \\ \frac{\partial^2 F}{\partial y \partial x}&=\frac{\partial}{\partial y}(3y^4+e^x\sin{y})=12y^3+e^x\cos{y}, \\ \frac{\partial^2 F}{\partial x \partial y}&=\frac{\partial}{\partial x}(12xy^3+e^x\cos{y})=12y^3+e^x\cos{y}. \end{align*}$$

The two mixed partial derivatives are equal, exactly as the theorem promised; recall from earlier that this $F$ is continuous since it is built from single variable continuous functions.

Example. Suppose that $F(x,y)=x^3y^2+e^{\sin{y}}\tan^{-1}{y}$. Find $\dfrac{\partial^2 F}{\partial x \partial y}$.
Following the notation literally, we would have to differentiate with respect to $y$ first; but $\dfrac{\partial F}{\partial y}$ requires a product rule on $e^{\sin{y}}\tan^{-1}{y}$, which is horrible. Instead, notice that $F$ and all its partial derivatives are continuous (compositions of single variable functions again), so the mixed derivative theorem lets us swap the order for free:

$$\begin{align*} \frac{\partial F}{\partial x}&=3x^2y^2 \quad (\text{the ugly term has no } x \text{, so it dies}), \\ \frac{\partial^2 F}{\partial y \partial x}&=6x^2y. \end{align*}$$

Therefore $\dfrac{\partial^2 F}{\partial x \partial y}=6x^2y$ as well. When one order of differentiation is much messier than the other, check continuity and then use the mixed derivative theorem to pick the easy order.

1.3 Tangent planes to surfaces

A tangent plane is just like a tangent line in $\mathbb{R}^2$, but for surfaces in $\mathbb{R}^3$. While deriving an equation for the tangent plane, a formula for the normal vector to the surface at a given point is also obtained.

Example. Suppose that $F(x,y) = x^2+y^2$. Find the Cartesian equation of the tangent plane to the surface $z=F(x,y)$ at the point where $(x,y,z)=(1,2,5)$. Find also a vector $\mathbf{n}$ that is normal to the surface at this point.

First, intersect the surface $z=x^2+y^2$ with the plane $x=1$ to get

$$\begin{cases} z = 1+y^2 \\ x = 1. \end{cases}$$

The gradient is $F_y(1,2)=4$. Using the point-gradient formula $(y-y_1)=m(x-x_1)$ but for $z=1+y^2$, a Cartesian equation for this tangent is given by

$$z-5=4(y-2), \quad x = 1.$$

If $y= \lambda + 2$, then the equation of the tangent line in parametric form is

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix}$$

whenever $\lambda \in \mathbb{R}$.
Similarly, we intersect the surface with the plane $y=2$ and repeat the previous steps to obtain

$$\begin{cases} z = 4+x^2 \\ y = 2. \end{cases}$$

If $x = \mu + 1$ then the parametric vector form is

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$$

whenever $\mu \in \mathbb{R}$.
Since the two direction vectors lie in the tangent plane and the vectors are nonparallel, the tangent plane to the surface $(1,2,5)$ is given by

$$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}.$$

We can now convert the parametric vector form to a point-normal form, where the normal vector is given by the cross product of the two directions,

$$\mathbf{n} = \begin{pmatrix} 0 \\ 1 \\ 4 \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -1 \end{pmatrix}.$$

Finally, you get the dot product of the normal and the coordinate vectors,

$$\begin{pmatrix} 2 \\ 4 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} x-1 \\ y-2 \\ z-5 \end{pmatrix} = 0.$$

By expanding the dot product, one obtains $z=5+2(x-1)+4(y-2)$, which simplifies to

$$2x + 4y - z = 5.$$

This is the cartesian form of the tangent plane.

Notice how $F_x(1,2) = 2$ and $F_y(1,2) = 4$; each number being the coefficient of the cartesian equation respectively as well as components of the normal vector. This is not an accident;

Example. Find a normal vector $\mathbf{n}$ and the Cartesian equation of the tangent plane to the surface $z=4x^2y$ at the point $(2,-1,-16)$.
The partial derivatives are

$$\begin{align*} F_x(x,y)&=8xy, \\ F_y(x,y)&=4x^2, \end{align*}$$

so at the point $(2,-1)$,

$$\begin{align*} F_x(2,-1)&=8(2)(-1)=-16, \\ F_y(2,-1)&=4(2)^2=16. \end{align*}$$

By the proposition, a normal vector is

$$\mathbf{n}=\begin{pmatrix} -16 \\ 16 \\ -1 \end{pmatrix}$$

and the tangent plane is

$$\begin{align*} z&=-16-16(x-2)+16(y+1) \\ &=32-16x+16y. \end{align*}$$

Therefore the tangent plane is $16x-16y+z=32$. Notice how much faster this is than the four-step vector geometry method; for a surface given explicitly as $z=F(x,y)$, jump straight to the proposition.

Another way to find the normal vector is as follows;
if a surface is given in the form of $g(x,y,z)=0$, then the vector $\mathbf{n}=\begin{pmatrix} g_x \\ g_y \\ g_z \end{pmatrix}$ evaluated at some point $P$ on the surface, is a vector normal to the surface at $P$.

Example. Find the equation of the tangent plane to the ellipsoid $\dfrac{x^2}{4}+\dfrac{y^2}{2}+\dfrac{z^2}{8}=1$ at the point $P(0, 1, 2)$.
To find $F_x(x,y)$, we use implicit differentiation and treat $z$ as a function of $x$.

$$\begin{align*} \frac{\partial}{\partial x}(\frac{x^2}{4}+\frac{y^2}{2}+\frac{z^2}{8})&=\frac{\partial}{\partial x}(1) \\ \frac{x}{2}+0+\frac{z}{4}F_x&=0 \\ F_x&=-\frac{2x}{z}. \end{align*}$$

Similarly,

$$F_y=-\frac{4y}{z}.$$

A vector normal to the ellipsoid at $P$ can be derived by two methods, either from the method described in the earlier proposition or the gradient-vector method.
Using the gradient vector method we obtain,

$$\mathbf{n}_1=\begin{pmatrix} \dfrac{x}{2} \\ y \\ \dfrac{z}{4} \end{pmatrix}_P = \begin{pmatrix} 0 \\ 1 \\ 1/2 \end{pmatrix}.$$

Using the proposition,

$$\mathbf{n}_2=\begin{pmatrix} -\dfrac{2 \cdot 0}{2} \\ -\dfrac{4 \cdot 1}{2} \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ -2 \\ -1 \end{pmatrix}$$

Notice that $\mathbf{n}_2 = -2\mathbf{n}_1$; demonstrating that both methods are valid.

Hence, the equation of the tangent has the form

$$0x + 1y + \frac{1}{2}z = d$$

for some constant $d$. Substituting $P$, the desired equation is $2y+z=4$.

Example. Find the equation of the tangent plane to the sphere $x^2+y^2+z^2=1$ at the point $P\left(\dfrac{1}{3},\dfrac{1}{2},\dfrac{\sqrt{23}}{6}\right)$.
First a quick sanity check that $P$ actually lies on the sphere: $\frac{1}{9}+\frac{1}{4}+\frac{23}{36}=\frac{4+9+23}{36}=1$.
We could rearrange for $z=\sqrt{1-x^2-y^2}$ and use the proposition, but the square root makes the partial derivatives messy; the gradient-vector method is much cleaner here. Writing $g(x,y,z)=x^2+y^2+z^2-1$,

$$\mathbf{n}=\begin{pmatrix} g_x \\ g_y \\ g_z \end{pmatrix}_P = \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}_P=\begin{pmatrix} 2/3 \\ 1 \\ \sqrt{23}/3 \end{pmatrix}.$$

Scaling by $3$ to clear the fractions, we may instead use $\mathbf{n}=(2,3,\sqrt{23})^T$. The tangent plane then has the form

$$2x+3y+\sqrt{23}z=d,$$

and substituting $P$,

$$\begin{align*} d&=\frac{2}{3}+\frac{3}{2}+\frac{23}{6} \\ &=\frac{4+9+23}{6} \\ &=6. \end{align*}$$

Therefore the tangent plane is $2x+3y+\sqrt{23}z=6$. If the surface is given implicitly, do not bother rearranging for $z$; use $\mathbf{n}=(g_x,g_y,g_z)^T$ directly.

Example. Find all points on the saddle $z=x^2-y^2$ where the tangent plane is horizontal.
A horizontal plane has normal vector $(0,0,-1)^T$ (up to scaling), so both partial derivatives must vanish:

$$\begin{align*} F_x(x,y)&=2x=0, \\ F_y(x,y)&=-2y=0, \end{align*}$$

which forces $(x,y)=(0,0)$. So the only such point is the origin, where the tangent plane is $z=0$. Notice however that the plane $z=0$ intersects the surface along the level curves $y=\pm x$, so the tangent plane actually passes through the surface. A horizontal tangent plane does not mean the point is a maximum or a minimum; here it is a saddle point.

1.4 The total differential approximation

Suppose that $f$ is a differentiable function of one variable. The equation of the tangent to the graph of $f$ at a point $x_0$ is given by

$$y = y_0 + m(x-x_0),$$

where $y_0=f(x_0)$. When $x$ is close to $x_0$, the tangent line is close to the graph of $f$, meaning

$$f(x) \approx f(x_0) + f'(x_0)(x-x_0).$$

Example. Let $\ell(x)$ be the tangent line to the function $y=p(x)=3x^2+4x+3$ at the point $[2,23]$. Find $\ell(1.9) \text{ and } p(1.9).$
Finding $\ell(x)$,

$$\begin{align*} y_0 = p(2) = 23 \\ f'(x) = 6x + 4 \\ f'(2) = 6(2) + 4 = 16 \\ \ell(x) = 23 + 16(x-2) \\ \ell(x) = 23 + 16x - 32 \\ \ell(x) = 16x - 9 \end{align*}$$

Evaluating $\ell(1.9) = 21.4$ whilst $p(1.9) = 21.43$. Therefore, it is clear that the tangent line for a point $x_0$ can be used to evaluate $p(x_1)$, where $x_1 \approx x_0$.

The image above depicts $p(x)$ in red and the tangent $\ell(x)$ in blue.

Rewriting $\Delta x = x - x_0$ and $\Delta f = f(x) - f(x_0)$, we have

$$\Delta f \approx f'(x_0) \Delta x.$$

This is called differential approximation to $\Delta f$. Extending this to functions with two variables,
suppose that a surface given by $z=F(x,y)$ has a tangent plane at the point $(x_0,y_0,z_0)$. The equation of the tangent is given by

$$z = z_0 + F_x(x_0,y_0)(x-x_0) + F_y(x_0,y_0)(y-y_0).$$

If $(x,y)$ are near $(x_0,y_0)$, and rewriting $z_0=F(x_0,y_0)$,

$$F(x,y) - F(x_0,y_0) \approx F_x(x_0,y_0)(x-x_0) + F_y(x_0,y_0)(y-y_0).$$

There's alot of iterating afterwards with $\Delta$ (image attached)

This formula is called the total differential approximation to $\Delta F$. By suppressing the point of evaluation, the total differential approximation may be written as

$$\Delta F \approx \frac{\partial F}{\partial x} \Delta x + \frac{\partial F}{\partial y} \Delta y.$$

The approximation improves as $\Delta x$ and $\Delta y$ get smaller.

Example. Consider the paraboloid $z=F(x,y) = \dfrac{x^2}{5^2} + \dfrac{y^2}{3^2}$ at the point $[5,1,10/9]$. Find the total differential approximation of $z=F(x,y)$ at $[5.1,1.1]$ and compare it to $F(5.1,1.1)$.

Evaluating $F_x(5,1)$:

$$\begin{align*} F(x,y)&=\frac{x^2}{5^2} + \frac{y^2}{3^2}, \\ F_x(x,y)&=\frac{2x}{5^2} + 0, \\ F_x(5,1)&=\frac{2(5)}{5^2} \\ &=\frac{2}{5}. \end{align*}$$

And $F_y(5,1) = \dfrac{2}{9}.$
Since they asked for the approximation for $z$ and not just the change, you would use $z = z_0 + F_x(x_0,y_0)(x-x_0) + F_y(x_0,y_0)(y-y_0).$
Plugging in $x_0=5, y_0=1, z_0=\dfrac{10}{9}$,

$$z = \frac{10}{9} + \frac{2}{5}(x - 5) + \frac{2}{9}(y - 1).$$

After simplifying,

$$z = \frac{2x}{5} + \frac{2y}{9} - \frac{10}{9}.$$

This is the equation for our tangent plane; plugging in $(5.1,1.1)$ we get $1.173$, whilst $F(5.1,1.1)$ is $1.175$.

Example. The ideal gas law asserts that the pressure $P$, volume $V$ and temperature $T$ of an ideal gas are related by the formula,

$$PV=kT,$$

where $k$ is a constant. If the temperature increased by $4$% and the volume is decreased by 5%, estimate the percentage increase in pressure.

Doing this the immediate way, which is simply plugging in $V=0.95V$ and $T=1.04T$ yields $P=\dfrac{1.04}{0.95} \dfrac{kT}{V} \approx 1.0947P, \Delta P \approx 0.0947P$, however, the question asks us to estimate, meaning we have to use the total differential approximation. Since we are calculating change, we use

$$\Delta F \approx \frac{\partial F}{\partial x} \Delta x + \frac{\partial F}{\partial y} \Delta y.$$

Rewriting,

$$\begin{align*} P(T,V) &= \frac{kT}{V}, \\ \Delta P \approx \frac{\partial P}{\partial T} \Delta T &+ \frac{\partial P}{\partial V} \Delta V. \end{align*}$$

Since we know that $\Delta T = 0.04T$ and $\Delta V = -0.05V$, we simply calculate $\dfrac{\partial P}{\partial T} \text{ and } \dfrac{\partial P}{\partial V};$

$$\frac{\partial P}{\partial T} = \frac{k}{V},$$

$$\frac{\partial P}{\partial V} = -\frac{kT}{V^2}.$$

Plugging in,

$$\begin{align*} \Delta P &\approx \left(\frac{k}{V}\right)(0.04T) + \left(-\frac{kT}{V^2}\right)(-0.05V) \\ &\approx 0.04\left(\frac{kT}{V}\right) + 0.05\left(\frac{kTV}{V^2}\right) \\ &\approx 0.04\left(\frac{kT}{V}\right) + 0.05\left(\frac{kT}{V}\right) \\ \text{Therefore, } \\ \Delta P &\approx 0.09P \quad (\text {since } P= \frac{kT}{V}) \end{align*}$$

Notice how the 9% we obtained from the estimation is close to the exact answer of 9.47%.

NOTE: Another way to calculate $z$ or find the linear approximation is to simply calculate $\Delta F$ and add it onto the original $F$ value, this might be a bit easier.

Example. Use the total differential approximation of $f(x,y)=\sqrt{x^2+y^2}$ to estimate $\sqrt{2.98^2+4.03^2}$.
Pick the nearby nice point $(x_0,y_0)=(3,4)$, so that $f(3,4)=\sqrt{9+16}=5$ and

$$\begin{align*} \Delta x &= 2.98-3=-0.02, \\ \Delta y &= 4.03-4=0.03. \end{align*}$$

The partial derivatives are

$$\begin{align*} f_x(x,y)&=\frac{x}{\sqrt{x^2+y^2}}, \\ f_y(x,y)&=\frac{y}{\sqrt{x^2+y^2}}, \end{align*}$$

so $f_x(3,4)=\dfrac{3}{5}$ and $f_y(3,4)=\dfrac{4}{5}$. Then

$$\begin{align*} \Delta f &\approx \frac{3}{5}\times(-0.02)+\frac{4}{5}\times0.03 \\ &=-0.012+0.024 \\ &=0.012. \end{align*}$$

Adding this onto the original value (as per the note above),

$$\sqrt{2.98^2+4.03^2}\approx 5+0.012=5.012.$$

A calculator gives $5.012115$, so the approximation is accurate to three decimal places. The whole trick with these 'estimate without a calculator' questions is choosing a nearby point where everything is clean.

Upper bounds on errors

No tools have absolute certainty in measurements; there is always a deviation from the exact value (i.e. a ruler can be off by $\pm 0.1$mm). Given this deviation, it is possible to use the total differential approximation to calculate how the maximum value of this deviation from the exact result. Given a function $F(x,y)$, one can interpret $\Delta F$ as the error in the output given errors $\Delta x$ and $\Delta y$. Typically, we don't know precise value of the deviation, but we can use bounds for the absolute errors $|\Delta x|$ and $| \Delta y |$.

$$\begin{align*} \text{ Error } &\approx |\Delta F| \\ |\Delta F|&\approx |\frac{\partial F}{\partial x} \Delta x + \frac{\partial F}{\partial y} \Delta y| \\ &\leq |\frac{\partial F}{\partial x}|| \Delta x| + |{\frac{\partial F}{\partial y}}| |\Delta y| \end{align*}$$

Example. The dimensions of a cylinder are measured to the nearest millimeter using a measuring tape. The circumference is measured to be 22.0 cm and height is measured to be 15.0 cm. Use these measurements to (a) estimate the volume of the cylinder, and (b) estimate an upper bound for the percentage error in your answer to part (a).

(a). Since $C = 2\pi r$,

$$V = \pi r^2 h = \pi (\frac{C}{2\pi}^2) h = \frac{C^2 h}{4 \pi}$$

By using $C =22$ and $h =15$,

$$V = \frac{1815}{\pi}.$$

And so the volume is estimated to be $\dfrac{1815}{\pi} \text{cm}^3$, which is $\approx 577.75 \text{cm}^3$.
It says estimate in the question here only because the measuring tools were also estimated values, hence we "estimate" the final result; we don't use linear approx here.

(b). Notice how the question says nearest millimeter, meaning $\pm 0.5\text{mm}$, or $0.05\text{cm}$. The 0.5mm is for reasons below ig:

We convert to cm to match the units of the given inputs and make our lives easier.
Since both the circumference and height were measured, we can say that

$$|\Delta C| \leq 0.05 \quad \text{ and } \quad |\Delta h| \leq 0.05.$$

We want to calculate $|\Delta V|$.

$$V = \frac{C^2h}{4\pi}, \frac{\partial V}{\partial C} = \frac{Ch}{2\pi} \text{ and } \frac{\partial V}{\partial h} = \frac{C^2}{4\pi},$$

subbing in,

$$\frac{\partial V}{\partial C} = \frac{165}{\pi}, \frac{\partial V}{\partial h}=\frac{121}{\pi}.$$

Finally, we put it all together,

$$\begin{align*} |\Delta V| &\approx |\frac{165}{\pi} \Delta C + \frac{121}{\pi} \Delta h| \\ &\leq \frac{165}{\pi} |\Delta C| + \frac{121}{\pi} |\Delta h| \\ &\leq \frac{165}{\pi} \times 0.05 + \frac{121}{\pi} \times 0.05 \\ &= \frac{286}{20\pi}. \end{align*}$$

So the upper bound for the absolute error in $V$ is approximately $\dfrac{286}{20\pi}$ (or about $4.55\text{cm}^3$). To get the percentage error, we can do

$$\begin{align*} \frac{\max{|\Delta V|}}{V} \times 100\% &\approx \frac{286}{20\pi}\cdot\frac{\pi}{1815} \times 100 \% \\ &=\frac{26}{33} \%. \end{align*}$$

Hence the percentage error is no more than about 0.79%.

Example. The volume of a football in the shape of an ellipsoid of revolution with semi-axes of length $a$, $b$ and $b$ is given by

$$V=\frac{4}{3}\pi ab^2.$$

The values of $a$ and $b$ are measured to be $12.0$ cm and $7.0$ cm respectively, each to the nearest millimetre. (a) Calculate the volume of the football, (b) estimate the maximum absolute error in the calculated value of $V$, and (c) hence estimate the percentage error.

(a) Substituting the measurements,

$$\begin{align*} V&=\frac{4}{3}\pi(12)(7)^2 \\ &=784\pi \\ &\approx 2463 \text{ cm}^3. \end{align*}$$

(b) Measuring to the nearest millimetre again means $|\Delta a| \leq 0.05$ and $|\Delta b| \leq 0.05$ (converted to cm). The partial derivatives evaluated at $(a,b)=(12,7)$ are