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Three-dimensional geometryPartial derivativesGradients and directional derivativesOptimization and constraintsDouble and triple integralsVector fields

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Visualizing surfaces and regions
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Knowing when a theorem or method applies

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We use sketches and geometric reasoning to make notation concrete, then build calculations with attention to coordinates, domains, and interpretation.

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Gradient and directional derivatives: one surface, two rates

Imagine a hill viewed from above. For the calculations below, $x$ and $y$ are the numerical values of horizontal map coordinates measured in metres, and $h(x,y)$ is the numerical value of height in metres above a reference level. Our hill model is $h(x,y)=20-x^2-y^2$ . We start at the map point $P=(2,1)$ . We will first find how quickly the height changes per metre in a chosen horizontal direction, then see how travelling faster changes the height gained per second.

1. Find the gradient at the starting point

A partial derivative measures the rate at which height changes when one coordinate changes and the other stays fixed. We write $h_x$ for the derivative with respect to $x$ , keeping $y$ fixed, and $h_y$ for the derivative with respect to $y$ , keeping $x$ fixed. The gradient, written $\nabla h=(h_x,h_y)$ , collects those two rates into a vector. Evaluate at $P$ : the height is $15\,\mathrm m$ , and the gradient is $(-4,-2)$ . Thus a small move in the positive direction of either coordinate initially lowers the height. The gradient points in the horizontal direction of steepest increase.

h_x=-2x,\qquad h_y=-2y

\nabla h(2,1)=(-4,-2),\qquad h(2,1)=15

2. Normalize for change per horizontal metre

Choose the horizontal direction vector $\mathbf v=(-3,4)$ : move toward decreasing $x$ and increasing $y$ , in the ratio three to four. Its length is $5$ . Divide by that length to obtain the unit vector $\mathbf u$ . The notation $D_{\mathbf u}h(P)$ means the height change per horizontal metre at $P$ in this unit direction. Compute it by taking the dot product $\nabla h(P)\cdot\mathbf u$ : multiply corresponding components, then add. The result is positive, so this direction is initially uphill. If $\Delta s$ is a small forward horizontal distance in metres, the approximate height gain in metres is $\frac45\Delta s$ .

\lVert\mathbf v\rVert=\sqrt{9+16}=5,\qquad\mathbf u=\left(-\frac35,\frac45\right)

D_{\mathbf u}h(P)=\nabla h(P)\cdot\mathbf u

=(-4)\left(-\frac35\right)+(-2)\left(\frac45\right)=\frac45\,\mathrm{m/m}

Circles from inner to outer:

h = 16

15

and

11

m.Dashed: travel directionSolid: steepest ascent

P = (2, 1)

;

h(P) = 15

This top view shows circles of equal height. Moving toward the centre raises the height. The dashed arrow is the chosen travel direction, and the solid arrow is the gradient direction. The route initially climbs, although it does not follow the steepest ascent. The arrows indicate directions; their drawn lengths do not represent speed or gradient magnitude.

3. Keep velocity for change per second

Now introduce time. Let $t\geq0$ be the number of seconds since departure, and let $\mathbf r(t)=(2-3t,1+4t)$ give the horizontal position. Its first component is $x(t)$ , and its second is $y(t)$ , so $\mathbf r(0)=P$ . Differentiating gives the horizontal velocity $\mathbf r^{\prime}(t)=(-3,4)$ in metres per second; its magnitude, the horizontal speed, is $5\,\mathrm{m/s}$ . The composite function $h(\mathbf r(t))$ is the height at time $t$ . The chain rule finds its time derivative by combining the gradient with this velocity. Use the full velocity vector without dividing by its length: normalizing it would remove the speed information.

\left.\frac{d}{dt}h(\mathbf r(t))\right|_{t=0}=\nabla h(P)\cdot\mathbf r^{\prime}(0)

=(-4)(-3)+(-2)(4)=4\,\mathrm{m/s}

\left(\frac45\,\mathrm{m/m}\right)\left(5\,\mathrm{m/s}\right)=4\,\mathrm{m/s}

4. Verify by substituting the path

Check the time rate without the gradient: insert $x(t)=2-3t$ and $y(t)=1+4t$ directly into the height formula, simplify, and differentiate with respect to $t$ . The initial rate is again $4\,\mathrm{m/s}$ . To check the distance rate, let $s\geq0$ be horizontal distance travelled from $P$ , in metres. The constant speed gives $s=5t$ . Replacing $t$ by $s/5$ in the position formula gives $\mathbf q(s)$ , the same route described by distance instead of time. Differentiate its height with respect to $s$ to recover $\frac45\,\mathrm{m/m}$ . The quadratic terms show why these are initial rates, rather than rates that stay constant.

h(\mathbf r(t))=20-(2-3t)^2-(1+4t)^2

=15+4t-25t^2

\frac{d}{dt}h(\mathbf r(t))=4-50t

\mathbf q(s)=\left(2-\frac35s,1+\frac45s\right)

h(\mathbf q(s))=15+\frac45s-s^2

\frac{d}{ds}h(\mathbf q(s))=\frac45-2s

Three questions to build your understanding

1. Foundation

Keep the same height function $h(x,y)=20-x^2-y^2$ and starting point $P=(2,1)$ , with the same metre-based numerical convention. For this check, take the horizontal direction $\mathbf v=(1,0)$ . Find $\nabla h(P)$ , check whether $\mathbf v$ is a unit vector, and calculate the initial height change per horizontal metre. Is this direction initially uphill or downhill? Retry without notes in the direction $\mathbf w=(0,-2)$ , normalizing if needed, before opening the answer.

Show the explained solution

The gradient is $\nabla h(P)=(-4,-2)$ . Since $\lVert\mathbf v\rVert=1$ , the initial slope is $(-4,-2)\cdot(1,0)=-4\,\mathrm{m/m}$ : initially downhill. For the retry, $\lVert\mathbf w\rVert=2$ , so use $\mathbf w/2=(0,-1)$ . The initial slope is $(-4,-2)\cdot(0,-1)=2\,\mathrm{m/m}$ : initially uphill. The unnormalized dot product would give 4, which is not the change per horizontal metre.

2. Apply the method

Let $\mathbf r_2(t)=(2-6t,1+8t)$ describe a second journey, with $t$ still measured in seconds and the coordinates in metres. Compare its initial direction, horizontal speed, height change per horizontal metre and height change per second with the original journey.

Show the explained solution

The velocity is $(-6,8)\,\mathrm{m/s}$ , with magnitude $10\,\mathrm{m/s}$ . Its unit direction is still $(-3/5,4/5)$ , so the initial slope stays $\frac45\,\mathrm{m/m}$ . The horizontal speed doubles, so the initial height rate doubles to $8\,\mathrm{m/s}$ . Direct substitution checks this: $h(\mathbf r_2(t))=15+8t-100t^2$ , whose derivative at zero is $8$ .

3. Challenge

Along the original path for $t\geq0$ , when does height stop increasing, and what is the greatest height?

Show the explained solution

The height rate is $4-50t$ . Setting it to zero gives $t=2/25$ , so the height stops increasing $\frac{2}{25}\,\mathrm{s}$ after departure. This rate is positive before that time and negative afterward, establishing the greatest height along this route for $t\geq0$ . Substitution into the position and height formulas gives $\mathbf r(2/25)=(44/25,33/25)$ and a height of $\frac{379}{25}\,\mathrm m$ . The horizontal distance travelled is $5(2/25)=2/5$ metres. This is the route’s maximum; the hill itself has height $20\,\mathrm m$ at the origin, which this route never reaches.

Original Nablio explanations and practice, not official exam questions.

McGill MATH 222: regions and double integrals UQAM MAT1115: multivariable calculus

Independent practice

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35 min

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Exam preparation can combine visualization, setup, and calculation across partial-derivative, optimization, and multiple-integral problems.

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Online Multivariable Calculus Tutoring

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McGill University

Concordia University

Topics covered

Common difficulties

How tutoring can help

Gradient and directional derivatives: one surface, two rates

1. Find the gradient at the starting point

2. Normalize for change per horizontal metre

3. Keep velocity for change per second

4. Verify by substituting the path

Three questions to build your understanding

1. Foundation

2. Apply the method

3. Challenge

Free study guides and practice

Constrained Optimization: Feasible Sets and Lagrange Multipliers

Constrained Optimization Practice: Six Questions and Solutions

Surface flux with oriented area vectors and closing surfaces

Six surface flux questions with explained solutions

Double integrals: regions and changing order

Concordia ENGR 233: Vector Calculus Practice

Line Integrals: Follow the Path and Its Direction

What to practise

Online Multivariable Calculus Tutoring

Courses by institution

Topics covered

Common difficulties

How tutoring can help

Gradient and directional derivatives: one surface, two rates

1. Find the gradient at the starting point

2. Normalize for change per horizontal metre

3. Keep velocity for change per second

4. Verify by substituting the path

Three questions to build your understanding

1. Foundation

2. Apply the method

3. Challenge

Free study guides and practice

What to practise

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