John Abbott · Science · 201-SNB-AB

Multivariate Calculus tutoring at John Abbott, 201-SNB-AB

Make sense of partial derivatives, surfaces and integrals with one-to-one help. Try the short tangent-plane example below, or bring a question, even if you do not know how to start.

Ask about a lesson Try the worked example

Working with several variables

201-SNB-AB, Multivariate Calculus, connects vector geometry with differentiation and integration in several variables. The example below uses partial derivatives to estimate a nearby value on a surface.

Optional course listed in the catalogue. It may not run every term.

Topics to work on

Vector functions: derivatives and integrals
Partial derivatives, directional derivatives and gradients
Vector fields, potential functions and line integrals
Double integrals and changes of variables, especially polar coordinates
Green’s theorem, curl and divergence

Before you calculate

Review the power rule, substitution into a function and equations of planes. For a partial derivative, name the variable that changes and the one that stays fixed. For an approximation, separate the starting coordinates from the small changes in those coordinates.

Two steps worth checking

Treating the other variable as a constant without dropping terms that still depend on the variable of differentiation
Using coordinate changes in the tangent-plane formula, then distinguishing an estimated value from an exact value

Choose the object you are working with

A surface near a point: tangent-plane approximation Recalculate the partial derivatives at the new starting point in the retry.
An area region: double-integral bounds Sketch the region and decide which variable describes each slice.
A directed curve: a line integral Identify the path and its direction before choosing the integral.

Worked example: turn partial derivatives into a tangent plane

Let $x,y$ be independent real coordinates with no units, and let $f(x,y)=x^2+xy+2y^2$ . The graph $z=f(x,y)$ is a surface, where $z$ is its height. Find the tangent plane at $P=(1,-1,2)$ , then use it to estimate $f(1.02,-1.01)$ . Check the estimate against direct substitution.

Show a hint

Find each partial derivative before substituting $(1,-1)$ . The tangent plane passes through $P$ and uses those two derivatives as its slopes in the coordinate directions. Multiply the slopes by the changes from the starting coordinates.

Show the worked solution

1. Hold one variable fixed
Write $f_x$ for the derivative with respect to $x$ , holding $y$ fixed, and $f_y$ for the derivative with respect to $y$ , holding $x$ fixed. In $f_x$ , the term $xy$ contributes $y$ , while $2y^2$ contributes zero:
$f_x(x,y)=2x+y,\qquad f_y(x,y)=x+4y.$
At the starting coordinates, $f(1,-1)=2$ , $f_x(1,-1)=1$ and $f_y(1,-1)=-3$ . In the slice with $y=-1$ , the surface has slope $1$ in the positive $x$ direction. In the slice with $x=1$ , it has slope $-3$ in the positive $y$ direction.
2. Build a plane through the surface point
This polynomial is differentiable, so its partial derivatives give a tangent-plane approximation. Let $L(x,y)$ be the plane’s height. Start at height $2$ , then add each slope multiplied by the corresponding coordinate change:
$L(x,y)=2+1(x-1)-3(y+1).$
The tangent plane is $z=L(x,y)$ . At $(1,-1)$ , both changes vanish and $L=2$ , so the plane passes through $P$ . Close to that point, $f(x,y)\approx L(x,y)$ .
3. Estimate, then check
Let $\Delta x$ and $\Delta y$ denote the changes from $(1,-1)$ . Here $\Delta x=0.02$ and $\Delta y=-0.01$ . The two first-order contributions to the height change are $0.02$ and $0.03$ : decreasing $y$ gives a positive contribution because the $y$ -slope is negative.
$L(1.02,-1.01)=2+0.02-3(-0.01)=2.05.$
For a check, evaluate the original function:
$\begin{aligned}f(1.02,-1.01)&=(1.02)^2+(1.02)(-1.01)\\&\quad+2(-1.01)^2\\&=2.0504.\end{aligned}$
The estimate is low by $0.0004$ . The plane captures the first-order change; the curved surface also has higher-order terms. Agreement at $P$ does not make the plane exact at every nearby point.

Your turn: move the point of tangency

Keep $f(x,y)=x^2+xy+2y^2$ , but start at $Q=(0,1,2)$ . (1) Find both partial derivatives at $(0,1)$ . (2) Write the new tangent plane and use it to estimate $f(0.01,0.98)$ . (3) Compare with direct substitution: is your estimate high or low?

Show a hint

The derivative formulas stay the same, but their values at the new point change. The coordinate changes are $0.01-0$ and $0.98-1$ . Keep the second change negative.

Check the three steps

1. Re-evaluate the slopes
$f_x(0,1)=1,\qquad f_y(0,1)=4,\qquad f(0,1)=2.$
The $y$ -slope is now positive. Reusing $-3$ from the first example would describe the wrong tangent plane.
2. Use the new starting coordinates
Call the new plane’s height $L_Q(x,y)$ . Then
$L_Q(x,y)=2+x+4(y-1).$
$L_Q(0.01,0.98)=2+0.01+4(-0.02)=1.93.$
The plane predicts a height decrease of $0.07$ . A positive $y$ -slope combined with a negative change in $y$ contributes a decrease.
3. Compare with the surface
$\begin{aligned}f(0.01,0.98)&=(0.01)^2+(0.01)(0.98)\\&\quad+2(0.98)^2\\&=1.9307.\end{aligned}$
The estimate is low by $0.0007$ . Check the two partial derivatives, the starting point and the coordinate changes before attributing a discrepancy to approximation error.

If your plane does not pass through its stated starting point, check the starting height and the coordinate differences. If only the nearby estimate is wrong, write $\Delta x$ and $\Delta y$ separately before substituting.

What to work through with a tutor

Bring a question, even if you have not started. You can choose the first step together or review any work you have tried. Your tutor can help you connect each partial derivative to a cross-section, build a tangent plane that passes through the right point, and explain the sign of each contribution. Then try a new point without copying the previous slopes.

For an integral, your tutor can help you sketch the region or parameterize the path before calculating. Together, you and your tutor can check bounds, orientation and the conditions for using a potential function or Green’s theorem against the problem you are working on.

Choose your review from your course outline

Choose practice from the topics announced for your assessment. For a partial derivative, state which variable stays fixed. For a tangent-plane question, check the base point and separate the coordinate changes from the coordinates. For an integral, sketch the region or path and explain the bounds before calculating. Redo a missed setup without its solution, and follow your section’s rules for permitted tools and assessed work.

Independent practice

Continue with a focused lesson

6 questions

Double integrals: regions and changing order

Sketch a region, choose its bounds and practise changing the order of integration.

18 min

Line Integrals: Follow the Path and Its Direction

Parameterize a path, check orientation and see how a potential function can simplify the integral.

35 min

How to Choose an Integration Method

Review single-variable integration methods when the setup is clear but the calculation stalls.

Course information

John Abbott: Science course descriptions

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