University Calculus III — Functions of Several Variables Worksheet

All 9 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Cylinders and Quadric Surfaces

   Consider the surface in space with equation $x^2+{\displaystyle \frac{y^2}{4}}-z^2=1$.

   1. Find the trace of the surface in the horizontal plane $z=k$, and say what kind of curve it is for every value of $k$.
   2. Find the traces in the coordinate planes $x=0$ and $y=0$, and name each curve.
   3. Name the surface, and state which coordinate axis is its axis.
   4. A second surface has equation $z=4-y^2$, with no $x$ in it. Describe this surface, and explain why it is called a cylinder although no slice of it is a circle.
2. Q2Domain and Range of a Function of Two Variables

   Let $f(x,y)=\sqrt{16-x^2-4y^2}$.

   1. Evaluate $f(2,-1)$ exactly.
   2. Write the domain of $f$ as a set defined by an inequality, describe it in words, and sketch it on the grid (each square is one unit). Show clearly whether the boundary belongs to the domain.
   3. Find the range of $f$, with a reason.

   A blank Cartesian grid for this question is on the printable PDF.

3. Q3Domain and Range of a Function of Two Variables

   Let $g(x,y)={\displaystyle \frac{\ln (x+y-2)}{\sqrt{16-x^2-y^2}}}$.

   1. State the two conditions a point $(x,y)$ must satisfy to be in the domain of $g$, and say why each inequality is strict.
   2. Sketch the domain on the grid (each square is one unit), using dashed curves for boundaries that are not included.
   3. Decide which of the points $A(2,1)$, $B(4,0)$, $C(0,2)$ and $D(3,3)$ are in the domain, and evaluate $g$ at each one that is.

   A blank Cartesian grid for this question is on the printable PDF.

4. Q4Level Curves and Level Surfaces

   Let $f(x,y)=y^2-x^2$.

   1. Find the equations of the level curves $f(x,y)=k$ for $k=-4$, $k=0$ and $k=4$, and name each curve.
   2. Sketch the three level curves on the grid (each square is one unit), labelling each with its value of $k$.
   3. The point $(1,3)$ lies on none of the three curves above. Find the value of $k$ for which the level curve $f(x,y)=k$ passes through $(1,3)$, and write the equation of that curve.
   4. Name the surface $z=y^2-x^2$.

   A blank Cartesian grid for this question is on the printable PDF.

5. Q5Level Curves and Level Surfaces

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

   1. Describe the level surface $F(x,y,z)=k$ for $k=3$, for $k=-1$ and for $k=-2$.
   2. Use part (a) to state the range of $F$.
   3. Find the level surface of $F$ that passes through the point $(2,1,3)$, and describe it.
6. Q6Limits Along Paths

   Work with each limit as directed.

   1. Evaluate ${\displaystyle {lim}_{(x,y)\to (2,-1)}\frac{x^{2}y+6}{x-y}}$, and say why direct substitution is allowed.
   2. Show that ${\displaystyle {lim}_{(x,y)\to (0,0)}\frac{3x^2-y^2}{x^2+2y^2}}$ does not exist, by finding the limit along each coordinate axis.
7. Q7Limits Along Paths

   Both limits below give $\frac{0}{0}$ on direct substitution.

   1. Evaluate ${\displaystyle {lim}_{(x,y)\to (4,1)}\frac{x-4y}{\sqrt{x}-2\sqrt{y}}}$, where $(x,y)$ is restricted to points with $x>0$, $y>0$ and $x\ne 4y$.
   2. Let $w(x,y)={\displaystyle \frac{4(x-1)y}{(x-1{)}^2+3y^2}}$. Find the limit of $w$ as $(x,y)\to (1,0)$ along the line $y=m(x-1)$. Then decide whether ${\displaystyle {lim}_{(x,y)\to (1,0)}w(x,y)}$ exists.
8. Q8Continuity of a Function of Several Variables

   Decide continuity from the way each function is built, naming the fact you use.

   1. Find the largest set on which $f(x,y)={\displaystyle \frac{x+y}{x^2-y}}$ is continuous, and describe the set of points where it is not.
   2. Find the largest set on which $g(x,y)=\ln (1+x-y^2)$ is continuous, and describe it in words.
   3. Use continuity to evaluate ${\displaystyle {lim}_{(x,y)\to (3,-1)}g(x,y)}$.
9. Q9Synthesis — drawing on several topics in this unit

   Let $f(x,y)=\sqrt{{\displaystyle \frac{x^2}{4}}+y^2-1}$.

   1. Find the domain of $f$ and describe it in words. Is $f$ continuous at every point of its domain?
   2. Find the equations of the level curves $f=0$ and $f=\sqrt{3}$, and name them.
   3. The graph of $f$ is the set of points with $z=f(x,y)$. Show that every point of the graph lies on a quadric surface, name that quadric and its axis, and say which part of the quadric the graph is.
   4. State the range of $f$, with a reason.

The 7 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

The rest of this unit

The worksheet is the practice, and it is free. Two more printable documents cover the same unit and come with the University Calculus III bundle: read the notes first, work this sheet, then sit the test closed-book. See what each one covers.

What the Functions of Several Variables notes cover  About the Functions of Several Variables unit test

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

A surface is recognised by slicing it

Q1 gives a surface as one equation in x, y and z and asks you to identify it without ever drawing it whole. The method is the one every quadric question uses: fix one variable at a constant, read the curve that is left in the plane, and let the family of curves tell you the shape. Part (a) slices with horizontal planes z = k, part (b) with the two vertical coordinate planes, and part (c) asks you to put the pictures together.

Part (d) is the other kind of surface in this sheet. When one variable is absent from the equation, a point satisfies it whatever that variable is, so the surface is the plane curve in the other two variables slid along lines parallel to the missing axis. Explaining the name means saying what a cylinder is in general — a surface swept out by parallel lines through a plane curve — rather than what it is in the circular case you met first.

The domain is a region, and its boundary is a decision

Q2 is a square root of an expression in x and y. The domain is every point where the radicand is non-negative, and the work is in turning that inequality into a region you can recognise: rearrange until the curve on the boundary is in a standard form, then decide which side of it the inequality keeps.

Part (c) asks for the range, and "with a reason" means two halves. First bound the values: ask how small and how large the radicand can be on the domain, and where, and that gives an interval the function cannot leave. Then show that every value in the interval is actually reached — restricting to one line through the domain turns that into a one-variable question. An interval with only the first half argued is a guess.

Q3 combines two restrictions of different kinds, a logarithm in the numerator and a square root in the denominator, and part (a) asks why each inequality is strict. Each has its own reason, and the reasons are different: one is about the logarithm, the other is about dividing. Part (b) is the intersection of the two regions those conditions describe, with each boundary drawn according to step 4 above.

Level curves: the contour map of a surface

A function of two variables has a graph in three dimensions, which is hard to draw. Its level curves f(x, y) = k live in the plane, and a labelled set of them carries the same information the way a contour map carries a landscape. Q4 asks for three of them for a difference of two squares, one for a negative k, one for zero and one for a positive k.

Part (c) runs the other way: given a point, find the level curve through it. There is nothing to solve — evaluate f at the point, and that number is the k of the curve that passes through it. Part (d) then asks you to name the surface whose contour map you have drawn, and the method is Q1's: its horizontal traces are the curves of part (a), and its vertical traces in x = c and y = c are what decide the name.

Level surfaces, and what an empty level set tells you

Q5 moves up one dimension. A function of three variables cannot be graphed at all, so its level surfaces F(x, y, z) = k are the only picture there is. Complete the square in the variable that carries a linear term, and every level set of this F reads as the same kind of equation with a different constant on the right.

For each value of k in part (a), look at the sign of the constant first and let it tell you which case you are in before you describe anything. Part (b) is the payoff: a value belongs to the range of F exactly when its level set is non-empty, so the range is read off from the sign condition, with no further calculation. Part (c) is Q4(c) in three dimensions — evaluate F at the point to find which level surface passes through it — and a quick distance check from the centre confirms the description.

A limit in the plane has infinitely many directions

In one variable a point can be approached from two sides. In the plane it can be approached along any line, any curve, any spiral, and a limit exists only if every one of those paths leads to the same number. That changes which arguments are available, and Q6 sets the two basic ones side by side.

Q6(b) is the standard way a limit fails. Approach the origin along the x-axis by setting y = 0, then along the y-axis by setting x = 0; each reduces the function to one variable, and the two one-variable limits are different numbers. Two paths with two different limits is a complete proof that the two-variable limit does not exist.

When both paths agree, try a family

Q7 opens with two limits that both give 0/0 on substitution and need different treatments. Part (a) is Calculus I in two variables: the numerator is a difference of squares in disguise, once each term is seen as the square of a root, and after the common factor cancels what is left is continuous at the point. Write down the restriction that makes the cancellation legal — the question states it for you.

Part (b) is built so that the method of Q6(b) does not decide it: the two coordinate axes give the same answer. So the question asks for the limit along a whole family of lines through the point at once, with the slope m left as a letter.

If the result still depends on m, different lines lead to different values, and the decision follows from Q6's rule. Write it with two specific values of m, so the two disagreeing paths are on the page as numbers, and do not stop at the two axes just because they happened to agree.

Continuity from the way a function is built

Q8 asks you to decide continuity without taking a single limit, by naming the fact that applies. Polynomials are continuous everywhere; a rational function is continuous wherever its denominator is non-zero; a composition of continuous functions is continuous wherever it is defined. Every function in the question is assembled from those pieces, so the work is in recognising the assembly and then finding where it breaks.

Part (b) is a logarithm of a polynomial, continuous on its whole domain, so the largest set of continuity is the domain, found the way Q3 found one and described in words. Part (c) is the reason continuity is worth knowing: at a point inside that set, the limit equals the value. Check that the point is in the set first — that check is the step that licenses the substitution.

The synthesis question, and why it is last

Q9 is one function, a square root, looked at four ways, and each part uses a different sheet. Part (a) is a domain as in Q2 — find the boundary curve, then test a point to see which side of it the inequality keeps, and do not assume it is the same side as before; the continuity half is Q8's composition argument. Part (b) is Q4's method, and it is worth writing down the level curve for a general k as well: part (d) will need it.

Part (c) is where the sheets meet. Setting z = f(x, y) and squaring both sides gives an equation in x, y and z, which you name by Q1's method of traces. But squaring lets in points the original equation did not have, so the graph is not the whole quadric: ask what sign z can take when it equals a square root, and say which part of the surface that leaves. Part (d) then reads the range the way Q5(b) did, from which level sets are non-empty.

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Getting the most out of it

Slice before you sketch

Every surface in this set is identified from its traces, never from a memory of a picture. Write the trace in z = k first, then in x = 0 and y = 0, name each curve, and only then name the surface. The traces are also the only way to draw a quadric by hand, so the work you do to identify it is the sketch.

Test a point on each side of a boundary

For a domain, pick one point that is plainly inside and one that is plainly outside each boundary curve and substitute them into the inequality. It takes seconds and it catches the most common error in this sheet, which is shading the wrong side of a correct curve.

Argue a range in two halves

A bound shows the function cannot go beyond an interval; a path or a level set shows it reaches every value inside it. Write both halves every time, and use a level set being empty or non-empty as the quickest test of whether a value is taken.

Keep the limit rule written at the top of the page

Two paths with different limits: the limit does not exist. Any number of paths with the same limit: nothing is settled. Continuity at the point: substitute, and say why. Most lost marks on limits in several variables come from running the first rule backwards.

Common questions