University Calculus III — Directional Derivatives and the Gradient Worksheet

All 10 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. Q1The Gradient Vector

   Find the gradient of each function, then evaluate it at the given point.

   1. $f(x,y)=y\ln x+x{y}^2$, for $x>0$, at the point $(1,2)$.
   2. $g(x,y,z)=x{z}^2+y^{2}z-3xy$ at the point $(1,-1,2)$.
2. Q2The Gradient Vector

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

   1. Find the point at which $\nabla f=(4,5)$.
   2. Describe the set of all points at which $\nabla f$ points in the direction of the positive $y$-axis, that is, $\nabla f=(0,c)$ with $c>0$.
3. Q3Directional Derivatives

   Find the directional derivative of $f(x,y)=x^2-3xy+y^3$ at the point $(1,2)$ in the direction of $\overrightarrow{v}=(5,-12)$. State whether $f$ is increasing or decreasing in that direction.

4. Q4Directional Derivatives

   The concentration of a dye in a tank, in $\text{mg}/\text{L}$, is $C(x,y,z)=x^{2}y-xz+y{z}^2$, with $x$, $y$, $z$ in centimetres.

   1. Find the rate of change of the concentration at the point $P(2,1,-1)$ in the direction from $P$ towards the point $Q(4,-1,0)$. Include units.
   2. A probe passes through $P$ heading towards $Q$ at $3\text{cm}/\text{s}$. At what rate, in $\text{mg}/\text{L}$ per second, is the concentration it reads changing at that instant?
5. Q5The Direction of Maximum Rate of Change

   Let $f(x,y)=\sqrt{x^2+2y}$, defined for $x^2+2y>0$. At the point $(1,4)$, find the unit direction in which $f$ increases most rapidly and the maximum rate of increase. Then state the unit direction in which $f$ decreases most rapidly, and the value of the directional derivative of $f$ in that direction (a signed number).

6. Q6The Direction of Maximum Rate of Change

   The temperature at the point $(x,y)$ of a heated plate is $T(x,y)=50-x^2-y^2-4y$, in ${}^{\circ }\text{C}$, with $x$ and $y$ in centimetres. An ant is at the point $(3,-2)$.

   1. In which unit direction should the ant set off to warm up as fast as possible, and what is that rate?
   2. The ant prefers to warm up at exactly $3{}^{\circ }\text{C}/\text{cm}$. Find every unit direction that achieves this.
   3. Explain why no direction lets the ant warm up at $7{}^{\circ }\text{C}/\text{cm}$.
7. Q7The Gradient and Level Curves

   Let $f(x,y)=\frac{1}{2}{x}^2-y$. On the grid below, the axes cross at the centre and each square is one unit, so the grid shows $-6\le x\le 6$ and $-6\le y\le 6$.

   1. Write the equation of the level curve of $f$ that passes through the point $(2,1)$, and sketch the part of it that fits on the grid.
   2. Compute $\nabla f(2,1)$ and draw it on the grid with its tail at $(2,1)$.
   3. Use the gradient to find the equation of the tangent line to the level curve at $(2,1)$, and draw the line.

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

8. Q8Tangent Planes and Normal Lines to a Level Surface

   Find an equation of the tangent plane, and parametric equations of the normal line, to the surface $x^2+2y^2-z^2=5$ at the point $(2,1,1)$.

9. Q9Tangent Planes and Normal Lines to a Level Surface

   Find the point on the paraboloid $z=x^2+3y^2$ at which the tangent plane is parallel to the plane $4x-6y-z=20$, and give an equation of that tangent plane.

10. Q10Synthesis — drawing on several topics in this unit

    The elevation of a hillside, in metres, is $h(x,y)=500-{\displaystyle \frac{x^2}{400}}-{\displaystyle \frac{y^2}{100}}$, where $x$ is the distance east and $y$ the distance north of a marker, both in metres. A hiker stands at the point above $(80,30)$.

    1. Find the hiker's elevation and $\nabla h(80,30)$.
    2. The hiker sets off towards the north-east, the direction of $(1,1)$. Is she climbing or descending at first, and at what slope (metres of elevation per metre travelled horizontally)?
    3. In which unit direction is the hill steepest uphill, and what is the slope in that direction?
    4. Give the two unit directions in which she could set off without changing elevation at first, and the equation of the tangent line at $(80,30)$ to the contour line she is standing on.

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 Directional Derivatives and the Gradient notes cover  About the Directional Derivatives and the Gradient 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.

The gradient vector: every partial derivative in one place

The gradient of f is the vector of its first-order partial derivatives, written in the order of the variables: two components for a function of x and y, three for a function of x, y and z. Q1 asks for the gradient of one function of each kind and then its value at a point, and there is nothing new in the calculus — each component is a partial derivative of the kind the previous set drilled. What is new is the bookkeeping.

Q2 runs the gradient backwards. Part (a) gives the value of the gradient and asks where it is taken: setting each component equal to its target gives two linear equations in x and y, a small system of the kind Linear Algebra solves. Substitute your point back into the gradient to confirm it.

Part (b) asks for every point where the gradient points along the positive y-axis, which the question translates for you into a vector (0, c) with c > 0. That is two conditions, not one: an equation for the first component and an inequality for the second. Solve the equation, substitute what it gives into the second component, and then decide where the inequality holds. A description that uses only the first condition answers a different question. Say also what happens at any point where the whole gradient is zero, since the zero vector has no direction at all.

Directional derivatives: a rate of change in any direction

The partial derivatives are the rates of change along the two axes. The directional derivative is the rate of change along any direction, and for a differentiable function it comes from the gradient by one dot product — provided the direction is a unit vector.

Q3 is those four steps with a direction vector whose length is a whole number, so the unit vector has simple fractions. It then asks whether f is increasing or decreasing in that direction, and step 4 answers it: the sign of the directional derivative is the whole verdict. A useful check on the size of your answer comes from the next sheet — no directional derivative can be larger in absolute value than the norm of the gradient.

Q4 puts the same computation in space, with a dye concentration in a tank and the direction given as "from P towards Q" rather than as a vector.

The direction of maximum rate of change

Writing the dot product with the angle θ between u and the gradient turns the directional derivative into a statement about geometry. Everything in this sheet follows from it.

Q5 asks for four things at one point: the unit direction of fastest increase and the rate there, then the unit direction of fastest decrease and the directional derivative in it. The first pair is the gradient normalised, and its norm. The second pair is the opposite direction, and the question insists on a signed number: the directional derivative in the direction of fastest decrease is negative, because the function is falling there, even though the size of the fall is the same norm.

Q6 gives the temperature of a plate and an ant that wants to warm up. Part (a) is Q5's first half in a context: the ant should head along the gradient, and the rate is its norm, in degrees per centimetre. Part (b) turns the question round — a fixed rate is required, and you must find every unit direction that delivers it.

The gradient and level curves

The directions in which f does not change at all are the ones at right angles to the gradient, and those are the directions along the level curve. So at every point where it is not zero, the gradient is perpendicular to the level curve through that point. Q7 makes you draw this on a grid.

Tangent planes and normal lines to a level surface

In three variables the same fact holds one dimension up: the gradient of F at a point is normal to the level surface of F through that point. The previous set wrote tangent planes only for graphs z = f(x, y); this one writes them for any surface given as an equation in x, y and z.

Q8 is that chain applied once. Name the function F whose level surface the equation is, check that the given point is on it, compute ∇F there, and use it twice: as the normal of the plane and as the direction of the normal line.

Q9 is the same idea run in reverse, on a paraboloid given as z equal to a function of x and y. Move everything to one side so the paraboloid becomes a level surface of a function of three variables, and its gradient gives the normal at a general point, in terms of x and y.

The synthesis question

Q10 puts the whole set on one hillside, with the elevation a function of distance east and distance north. Part (a) is Q1: the elevation at the hiker's position and the gradient there. Part (b) is Q3 — normalise the north-east direction, take the dot product, and let the sign say whether she climbs or descends; the slope is metres of elevation per metre travelled horizontally. Part (c) is Q5: the steepest uphill direction is the gradient normalised, and the slope in it is the gradient's norm. Part (d) joins the last two sheets: the directions in which the elevation does not change at first are the two unit vectors perpendicular to the gradient, and the contour line she stands on is the level curve through her position, so its tangent line has the gradient as its normal, exactly as in Q7.

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Getting the most out of it

Normalise the direction before anything else

The formula D_uf = ∇f · u is only true for a unit vector. Make it a habit to divide the given direction by its norm the moment you read it, and to write u with a small hat or the word "unit" beside it. Most lost marks in this topic are a missing division.

Check every rate against the norm of the gradient

Once you have ∇f at a point, you know the range every directional derivative there must lie in: from minus its norm to plus its norm. A directional derivative outside that range is wrong, and one exactly at an end of it means your direction is along the gradient or against it. It is a free check, so use it.

Draw the gradient on the level curve

Even when the question does not ask for a sketch, a rough picture of the level curve through the point with the gradient drawn from it shows at a glance which directions climb, which fall and which stay level. It also catches a gradient with a sign error, which will point downhill instead of up.

Write a surface as a level surface before looking for a normal

Whatever form the surface is given in, move everything to one side and name the function F. Then the normal is ∇F, with no special cases to remember, and the same recipe gives the tangent plane and the normal line.

Keep the units on every rate

A directional derivative is a rate per unit of distance. Writing the units forces you to notice when a question wants a rate per unit of time instead, as a moving probe or a walking hiker does, and to multiply by the speed.

Common questions