Concordia University · MAST 218

Concordia MAST 218 tutoring

Multivariable Calculus I: one-to-one online tutoring and exam preparation.

Online 60 min: $55 CAD · Pricing Midterm & final exam help

Listed course

Course overview

MAST 218 develops multivariable differential calculus after work on curves and vector geometry. Use your section outline to connect partial derivatives, tangent planes, directional rates and constrained optimization. Send your chapter and attempt so we can confirm support before booking.

MAST 218 topics to review

Parametric and polar curvesVectors and space curvesPartial derivatives and tangent planesGradients and directional derivativesExtrema and Lagrange multipliers

Common difficulties

Normalizing a direction before measuring change per unit distance
Distinguishing zero first-order change from a constant function
Keeping constraints separate from unconstrained optimization

A direction needs a unit vector

Use the numerical coordinates $x,y$ in centimetres. The original model $T(x,y)=20+3x^2-2y^2$ gives temperature in degrees Celsius. At $P=(1,1)$ , find the instantaneous temperature change per centimetre travelled in the direction $\mathbf v=(4,3)$ . Predict whether temperature initially rises or falls. This is a mathematical practice model, not experimental data.

Hint: separate direction from distance

The gradient contains the two partial rates. A direction vector specifies a direction but its length need not be one. Normalize it before asking for change per unit distance.

Work through the direction and rate

Holding the other coordinate fixed gives $\nabla T=(6x,-4y)$ , so $\nabla T(P)=(6,-4)$ , in °C/cm. The vector has length $\lVert\mathbf v\rVert=5$ ; its unit direction is $\mathbf u=(4/5,3/5)$ .

D_{\mathbf u}T(P)=\nabla T(P)\cdot\mathbf u=\frac{24-12}{5}=\frac{12}{5}\,\mathrm{^{\circ}C/cm}.

The positive result means an initial rise. For a small positive distance $s$ in centimetres, the first-order change is approximately $(12/5)s$ °C. The gradient itself is a vector; the directional derivative is a signed scalar rate.

Using $\nabla T(P)\cdot\mathbf v=12$ without normalization does not give the rate per centimetre. It gives the rate per unit of the parameter in $P+t\mathbf v$ .

Retry: zero rate does not mean a constant function

Keep the same model and point, but use direction $\mathbf w=(2,3)$ . Find the directional derivative. Then substitute the unit-distance path $\mathbf q(s)=(1+2s/\sqrt{13},1+3s/\sqrt{13})$ . Is temperature constant along this path?

Check the independent retry

The unit direction is $(2,3)/\sqrt{13}$ , so the rate is $(12-12)/\sqrt{13}=0$ °C/cm. Direct substitution gives $T(\mathbf q(s))=21-6s^2/13$ . For every sufficiently small nonzero displacement $s$ , the temperature is below its starting value $T(P)=21$ °C, although its first derivative at $s=0$ is zero. The directional derivative describes the first-order change at the point; it does not describe the whole path.

For a longer explanation of distance rates versus time rates, use the existing gradient and directional-derivative lesson. It is topic practice, not a MAST 218 assessment.

From directional rates to constrained extrema

For $f(x,y)=x^2+4y^2$ with $x+y=5$ , first ask: is $(0,0)$ allowed, and what values may $x$ take? A stationary point from the unconstrained problem is useful only if it fits the actual constraints.

Check the feasible set before differentiating

The point $(0,0)$ is infeasible because its coordinates sum to zero. On the whole line, $y=5-x$ and $x$ may be any real number. Adding $x\ge0$ , $y\ge0$ would instead restrict $x$ to $[0,5]$ , so endpoints would need checking.

Try the guided questions to compare substitution with Lagrange multipliers, include interior and boundary candidates, and justify whether global extrema exist. These are later-course topic exercises; use your section outline to choose what to review.

Six constrained-optimization questions · Read the feasible-set and Lagrange lesson

How tutoring can help

Bring a surface sketch, the point and your proposed direction. Your tutor can check which quantity is held fixed, whether the direction has unit length and what units the rate should have. Then test a direction with zero first-order change and explain what that does and does not imply.

For constrained optimization, also bring the allowed set, your full candidate list and your global or local conclusion. Your tutor can check whether a missed boundary point, a regularity assumption or a feasibility condition changes the answer.

Exam preparation for MAST 218

Combine vector geometry and curve problems with partial derivatives, tangent planes and optimization from your section’s outline. Sketch the geometry, identify what stays fixed when differentiating, and check constraints before using Lagrange multipliers.

Use your section’s outline to choose the chapters to review. We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

Need help with MAST 218?

Send your topic and preferred times. We’ll confirm whether we can help and, if so, agree on a time by text or email before payment.

Request MAST 218 tutoring

Nablio is an independent service and is not affiliated with this institution.

MAST 218 · official course reference MAST 218 · official course outline