Concordia ENGR 233: Vector Calculus Practice

How to use these questions

Attempt each question with its solution closed. Write the method and one reason for choosing it.
After checking, label the first error: geometry, derivative, algebra, field test or orientation. Follow the relevant repair link, then retry without notes.
Questions 2 and 3 share a motion model; Question 3 restates the values it needs. This is a progressive selection, with no suggested score or exam timing.

All variables and parameters are real. Coordinates are Cartesian; vector components follow the x, y, z order. The dot and cross symbols denote the usual dot and right-handed cross products. A prime denotes differentiation with respect to the displayed parameter. All quantities are dimensionless except the motion quantities in Questions 2–3, where units are specified. The symbols e, ∇ and ‖ · ‖ denote the natural exponential base, the gradient operator and the Euclidean norm.

1. Build a plane from two directions

Three points in space are $A=(1,-1,0)$ , $B=(3,0,0)$ and $C=(1,-1,2)$ . Find a normal vector and a Cartesian equation for their plane. Then find where the line

\begin{gathered}x=s,\qquad y=-2s,\\z=1,\qquad s\in\mathbb R\end{gathered}

meets the plane. Explain why the intersection is unique.

Hint

Form two directions lying in the plane, $\overrightarrow{AB}$ and $\overrightarrow{AC}$ . Their cross product is perpendicular to both. Substitute the line into the plane equation only after finding that equation.

Explained solution and checks

The plane directions are $\mathbf u=(2,1,0)$ and $\mathbf v=(0,0,2)$ . Their cross product is $\mathbf u\times\mathbf v=(2,-4,0).$ It is nonzero, so the directions are independent and the three points determine a plane. Divide by 2 to use the simpler normal $\mathbf n=(1,-2,0)$ .

A point $(x,y,z)$ is in the plane precisely when its displacement from A is perpendicular to this normal:

\begin{gathered}\mathbf n\cdot(x-1,y+1,z)=0,\\(x-1)-2(y+1)=0,\\x-2y=3.\end{gathered}

Substitute the line: $s-2(-2s)=3$ , so $s=3/5$ . The intersection has coordinates

x=\frac35,\qquad y=-\frac65,\qquad z=1.

The line direction is $\mathbf d=(1,-2,0)$ . Since $\mathbf n\cdot\mathbf d=5\ne0$ , the line crosses the plane once; it is neither parallel to it nor contained in it.

Check the result

All three original points satisfy $x-2y=3$ . At the intersection, $3/5-2(-6/5)=3$ . Reversing the cross-product order gives the opposite normal and an equivalent plane equation.

Common errors

Using the position vectors of B and C instead of displacements from a common point.
Using a direction that lies in the plane as its normal.
Losing the +1 in the displacement’s y-component.

Choose your next step

If the plane equation feels arbitrary, review the normal-vector interpretation in plane intersections and geometry. If vector components or dot products caused the error, start with matrix and vector fundamentals.

2. Recover position and read the velocity

A particle moves in three-dimensional space. In the formulas below, t is the numerical time in seconds, position components are in metres, velocity components in metres per second, and acceleration components in metres per second squared. For $0\le t\le2$ ,

\begin{gathered}\mathbf a(t)=(0,2,-2),\\\mathbf v(0)=(1,-1,2),\\\mathbf r(0)=(0,1,0).\end{gathered}

Find $\mathbf v(t)$ and $\mathbf r(t)$ . At $t=1$ , find the position, velocity, speed and unit tangent in the direction of motion. Explain why the particle is still moving when its z-velocity is zero.

Hint

Integrate each acceleration component and use the initial velocity. Integrate again and use the initial position. Speed is the norm of the complete velocity vector.

Explained solution and checks

Because $\mathbf v^{\prime}=\mathbf a$ , componentwise integration and the velocity at t=0 give

\begin{aligned}v_x(t)&=1,\\v_y(t)&=2t-1,\\v_z(t)&=2-2t.\end{aligned}

Now use $\mathbf r^{\prime}=\mathbf v$ and the initial position:

\begin{aligned}x(t)&=t,\\y(t)&=t^2-t+1,\\z(t)&=2t-t^2.\end{aligned}

Each integration has its own constant in each component. The two sets of initial conditions serve different integrations.

At t=1, the position is $(1,1,1)$ metres and the velocity is $(1,1,0)$ metres per second. Thus

\begin{gathered}\|\mathbf v(1)\|=\sqrt2\ \mathrm{m/s},\\\mathbf T(1)=\frac1{\sqrt2}(1,1,0).\end{gathered}

The unit tangent T has no units. A zero z-component means there is no instantaneous motion in the z-direction; the x- and y-components still give nonzero speed. In fact $v_x=1$ throughout the interval, so the velocity never vanishes.

Check the result

Differentiate the three position components to recover velocity, differentiate velocity to recover acceleration, and substitute t=0 in both. Also $\|\mathbf T(1)\|=1$ .

Common errors

Applying the initial position to the velocity constants.
Adding velocity components to get speed.
Confusing a zero component with a zero vector.

Choose your next step

If differentiation or antidifferentiation caused the error, check those component calculations before interpreting the motion. Use the derivative-rule check for a derivative repair, or integration methods when selecting an antiderivative. Then redo the two initial-condition steps here.

3. Separate turning from speeding up

For the particle in Question 2, at t=1 the velocity is $\mathbf v=(1,1,0)$ m/s and the acceleration is $\mathbf a=(0,2,-2)$ m/s². Find the curvature κ, the signed tangential acceleration $a_T$ , and the nonnegative normal acceleration magnitude $a_N$ . Is the speed increasing at that instant? Does a positive curvature tell you the sign of $a_T$ ?

Hint

At a point where velocity is nonzero, use

\begin{gathered}\kappa=\frac{\|\mathbf v\times\mathbf a\|}{\|\mathbf v\|^3},\\a_T=\frac{\mathbf v\cdot\mathbf a}{\|\mathbf v\|},\\a_N=\frac{\|\mathbf v\times\mathbf a\|}{\|\mathbf v\|}.\end{gathered}

The tangential component changes speed; the normal component changes the direction of motion.

Explained solution and checks

First calculate the numerical values in the SI units stated in Question 2:

\begin{gathered}\|\mathbf v\|=\sqrt2,\qquad\mathbf v\cdot\mathbf a=2,\\\mathbf v\times\mathbf a=(-2,2,2),\\\|\mathbf v\times\mathbf a\|=2\sqrt3.\end{gathered}

Using their physical units gives

\begin{aligned}\kappa&=\frac{2\sqrt3}{(\sqrt2)^3}=\frac{\sqrt6}{2}\ \mathrm{m}^{-1},\\a_T&=\frac2{\sqrt2}=\sqrt2\ \mathrm{m/s^2},\\a_N&=\frac{2\sqrt3}{\sqrt2}=\sqrt6\ \mathrm{m/s^2}.\end{aligned}

Since $a_T=d\|\mathbf v\|/dt>0$ , the speed is increasing. The positive curvature describes turning per unit distance travelled. Its nonnegative value alone says nothing about whether speed is increasing, constant or decreasing.

Check the result

The acceleration components are perpendicular, so their squared magnitudes must add:

\begin{gathered}a_T^2+a_N^2=2+6=8,\\\|\mathbf a\|^2=0^2+2^2+(-2)^2=8.\end{gathered}

The second check $a_N=\kappa\|\mathbf v\|^2$ gives $(\sqrt6/2)(2)=\sqrt6$ .

Common errors

Using $\|\mathbf v\|^2$ rather than its cube in the curvature formula.
Replacing signed $a_T$ with $\|\mathbf a\|$ .
Calling curvature an acceleration; its units are inverse length.

Choose your next step

If only the cross product was wrong, return to the perpendicular-vector calculation in Question 1. If the distinction between direction and speed was unclear, redo the unit tangent in Question 2, then explain the two acceleration components in words before recomputing them.

4. Keep partial and path derivatives separate

Define $f:\mathbb R^2\to\mathbb R$ by

f(x,y)=x^2y+e^{xy}.

Find $f_x$ , $f_y$ and $f_{xy}$ , where $f_{xy}$ means first differentiate with respect to x, then with respect to y. A path is $x(t)=1+t$ , $y(t)=2t$ for real t. Find the derivative of $h(t)=f(x(t),y(t))$ at t=0. Explain why this is not generally just $f_x$ evaluated on the path.

Hint

In a partial derivative, hold the other independent variable fixed. Along the path, both coordinates change:

h^{\prime}(t)=f_x\,x^{\prime}(t)+f_y\,y^{\prime}(t),

with both partial derivatives evaluated at $(x(t),y(t))$ .

Explained solution and checks

Differentiating one coordinate at a time gives

\begin{aligned}f_x&=2xy+ye^{xy},\\f_y&=x^2+xe^{xy}.\end{aligned}

The exponential needs the chain rule: the x-derivative of its exponent is y, and the y-derivative is x. Differentiate $f_x$ with respect to y, using the product rule on $ye^{xy}$ :

f_{xy}=2x+e^{xy}+xy e^{xy}.

At t=0, the path passes through $(1,0)$ , where $f_x=0$ and $f_y=2$ . Its coordinate rates are $x^{\prime}=1$ and $y^{\prime}=2$ . Therefore

h^{\prime}(0)=0(1)+2(2)=4.

A partial derivative changes x while keeping y fixed. Here y also changes, so the y contribution is essential.

Check the result

Substitution first produces

h(t)=2t(1+t)^2+e^{2t(1+t)}.

At t=0, the derivatives of these two terms are 2 and 2, giving 4. Also, differentiating $f_y$ with respect to x gives the same mixed derivative; all these derivatives are continuous on the plane.

Common errors

Differentiating y with respect to x when taking a partial derivative of the stated two-variable function.
Omitting the y or x factor from the exponential’s chain rule.
Using the path’s position coordinates as its velocity components.

Choose your next step

If the exponential or product caused trouble, review chain-rule decisions. If the two path contributions were the issue, write $x^{\prime}$ and $y^{\prime}$ separately before substituting in the multivariable chain rule.

5. Use a gradient for direction and a tangent plane

Let $f(x,y)=x^2+y^2$ on $\mathbb R^2$ , and let $P=(1,2)$ . Find the directional derivative at P in the direction of the vector $\mathbf w=(3,-4)$ . Find the largest possible directional derivative at P and a unit direction that attains it. Then find the tangent plane and a normal line to the graph $z=f(x,y)$ at $(1,2,5)$ .

Hint

A directional derivative uses a unit vector. The gradient of f lies in the xy-plane; a normal to its graph in three dimensions can be obtained from $G(x,y,z)=f(x,y)-z$ .

Explained solution and checks

The gradient is $\nabla f=(2x,2y)$ , so $\nabla f(P)=(2,4)$ . Normalize the supplied direction:

\mathbf u=\frac{\mathbf w}{\|\mathbf w\|}=\left(\frac35,-\frac45\right).

Hence

D_{\mathbf u}f(P)=2\frac35+4\left(-\frac45\right)=-2.

For a unit direction, the largest gradient dot product is its norm. Thus the maximum is $2\sqrt5$ , attained in the direction $(1,2)/\sqrt5$ . The negative derivative in the requested direction means f decreases initially along that direction.

The tangent plane uses the value and the two partial derivatives at P:

z-5=2(x-1)+4(y-2).

For the normal line, $\nabla G(1,2,5)=(2,4,-1)$ . A convenient parametric form is

\begin{gathered}x=1+2s,\\y=2+4s,\\z=5-s,\qquad s\in\mathbb R.\end{gathered}

The planar vector $\nabla f$ supplies two slopes. The three-component vector $\nabla G$ is normal to the graph. Reversing the normal gives the same line.

Check the result

The base point satisfies the plane. The graph’s tangent vectors $(1,0,2)$ and $(0,1,4)$ both have zero dot product with $(2,4,-1)$ . Also $|-2|\le2\sqrt5$ , as every unit-direction derivative must be bounded by the gradient norm.

Common errors

Using w directly, obtaining −10 instead of the change per unit distance −2.
Calling $(2,4,0)$ a normal to the graph.
Writing the tangent-plane equation through the origin instead of through $(1,2,5)$ .

Choose your next step

For the normalization decision, use gradients and directional derivatives. For the plane/normal distinction, compare the role of the normal in Question 1.

6. Compute curl and divergence, then judge a claim

On all of $\mathbb R^3$ , let

\mathbf F(x,y,z)=(xy,yz,zx).

Compute its divergence and curl. A student evaluates the curl at the origin, obtains zero, and concludes that F is conservative on $\mathbb R^3$ . Is that conclusion justified? Give a decisive check. Here conservative means that F is the gradient of a scalar potential on the stated domain.

Hint

For $\mathbf F=(P,Q,R)$ , divergence is $P_x+Q_y+R_z$ . The three curl components are $R_y-Q_z$ , $P_z-R_x$ and $Q_x-P_y$ . A condition at one point is not a condition throughout a domain.

Explained solution and checks

The divergence is a scalar:

\nabla\cdot\mathbf F=y+z+x.

The curl is a vector. Compute one component at a time:

\begin{aligned}(\nabla\times\mathbf F)_x&=0-y=-y,\\(\nabla\times\mathbf F)_y&=0-z=-z,\\(\nabla\times\mathbf F)_z&=0-x=-x.\end{aligned}

Thus the curl is zero at the origin, but at $(1,0,0)$ it equals $(0,0,-1)$ . The components of F are continuously differentiable everywhere. If F were a gradient on $\mathbb R^3$ , its curl would vanish everywhere there, by equality of the potential’s mixed partial derivatives. This one nonzero value rules out a potential on the whole domain.

Check the reasoning

Divergence does not decide whether a field is conservative. A continuously differentiable field with zero curl throughout an open, simply connected domain has a potential there. Simply connected means every closed loop can shrink continuously to a point within the domain. Here a nonzero curl value already rules out a potential. Question 7 establishes a potential directly by constructing and verifying it.

Common errors

Adding the three curl components and reporting one number.
Reversing the subtraction in the middle curl component.
Treating a successful check at one point as a statement about every point.

Choose your next step

If the derivatives are incorrect, repeat them while explicitly holding the other two coordinates fixed. If the conclusion was the issue, review the potential and domain check, then identify which part of the student’s claim fails.

7. Construct and verify a potential

On $\mathbb R^3$ , define F by its components:

\begin{aligned}F_x&=2xy+z,\\F_y&=x^2+2y,\\F_z&=x+2z.\end{aligned}

Find a scalar potential φ with $\nabla\varphi=\mathbf F$ , and verify all three partial derivatives. Use it to evaluate $\int_C\mathbf F\cdot d\mathbf r$ for any piecewise continuously differentiable path C from $A=(0,1,0)$ to $B=(1,0,2)$ . What is the integral along the same path in reverse?

Hint

Integrate $\varphi_x=2xy+z$ with respect to x. The integration “constant” may depend on y and z. Determine that function by comparing the other two partial derivatives.

Explained solution and checks

Start with

\varphi=x^2y+xz+g(y,z).

Comparing the y-derivative with $F_y$ gives $g_y=2y$ , so $g=y^2+h(z)$ . Comparing the z-derivative with $F_z$ then gives $h^{\prime}(z)=2z$ . Therefore one potential is

\varphi=x^2y+xz+y^2+z^2.

An arbitrary additive real constant is also allowed and cancels from endpoint differences. Verify, rather than assume:

\begin{aligned}\varphi_x&=2xy+z=F_x,\\\varphi_y&=x^2+2y=F_y,\\\varphi_z&=x+2z=F_z.\end{aligned}

The potential is continuously differentiable on all of the domain, so the line integral depends only on its endpoint values:

\begin{gathered}\varphi(A)=1,\qquad\varphi(B)=6,\\\int_C\mathbf F\cdot d\mathbf r=6-1=5.\end{gathered}

Reversing C exchanges its endpoints and gives −5. No parameterization is needed once the potential has been established.

Check the result

For an independent calculation, take the broken path from $(0,1,0)$ to $(0,0,0)$ , then to $(1,0,0)$ , then to $(1,0,2)$ . Only one coordinate changes on each segment:

\begin{gathered}\int_1^0 2y\,dy=-1,\qquad\int_0^1 0\,dx=0,\\\int_0^2(1+2z)\,dz=6.\end{gathered}

The contributions add to 5, agreeing with the endpoint calculation.

Common errors

Writing a numerical constant after the first integration with respect to x and losing the y- and z-dependent terms.
Checking only $\varphi_x$ .
Subtracting the final potential from the initial potential.

Choose your next step

If endpoint subtraction was correct but the potential was incomplete, follow the verified-potential method. Then reconstruct g and h here with the solution closed.

8. Compare two routes and check their orientation

Let $\mathbf H(x,y)=(-y,x)$ on $\mathbb R^2$ . The oriented quarter-ellipse C is

\begin{gathered}x(t)=2\cos t,\qquad y(t)=\sin t,\\0\le t\le\frac\pi2,\end{gathered}

with t measured in radians. Compute $\int_C\mathbf H\cdot d\mathbf r$ . Compare it with the integral along the straight segment L joining the same start and end in the same direction. Then give the integral along C in reverse. Explain whether the unchanged endpoints let you use path independence.

A quarter ellipse C and a straight chord L both point from A at (2,0) to B at (0,1). The ellipse curves above the chord in the first quadrant. The coordinate axes have equal unit scales. — Both routes start at A and end at B. Their arrows agree on endpoint order; the routes themselves differ. Reversing C changes the arrow direction.

Hint

For each route, write its endpoints, differentiate its parameterization and form $\mathbf H(\mathbf r(t))\cdot\mathbf r^{\prime}(t)$ . A segment can use $\mathbf q(s)=(2-2s,s)$ , $0\le s\le1$ .

Explained solution and checks

The ellipse starts at $A=(2,0)$ and ends at $B=(0,1)$ . Along it,

\begin{aligned}\mathbf r^{\prime}(t)&=(-2\sin t,\cos t),\\\mathbf H(\mathbf r(t))&=(-\sin t,2\cos t).\end{aligned}

Take the dot product before integrating:

\begin{gathered}\mathbf H(\mathbf r(t))\cdot\mathbf r^{\prime}(t)\\=2\sin^2t+2\cos^2t=2.\end{gathered}

Thus $\int_C\mathbf H\cdot d\mathbf r=\int_0^{\pi/2}2\,dt=\pi$ .

For the straight segment, use $\mathbf q(s)=(2-2s,s)$ , $0\le s\le1$ . Its derivative is $(-2,1)$ , and its field value is $(-s,2-2s)$ . Therefore

\begin{gathered}\mathbf H(\mathbf q(s))\cdot\mathbf q^{\prime}(s)\\=2s+(2-2s)=2,\\\int_L\mathbf H\cdot d\mathbf r=\int_0^1 2\,ds=2.\end{gathered}

The integrands both simplify to 2, but their parameter intervals differ. The two integrals are unequal, so endpoints alone cannot determine this field’s integral.

Traversing the ellipse in reverse gives −π. For example, on $0\le t\le\pi/2$ , the reversed path is

\begin{aligned}x_{\rm rev}(t)&=2\sin t,\\y_{\rm rev}(t)&=\cos t.\end{aligned}

It starts at B and ends at A. Its derivative is $(2\cos t,-\sin t)$ , giving the constant integrand −2.

Check the field and the orientation

Write $\mathbf H=(P,Q)$ . Here $P_y=-1$ and $Q_x=1$ , so the necessary equality for a continuously differentiable planar conservative field fails everywhere. A potential shortcut is unavailable. This calculation uses direct line integrals, without Green’s theorem.

Changing how fast a parameter traces C once in the same direction preserves the integral because the derivative and bounds change together. Reversing the traversal changes its sign; choosing a different route can change its magnitude as well.

Common errors

Integrating the field components without the path derivative.
Replacing $\mathbf r^{\prime}$ by a unit tangent while still using dt; a unit-tangent formula needs arc length ds.
Using the same bounds for different parameters without checking them.
Assuming all fields are path independent because the endpoints match.

Choose your next step

If the endpoints or reversed path caused the error, review parameterization and direction. If the integrand was wrong, use the line-integral setup, then compute the ellipse again without opening this solution.

Turn an error into the next attempt

Geometry: redraw the point, direction and normal from Question 1. Check the plane with all three points.
Motion: differentiate your position function in Question 2. In Question 3, explain what each acceleration component changes before using a formula.
Several variables: in Questions 4–5, write what is held fixed, what changes and whether the direction has length 1.
Fields and paths: in Questions 6–8, distinguish a local calculation, a verified potential on a domain and an oriented path.

If Question 8 is too large a first step, use the guided line-integral question, then return to the ellipse and chord.

For a second pass, mix one geometry question, one derivative question and one field/path question. Explain your first decision aloud. Use the ENGR 233 study path to plan the remaining course practice.

What to bring to a tutoring session

Bring your attempted solution and mark the first step you cannot justify. A tutor can watch you choose the parameter, normal or derivative; ask you to explain that choice; and give a targeted hint before you calculate again. For a line integral, compare your direct calculation with a verified potential when one exists. Finish by setting up a fresh question yourself and checking it without the worked solution.

ENGR 233 course fit

The Fall 2026 outline places a 60-minute midterm in the first lecture of teaching week 7, covering sections 7.1–9.9. Its October 27 label identifies that week’s Tuesday, not every section’s exam day. Confirm current section announcements. Double integrals and the later integral theorems are outside this stated cutoff.

These questions select topics within that scope. They are independent practice, not a complete mock exam or a prediction of questions, weighting or difficulty.

Scope checked October 4, 2026: Concordia ENGR 233 Fall 2026 outline, pages 2–3.

Concordia ENGR 233: Vector Calculus Practice

1. Build a plane from two directions

Check the result

Common errors

Choose your next step

2. Recover position and read the velocity

Check the result

Common errors

Choose your next step

3. Separate turning from speeding up

Check the result

Common errors

Choose your next step

4. Keep partial and path derivatives separate

Check the result

Common errors

Choose your next step

5. Use a gradient for direction and a tangent plane

Check the result

Common errors

Choose your next step

6. Compute curl and divergence, then judge a claim

Check the reasoning

Common errors

Choose your next step

7. Construct and verify a potential

Check the result

Common errors

Choose your next step

8. Compare two routes and check their orientation

Check the field and the orientation

Common errors

Choose your next step

Turn an error into the next attempt

What to bring to a tutoring session

ENGR 233 course fit

Useful for

Work on the step you’re missing.

Free study guides and practice

Line Integrals: Follow the Path and Its Direction

Plane Intersections: Row Reduction and Geometry

Derivative Rules: Choose the First Step